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f(b) =/: f(c)))
This in turn can be translated into ( +) There is no f in M such that 9Jt
F
f: a
->
w /\ 'rIb,c E a (b
=/: c -> f(b) =/: f(c))
On the other hand, our assertion that the set A = {b EM: 9Jt F b E a} is countable translates into: (++) There exists a one-to-one function F from the set A = {b EM: b E a} into w. The two assertions (+) and (++) are not the same. Let us ignore the questions of whether 9Jt's version of w is the real w, and whether 9Jt's idea of a one-to-one function bears any resemblance to the real thing. The important point is that (++) asserts the existence of a certain function in the real world, whereas (+) declares that no such function corresponds to an element of M. Do you see any contradiction between (+) and (++)? We don't. EXERCISE 27(PG): In Chapter 4, we mentioned that the membership relation E is strictly wellfounded. This property of E will be reflected by the axiomatic
version ZFC, i.e., if (M, E) F ZFC, then (M, E) F "E is strictly wellfounded." Does this necessarily mean that E has to be a strictly wellfounded relation on M whenever (M, E) F ZFC? Hint: Use Theorem 2.2. We have seen two examples of complete theories. Both are special cases of a more general phenomenon. Let 21 be a model of a language L. By Th(21) we denote the set of all sentences 'P such that 21 F 'P. We call Th(21) the theory of the model 21 or the theory of the structure 21. Note that TGTf- = Th(({e},*,e)) and DLONEf- = Th( (Q,:5)). This is no accident. EXERCISE 28(PG): Show that a theory Tin L is complete iff Tf- = Th(21) for some model 21 of L. Why bother to write down lists of sentences like DLONE? Would it not be better to just talk about ~'the sentences in Th( (Q,:5) )?" Recall Postulate 2 made on the rules of inference .. The list of sentences in DLONE is short, and so is the list of sentences in GT (at any rate, much shorter than the lists of sentences in Th( (Q,:5)) or GTf-). More importantly, since the sets GT and DLONE are finite, it is easy to design a mechanical procedure for checking whether or not a given sentence is on
92
6. POWER AND LIMITATIONS OF THE AXIOMATIC METHOD
the list. The latter property is also shared by some infinite lists of sentences; we will refer to it by saying that a given list of sentences is recursive. If a list of sentences is recursive, then it can serve as a set of axioms for a theory. For example, the list ZFC of axioms of set theory is infinite, and yet recursive. Postulate 2 implies that there must be a purely mechanical way of verifying that a sentence rpi' appearing in the proof of rp by virtue of being an element of T is in fact an element of T. Recursive sets of axioms have this property.16 We call a theory T in a language L axiomatizable if it has a set S of axioms, i.e., if there exists a set S of sentences of L such that Sf- = Tf- and membership of a sentence in S can be determined in a purely mechanical way.
6.2. The Incompleteness Phenomenon . "... , and the more he looked inside the more Piglet wasn't there." A. A. Milne, The House at Pooh Corner We have seen that Th( (Q, ~)) is axiomatizable; how about theories of other familiar structures? Like, for instance, Th( (N, X, the equality 8(0) = 8(1) holds if and only if for all 8* : N -> X, the equality 8*(0) = 8*(1) implies that (8*(2)E8*(0) iff 8*(2)E8*(I)). EXERCISE 3(PG): Recall the definition of F from Chapter 5. Convince yourself that we correctly translated the statement "(X, E) F (AI)" into the terminology of valuations. Now observe that since we care only about 8(0),8(1), and 8(2), quantifying over entire valuations amounts to carrying an excessive amount of unnecessary luggage. When we defined the satisfaction relation Fs in Chapter 5, it was convenient to quantify over arbitrary valuations. This allowed us to minimize the number of symbols used in the definition. But whenever you analyze the relation Fs cp for a specific formula cp, you will need to consider only the values 8(n) for the finitely many n for which Vn actually appears in cp (see Exercise 5.5 for a variation on this theme). Therefore, you may want to translate the symbol 9J1 Fs cp (or 9J1 F cp) into a statement that involves quantification over elements of the universe of 9J1 instead of quantification over valuations. Here is such a translation for "(X, E) F (AI)": (+) For all xo, x 1 in X, Xo is equal to x 1 if and only if for all X2 in X (X2EXO iff X2 EX l). In our translations, we have been very careful about the style of symbols that we used for logical connectives and the membership relation. Symbols like V, +-+ and E were used only for the language of the model. Plain English was used for the metalanguage. Moreover, we used E. For good reason. EXERCISE 4(G): Convince yourself that if you replace in (+) all occurences of E by E, then you get a translation of (X, E) F (AI) for an arbitrary (possibly nonstandard) model (X, E) of Ls. Now we prove (+). Let X be a transitive set, and let xo, Xl EX. If Xo is equal to Xl, then Xo and Xl have the same elements. Therefore, for all X2 in X, x2Exo iff X2EXl' For the other implication, assume Xo is not equal to Xl' Then there exists some X2 which is in the symmetric difference of Xo and Xl' To be specific, assume X2 is in Xo but not in Xl (the other case is symmetric). Since Xo is in X and X is transitive, X2 is also an element of X. Thus there exists an X2 in X such that the biconditional "x2Exo iff X2Exl" fails. This proves the "if"-direction of (+). The proof of Theorem 1 is complete. 0 Exercise 2(a) shows that transitivity is not a necessary condition for (X, E) to satisfy the Axiom of Extensionality: For n > 0, the set w\n does not contain 0, but by Exercise 4.4(b), 0 E TC(n + 1) ~ TC(w\n). Therefore, the set w\n is not transitive. Nevertheless, (w\n, E) F (AI). Now let E = {(n + 1, n) : nEw}. Then (w, E) is a nonstandard model of Ls. EXERCISE 5(PG): Show that (w,E)
F --,(A2).
Can the Empty Set Axiom also fail in a standard model (X, E)? No. In Chapter 4 we mentioned that the relation E is strictly wellfounded.
7. THE AXIOMS
110
EXERCISE 6(PG): (a) Show that if E is a strictly wellfounded binary relation
on a set X, then (X,E) F (A2). (b) Conclude that the Empty Set Axiom holds in every standard model.
Most of the remaining axioms will specify what procedures can be used to construct new sets from given ones. This should not come as a surprise. Recall that Russell's Paradox arises from an attempt to collect too many sets into another set. We are trying to eliminate this paradox by adopting the architect's view that sets can only be assembled from ready-made building blocks (i.e., given sets) according to sound construction methods. The remaining axioms (except the Axiom of Infinity and the Foundation Axiom) describe sound construction methods. EXERCISE 7(G): What is the simplest possible way of forming a new set from given ones? Try to formulate an axiom that describes it. To help you in resisting the temptation to read on before doing Exercise 7, we delay the answer and first discuss another axiom. (A3) Foundation Axiom:
'v'x(x i- 0 -+ :Jy E x'v'z E x(z ¢. y)). This axiom expresses the fact that E is strictly wellfounded. As you already know, wellfoundedness of E implies that no set x contains itself as an element, that there are no finite sequences (XO, Xl, ... , Xk) with XOEXI E ... EXkExO, and no infinite sequences (xn : nEw) such that Xn+l EX n for all nEw. If you paid close attention to our distinction between E and E, you are probably puzzled why we used E instead of E. Of course, the axiom should express some truth about E. This it does. Does it also imply wellfoundedness of every relation E on a set X such that (X, E) F (A3)? EXERCISE 8(PG): Let (X, E) be the model of Exercise 5. Show that E is not wellfounded, and yet (X, E) F (A3). How come? The problem may seem to arise from the fact that (X,E) is not a model of all the axioms of ZFC, but this is not the crux of the matter. In Chapter 13 we shall show how it is possible to construct models (Y, F) for all of ZFC so that F is not a well founded relation. This phenomenon can occur because wellfoundedness is not a first-order property. In Exercise 6.27, you were asked to convince yourself that the statement (Y, F) F "E is wellfounded" does not necessarily imply that F is a wellfounded relation on Y. Let us now see how this phenomenon arises in the special case of the model (x, E) of Exercise 5. While you, looking at (X, E) from the outside, see that the universe X of this model is not a wellfounded set, the model does not "see" this. The model "sees" only the elements of its underlying set, and fo.r each x that it "sees," it also "knows" 2 of some yEx such that -'zEy holds for all zEx (if x = n, then the y is n + l. Note 2This anthropomorphic language is not one of our idiosyncrasies. Many logicians speak (but less frequently write) of models that see, know, or believe.
111
7. THE AXIOMS
that the unique z such that zEy is n + 2, and this z does not satisfy zEx). Let us contrast the second-order property of strict wellfoundedness with a firstorder consequence of it. The statement
[0, +00] whatsoever such that m3 = mlL3 and m is invariant under isometries of IR3. This conclusion from the Banach-Tarski Paradox is certainly somewhat counterintuitive; but as far as we know, it does not lead to a mathematical contradiction. To show you how the Axiom of Choice is used in the formation of nonmeasurable sets, let us now present one such construction. One-dimensional Lebesgue measure is a certain measure ml that is defined on a a-algebra Ll of subsets of R This measure vanishes on the points, is invariant under isometries of IR (in particular, ml is translation-invariant), and assigns to each interval in IR its length. A measure m is 'said to extend ml if m is defined on some a-algebra of subsets of IR and miLl = mI. A set V c IR is called a Vitali set if V is not m-measurable for any translation-invariant measure m that extends mI'
Theorem 23: There exists a Vitali set.
9.2. SET THEORY WITHOUT THE AXIOM OF CHOICE
145
Proof: On [0,1), define an equivalence relation", by: x'" y iff x - y E Q. By the Axiom of Choice, there exists a selector V for "', i.e., a set V e [0,1) such that IV n xl ~I = 1 for every equivalence class x/~. We show that V is as required. For every q E Q, consider the set q+ V = {q+x : x E V}. EXERCISE 39(G): Show that if q, r E Q and q =I- r, then (q + V) n (r + V) =
0.
Now let m be a translation-invariant measure that extends mI' Then m( q + V) = m(V) for every q E Q. Therefore, if W = UqE[-I,ljnQ(q + V), then
m(W)
L
=
m(q+ V)
qE[-l,IlnQ Thus, if m(V) = 0, then m(W) [0, 1) ~ W ~ [-1,2), and hence 1 = mI([O, 1))
= OJ
=
L
m(V).
qE[-I,IlnQ and if m(V) > 0, then m(W)
= +00.
But
= m([O, 1)) S m(W) S m([-I, 2)) = mI([-I, 2)) = 3,
a contradiction.
0
In particular, Theorem 23 implies that not all subsets of the real line are Lebesgue measurable. Now the question arises: Can one-dimensional Lebesgue measure be extended to any (not necessarily translation-invariant) measure defined on all subsets of R? We shall return to this intriguing question in Chapter 23. Let us briefly mention another important construction of nonmeasurable sets. Definition 24: Let X be a topological space. A set B e X is called a Bernstein set if F n B =I- 0 and F\B =I- 0 for every uncountable closed set F e X. Theorem 11.4 shows that for every positive integer n there is a Bernstein set BeRn. Lebesgue measure (of any dimension n) is regular, which means that if A e Rn is Lebesgue-measurable, then mn(A) = sup{mn(K) : K ~ A /\ K is closed} = inf{mn(G) : A ~ G /\ G is open}. It follows that if B is a Bernstein set in Rn, then B is not Lebesgue measurable. To see this, note that if K is a closed subset of B, then K is countable, hence mn(B) = 0 (recall that mn vanishes on the points). Similarly, if G is an open superset of B, then Rn\G is countable, hence of measure zero. Thus, sup{mn(K) : K ~ A /\ K is closed} = 0 < inf{mn(G) : A ~ G /\ G is open} = +00. It can be shown that if ZF is consistent, then so is the theory ZF + "every subset of the real line is Lebesgue measurable." Since the existence of nonmeasurable sets of reals is somewhat counterintuitive, it makes sense to try to develop mathematics without the Axiom of Choice. As it turns out, many useful mathematical theorems are indeed provable in ZF alone. So the questions arises: Why do most mathematicians accept the Axiom of Choice? In our opinion, there are three reasons for this: Reason 1: The Axiom of Choice is necessary for the development of many mathematical theories. Reason 2: In many cases, even if a theorem can be proved in ZF, a clever use of (AC) can save a lot of effort. Reason 3: The consistency of the theory ZFC is not more dubious than the
consistency of the theory ZF.
146
9. VERSIONS OF THE AXIOM OF CHOICE
Let us discuss these three points one by one. As for Rea.son 3; In 1938, Kurt Gadel proved that if ZF is a consistent theory, then so is ZFC. In Chapter 12, we shall give an outline of his proof. Concerning Reason 2, let [X]2 denote the set of unordered pairs of elements of X, and consider the following two statements: (HSO) (HSU)
For every infinite set X, there exists an injection f : X x X ---; Xi For every infinite set X, there exists an injection f : [X]2 ---; X.
Using the Axiom of Choice, it is trivial to show that (HSO) implies (HSU): Given X, let f : X x X ---; X be an injection. For each unordered pair {x,y} choose g({x,y}) from the set U((x,y)),f((y,x))}. The implication (HSO) ---; (HSU) is also a theorem of ZF: Given any set X, the function h : X ---; X x X defined by h(x) = (x, x) is an injection. Thus, by the Cantor-Schrader-Bernstein Theorem, (HSO) implies that X ~ X x X for every infinite set X. By a result of Tarski l l , the latter implies the Axiom of Choice. Once we have proved (AC), we can deduce (HSU) as above. This is of course a rather roundabout way of proving the implication (HSO) ---; (HSU). Here is a challenge for you: EXERCISE 40(X): Find a simpler proof of the implication (HSO) ---; (HSU) in ZF. By far the most important argument in favor of the Axiom of Choice is Reason 1. If ZF were the official framework of mathematics, then topologists would have to live without Tychonff's Theorem, algebraists would have to consider vector spaces without bases, and functional analysts could not prove the Hahn-Banach Theorem. To illustrate what can happen to set theory itself without the Axiom of Choice, consider the notion of a cardinal. In Chapter 11, we will show how one can (in ZFC) assign to each set X a set IXI (its cardinal number) in such a way that X ~ Y implies IXI = WI. An alternative construction of cardinal numbers can be carried out in ZF alone (see Definition 12.12). However, if cardinal numbers are supposed to conform to our basic intuitions about "numbers," then any two cardinals should be comparable. By Theorem 19, in the absence of AC, not all cardinals are pairwise comparable (by the discussion preceding Fact 18, even the word "comparable" is ambiguous in ZF). This phenomenon may lead to rather bizarre situations. For example, it is relatively consistent with ZF that there exists a set A such that n < IAI for every natural number n, and yet there is no injection f : w ---; A. EXERCISE 41(PG): Show (in ZF) that the following are equivalent for every set A: (a) There is no injection f : w ---; Ai (b) Every injection f : A ---; A is a surjection. A set A that satisfies conditions (a) and (b) of the above exercise is called Dedekind-finite. 12 It is a matter of taste which one considers more counterintuitive: the existence of infinite, Dedekind-finite sets or the Banach-Tarski Paradox. But, definitely, 11 See: A. Tarski, Sur quelques theoremes qui equivalent a l'axiome du choix, Fund. Math. 5 (1924), 147-154. 12 A discussion of various notions of finiteness in the absence of (AC) can be found in J. Truss, Classes of Dedekind finite cardinals, Fund. Math. 84 (1974) 187-208.
9.2. SET THEORY WITHOUT THE AXIOM OF CHOICE
147
both situations are paradoxical. Could one perhaps add an axiom to ZF that would not imply the most paradoxical consequences of the Axiom of Choice, and yet would yield all its benefits? No. As Theorem 17 shows, if you want to have Tychonoff's Theorem, then you have to live with all consequences of (AC). Similarly, by Theorem 19, if you want all cardinal numbers to be comparable, then you must admit the Axiom of Choice. You cannot eat your cake and have it too. But you can eat some of it and leave the rest for later. Mathematicians have studied a number of weaker versions of the Axiom of Choice. Some of these are strong enough to prove that every Dedekind-finite set is finite, but do not imply the existence of nonmeasurable subsets of the real line. T. Jech's book The Axiom of Choice, North-Holland 1973, gives a thorough exposition of weaker versions of (AC). Let us discuss just two of them here. The most natural way of weakening the Axiom of Choice is by restricting the cardinalities of the families involved. As an example of this procedure, let us formulate the Countable Axiom of Choice. Every countable family A of nonempty, pairwise disjoint sets has a selector. EXERCISE 42(G): (a) Return to the proof of Theorem 3.17 and convince yourself that the Axiom of Choice is tacitly used in it. 13 (b) Convince yourself that Theorem 3.17 is provable in ZF + (ACh~o' EXERCISE 43(R): Prove in
(ACh~o
that every Dedekind-finite set is finite.
Your first impulse in attacking Exercise 43 was probably to go to the proof of Theorem 4.11 and convince yourself that it can be carried out in ZF + (ACh~o' This is a reasonable thing to try, but it won't work. Our proof of Theorem 4.11 involves a recursive construction, and unfortunately, ZF + (ACh~o is not strong enough to prove the Principle of Recursive Constructions (Theorem 4.24). The latter theorem is equivalent to a weak form of the Axiom of Choice known as the Principle of Dependent Choices.
(DC)
If R is a binary relation on a nonempty set X such that for every x E X there exists ayE X with (x, y) E R, then there exists a sequence (Xn)nEw of elements of X such that (Xn' Xn+l) E R for every nEw.
EXERCISE 44(G): Prove that the Principle of Recursive Constructions and the Principle of Dependent Choices are equivalent in ZF. EXERCISE 45(PG): (a) Dig out your solution of Exercise 2.10 and convince yourself that Theorem 2.2 is provable in ZF + (DC). (b) Show that ZF + Theorem 2.2 r- (DC). EXERCISE 46(G): Show that ZF
+ (DC) r-
(ACh~o'
The last three exercises demonstrate that the theory ZF + (DC) is a natural candidate for a foundations of mathematics. Another advantage of this framework is that it is appare'ntly compatible with the most interesting alternative to the full Axiom of Choice proposed so far. This alternative is the topic of our next section. 13It
can be shown that Theorem 3.17 is not a theorem of ZF.
9. VERSIONS OF THE AXIOM OF CHOICE
148
9.3. The Axiom of Determinacy J. Mycielski and H. Steinhaus proposed in 1962 an alternative to the Axiom of Choice which is called the Axiom of Determinacy and usually abbreviated (AD). It can be described in the form of infinite games. Let A S;; Ww. Imagine a game r A played by two persons, I and II, who take turns in picking natural numbers. Player I starts and chooses ao E w, then Player II chooses bo E w, then I chooses a1 E w, next II chooses b1 E w, and so on. If the resulting sequence (ao,b o,a1,b1, . .. ) is a member of A, then I wins, otherwise II wins. The sequence (aO,b O,a1,b1, . .. ) is called the outcome of the game. For s E ww, let sl = (S2n)n<w and sl! = (S2n+l)n<w' If s is the outcome of the game, then sl is the sequence of numbers chosen by I, and sII is the sequence of numbers chosen by II. A strategy (for I or II) is a rule that tells the player which number to choose. The n-th move prescribed by a strategy depends on the moves made by the opponent up to this stage of the game. More specifically, a strategy for I is a function F : UnEw nw -+ w. Player I is said to follow the strategy F if she always chooses the number F(b o, . .. , bn-d for her n-th move an. Similarly, a strategy for II is a function G : UnEw\{O} nw -+ w. Player II follows this strategy if he always chooses bn = G(ao, ... , an). A strategy is a winning strategy, if the player who follows it always wins, independently of what the opponent does. The game r A is determined if one of the players has a winning strategy. EXERCISE 47(G): Convince yourself that in any given game of the players can have a winning strategy.
r A at most
one
EXERCISE 48(G): Let A = {Si : i E w} be a countable subset of Ww. Let G(ao, ... , an) = sn(2n) + 1 for all (ao, ... , an). Show that G is a winning strategy for player II in r A. EXERCISE 49(G): Let B = {s E Ww : "in < w (S2n+l + S2n+2 is even)}. Show that player I has a winning strategy in the game r B. The Axiom of Determinacy asserts the following:
(AD)
For every A S;; W w the game
rA
is determined.
Let n < wand t E nw. We define Ut = {s E Ww : t S;; s}. The family B = {Ut : t E U nw} is a basis for a topology 7 on Ww. This means that a set X is in 7 if and only if X is a union of elements of B. EXERCISE 50(PG): Show that (W w , 7) is a second countable completely metrizable space. Hint: For r,s E Ww such that r -=f:. s, let e(r,s) = 1/min{n: r(n) -=f:. s(n)}. Moreover, let e(r,r) = O. Show that e is a complete metric on Ww that induces the topology 7. If player I follows a strategy F, then the outcome depends only on F and the sequence b = (bn)n<w of moves played by II. We denote the outcome of such a game by F * b. If player II follows a strategy G, then the outcome depends only on G and the sequence a = (an)n<w of player I's moves. We denote the outcome of such a game by a * G. If player I follows a strategy F and player II follows a strategy G, then we denote the outcome of the resulting game by F * G.
9.3. THE AXIOM OF DETERMINACY
149
Let F be a strategy for player 1. By F we denote the set {F * b: b E W w} of all outcomes compatible with F. Similarly, if G is a strategy for player II, then we let G={a*G:aEWw}. EXERCISE 51(G): Show that if F is a strategy for player I and G is a strategy for player II, then F and G are uncountable closed subsets of (X,7).
Theorem 35 (ZF
+ (AD»:
There exists no Bernstein set in (W W,7).
Proof: Let A C Ww. By (AD), one of the players has a winning strategy in r A. If F is a winning strategy for player I, then F ~ A. If G is a winning strategy for player II, then Gn A = 0. In either case, by Exercise 51, A is not a Bernstein set. 0 Corollary 36 (ZF): (AD)
--+
...,(AC).
Proof: By Theorem 11.4 and Exercise 50, (AC) implies the existence of a 0 Bernstein set in (WW,7). Although (AD) is incompatible with the full Axiom of Choice, it implies a weak version of the latter.
+
Lemma 37 (ZF (AD»: Let X = {Xi: i < w} be a countable family of nonempty subsets ofw w. Then TIi<w Xi ¥- 0. Proof: Let A be the set of all s ~ Ww with SII E Xs(O)' Let B = Ww \ A, and consider the game G B. Player I cannot have a winning strategy in this game: Suppose F is a strategy for I, and suppose F prescribes ao as I's first move. If b E X ao ' then F * b E A, and player II wins the game. Since Xao is not empty, it is possible for player II to defeat F. Since player I does not have a winning strategy in r A, it follows from (AD) that player II has a winning strategy G. We fix such G and define a choice function f E ITi<w Xi by f(i) = ((i, 0, 0, ... ) * G)II. EXERCISE 52(G): Convince yourself that the function f defined in the last line 0 of the above proof is an element of TIi<w Xi' The definition of r A can be generalized in a straightforward manner to a situation where X is an arbitrary nonempty set and A ~ W X. The Axiom of Determinacy implies that for each countable set X and each A ~ W X, the game r A is determined.
Theorem 38 (ZF
+ (AD»:
There is no free ultrafilter on w.
Proof: Assume toward a contradiction that U
~
P(w) is a free ultrafilter on
w. We shall describe a game rAin which neither of the two players has a winning
strategy. Let X = [W] X; be an isometry with /i[X i ] = X;. Let f = Ui ,8 be a one-to-one function that maps a onto an initial segment I of,8 and such that ~
E T/ iff F(~) E F(T/) for all ~,T/ E a.
By Corollary 5, rng(F) is an ordinal. In particular, the range of F is a transitive set. It is not hard to see that condition (*) is exactly the same as condition (M) for the collapsing function of (a, E). Thus, rng(F) is the Mostowski collapse of (a, E). By Exercise 2(b), rng(F) = a. If F is a surjection, then a = rng(F) =,8. If not, then by Exercise 4.42(d), a = rng(F) = I E (() = ( for some ( E,8. In other words, in the latter case a E ,8. 0 EXERCISE 5(PG): Prove Lemma 6 without reference to the Mostowski collapse. Corollary 7: The relational class E n (ON x ON) strictly wellorders ON.
EXERCISE 6(PG): Recall from Chapter 8 what is meant by the phrase "a relational class W wellorders a class X." Derive Corollary 7 from Lemma 6. 0 If a and ,8 are ordinals, we shall often write a < ,8 instead of a E,8. This terminology is justified by Corollary 7. Thus, a ~ ,8 if and only if a is an initial segment of,8. (Compare this with Definition 4.33.)
Claim 8: Let a,,8 E ON. Then a
~
,8 iff a
~
,8.
Proof: Let a,,8 E ON. In terms of the notation O.
Lemma 23: Let 0:,,8 E ON. Then
(1)
0: + ,8
= 0: U {o: + 'Y : 'Y < ,8}.
Proof: We prove the lemma by induction over ,8. If,8 = 0, then the right hand side of (1) is o:u0 = 0:. The left hand side is 0:+0, which is also equal to 0: by Definition 22(i). Assume that ,8 = 8(-y), and that the equation (1) holds for 'Y. Then 0: U {o: + 0 : 0 < ,8} = 0: U {o: + 0 : 0 < 'Y} U {o: + 'Y}
=
= O:+'YU {O:+'Y} = 8(0: + 'Y) ==
(by inductive assumption)
(by Definition 22(ii))
= 0: +,8. Now assume that ,8 is a limit ordinal and that (1) holds for all 'Y < ,8. Then for each 0 < ,8 there exists a 'Y such that 0 < 'Y < ,8, and hence: 0: U {o: + 0 : 0 < ,8} = 0: U
U {o: + 0 : 0 < 'Y} ")' 1. As the next theorem shows, the same is true for arbitrary ordinals, Theorem 35 (Cantor's Normal Form Theorem): Let a, /3 be ordinals such that 1 < a $ /3. Then there exist a unique k and unique 'Yo,' .. ,'Yk-l and 80, ... ,8k-1 with 'Yo > 'Y1 > ... > 'Yk-l and 0 < 8i < a for i < k such that
/3 =
(5)
a'''(O •
80
+
a'''(1 •
81 + ... + a,"(k-1
•
15k-I.
Proof: Let a > 1 be fixed. By induction over /3 2:: 0 we show the existence of an expansion as in (5). If /3 = 1, then a'o . 1 is the desired expansion. Now assume that /3 > 1, and that an expansion as in (5) exists for every 1J such that 1 $ 1J < /3. Let r = {'Y : a'''( $ /3}. Claim 36: The set
r
has a largest element.,
Proof: Suppose not, and let 'Yo = Ur. Since the assumptions of Theorem lO(b) are satisfied, 'Yo is a limit ordinal, and 'Yo rf; r, On the other hand, by Definition 34(iii), a'''(O = U{a'''( : 'Y E 'Yo} = U{a'''( : 'Y E r} ~ /3.
By Claim 8,
a'''(o
$ /3, and thus 'Yo E
r.
We have reached a contradiction.
Now let 'Yo be the largest ordinal in r. By Corollary 31, there exist ordinals 80 > 0 and /31 < /3 such that /3 = a'''(O • 80 + /31. Note that 80 < a: If not, then /3 = a'''(o • 80 + /31 2:: a'''(O • 80 2:: a'''(O • a = a,("(oH), which contradicts the choice of
'Yo·
Now, if /31 = 0, then a'''(O • 80 is the desired expansion, If /31 > 0, then, by the inductive assumption, /31 has an expansion as in (5); say, /31 = a'''(1 • 81 + ... + a,"(k-1 • 8k - 1 with 'Y1 > 'Y2 > ... > 'Yk-1 and 0 < 8i < a for 0 < i < k . EXERCISE 31(PG): Show that, in the above notation, 'Yo > 'Y1. Thus, adding the term a'''(O • 80 from the left to the expansion of /31 yields the desired expansion for /3. Now we show uniqueness. Suppose towards a contradiction that /3 is a counterexample, and let: /3 = a'''(o • 80 + a'''(1 • 81 + ... + a,"(k-1 • 15k-I,
/3 =
a'''(o •
8b + a'''(1
•
8~
+ ... +
a,"(k-1 •
8~_1
be two different expansions of /3 such that 'Yo > 'Yl > ... > 'Yk-l, 8i , 8~ < a for < k, and max{ 80 , 8b} > O. (By allowing 8i , 8~ = 0, we may conveniently assume that the sequence of exponents is the same for both expansions of /3.) Moreover, if 80 = 8b, then /31 = a'''(1 • 8~ + ... + a,"(k-1 • 81._1 is a smaller ordinal with two different expansions. But the latter contradicts the choice of /3, so we must have 80 =I- 8b, Without loss of generality, assume 80 < 8b. But then:
i
169
10,2, ORDINAL ARITHMETIC
(3 =
< a'')'o =
80 + a'')'1 • 81 + ... + a'(')'k-1 +1) • 8 + a'')'1 • 8 + ... + a,')'k-2 • 8k-2 + a,')'k-2 0 1
a'')'O •
::; a'')'o
=
80 + a'')'1 • 81 + ... + a,')'k-1 • 8k-1 • 8 + a'')'1 • 8 + ... + a,')'k-1 • a 0 1
0(')'0 ,
80 • 8 0
a'')'o •
::; a'')'o
+ +
a'')'1
::; . . . ::; a'')'o • ::; a'')'o •
8~
+
81 + ... + a,')'k-2 • (8k-2 • 8 + ... + a,')'k-2 • a 1
a'')'1 •
(80
+ 1) ::; 8~ + ... +
a'')'o •
a'')'1 •
+ 1)
8~
a,')'k-1 •
8k- 1 = (3,
o
a contradiction.
We conclude this section with the proof of Goodstein's Theorem, which is an application of infinite ordinal arithmetic to number theory. Before we can formulate the theorem, we need to describe the expansion of a natural number in superbase n. Let us begin with an example. If we expand the number 27 in base 2 we get: 27 = 24 + 23 + 21 + 20 . This expansion still contains a four and a three which we would like to eliminate. We can do this by expanding these two numbers in base 2 as well: 2 1 27 = 2(2 ) + 2(2 +1) + 21 + 20 • For larger numbers we may have to repeat this procedure more often. For example, the process of expanding the number 521 in base 2 and eliminating all exponents larger than 2 yields: 521 = 29 + 23 + 1 = 2(2 3 +1) + 2(2+1) + 1 = 2(2(2+1)+1) + 2(2+1) + 1. This representation is called expansion in s'Uperbase 2. Similar expansions in superbase n exist for every natural number n > 1. For example, expanding 521 in superbase 3 we get: 521 = 2 . 35 + 33 + 2 . 3 + 2 = 2 . 3(3+2) + 33 + 2 . 3 + 2. Now let n be an arbitrary natural number ?: 2. We want to describe a function 8 n that assigns to a natural number k the number one obtains by expanding k in
superbase n and replacing all occurrences of n in this expansion by n a formal definition of 8 n :
Definition 37: For each n < w with n ?: 2, let 8 n defined recursively by:
8 n (k· nt
+ b)
= k(n
8 n (k)
=k
+ 1)8
(t)
n
Let us consider an example: 8 3(125) = 8 3(34 + 44) = 483 (4)
:
w
-+
+ 1.
Here is
w be the function
for k < n;
+ 8 n (b)
for k < n, b < nt, t?: 1.
+ 8 3(44) = 483 (3+1) + 8 3(44) = 44+1 + 8 3(44) = 45 + 8 3(33 + 17) = 45 + 483 (3) + 8 3(9 + 8) = 45 + 44 + 42 + 8 3(2 . 3 + 2) = 45 + 44 + 42 + 2,4 + 2 = 1306.
170
10. THE ORDINALS
Now we define functions dinal.
In that assign to each natural number a countable or-
Definition 38: For each n defined recursively by:
< w with n 2: 2, let In : W
In(k) = k In(k· nt
+ b) =
for k
win(t) . k + In(b)
--+ WI
be the function
< n;
for k
< n, b < nt, t 2:
l.
Roughly speaking, In(k) is the ordinal obtained by replacing in the expansion of k in superbase n all occurrences of n by w (and reversing the order of the factors). Thus, for example, /3(521) = w(w+2) ·2 + WW + W· 2 + 2. EXERCISE 32(R): (a) Show that if k < m then In(k) < In(m). (b) Show that for all n,k < w, n 2: 2 the equality In+I(8n (k)) = In(k) holds.
Definition 39: For each n < w with n 2: 1 let gn : w --+ w be given by gl(m) = m; if gn(m) > 0; otherwise.
o.
Theorem 40(GooCtstein): For each m there exists an n 2: 1 such that gn(m) =
Before proving Theorem 40, we illustrate by means of an example how the sequences (gn(m)h' = I/'i, x {O} U>. x {1}1; /'i,·>'=I/'i,x>'l; /'i,A = IA4 Definition 7: Let
/'i"
Unfortunately, in the set-theoretic literature, the same symbols + and· are used for both cardinal and ordinal addition and multiplication. Since this may at times be confusing, we shall often explicitly state whether a particular addition/multiplication operation is supposed to be cardinal or ordinal addition/multiplication. Another device that helps eliminate ambiguity is to denote the n-th infinite cardinal by ~a instead of Wa. Thus, when we write ~a . ~i3' we refer to cardinal multiplication, whereas Wa . wi3 refers to ordinal multiplication. While we shall stick to this convention throughout this text, it is not universally accepted in the literature. Many authors write Wa ·wi3 when they want to refer to multiplication of cardinals. Our first goal in this section is to prove Theorem 3.19. We split this into two parts.
11.2. CARDINAL ARITHMETIC
177
Theorem 8: Let "-, A be cardinals such that "- ::; A and A is infinite. Then ,,-+A=A. Proof: Earlier in this section, we already proved Theorem 8 for a special case: Exercise 10 shows that 1 + A = A for every infinite cardinal A. This observation has an important corollary. Lemma 9: Every infinite cardinal is a limit ordinal. EXERCISE 13(G): Derive Lemma 9 from the result of Exercise 10. We need one more basic property of ordinal addition.
Lemma 10: II Q is any ordinal, then there exist a unique limit ordinal {3 and a unique nEw such that Q = {3 + n. Proof: Let Q E ON. Recursively define a sequence (Qk : k E w) as follows: First, set Qo = Q. Then suppose Q n has already been defined. If Q n is a successor ordinal, then let Q n +l be the unique ordinal {3 such that Q n = (3 + 1. If Q n is a limit ordinal, then let Q n +! = Qn. By Theorem 2.2 and wellfoundedness of E, it is not possible that Q n +! E Q n for all nEw. Therefore, the equality Q n = Q n +! must hold for some nEw. Let no be the smallest such n. EXERCISE 14(G): Convince yourself that Q no is a limit ordinal, and that Q no + no· Hint: For a formal proof, use induction over no.
Q
=
Q
= {3 + n = {3'
EXERCISE 15(G): Show that if {3, {3' E Lim and n, n' E w are such that + n', then {3 = (3' and n = n'.
Now let us return to the proof of Theorem 8. Let "-, A be as in the assumptions. By Theorem 3.13(£1), A = 0 + A ::; "- + A ::; A + A. Therefore, by the CantorSchroder-Bernstein Theorem, it suffices to prove that A + A ::; A. By Lemma 9, A is a limit ordinal. In particular, if Q E A and mEw, then .Q + mEA. We define a function I : A x {O} U A x {1} -+ A as follows: Suppose Q E A, {3 E Lim, and nEw are such that Q = (3+n. Then we let I(Q, i) = {3+2n+i, where i E {O, 1}. The following exercise concludes the proof of Theorem 8. EXERCISE 16(G): Convince yourself that tion from A x {O} U A x {1} into A.
I is a well defined, one-to-one func0
Theorem 11 (Hessenberg): Let "-, A be cardinals such that 1 ::; "- ::; A and A is infinite. Then "- . A = A.
Proof; Again, it suffices to prove that A . A
:s A for every infinite cardinal A.
178
11. THE CARDINALS
EXERCISE 17(G): Why? We prove this by induction over infinite cardinals >.. We have already shown that No' No = No, but the present proof subsumes this special case. All we need is the inductive assumption that>. is an infinite cardinal, and the fact that K, • K, < >. for every cardinal K, < >.. The latter is of course true if >. = No. We define a strict wellorder relation <won>. x >. as follows: (a,f3) <w (r,8) iff a + 13 < , + 8 or
a
+ 13 = , + 8 and a < ,
EXERCISE 18(G): Convince yourself that (>. x >., <w) is a strict 1.0. We show that <w is a wellfounded relation. Let A be a nonempty subset of>. x >., and let,· = min{a + 13: (a,f3) E A}. Consider B = {(a, 13) E A: a + 13 = ,*}, and let a· = min{ a : 313 ((a, 13) E By Corollary 10.25, there exists a unique 13· such that a· + 13· = The pair (a*, 13*) is the smallest element of A. Now let 8 = ot((>. x >., <w)). Since 181 = I>' x >'1, it suffices to prove that 8:::; >..
Bn.
,*.
EXERCISE 19(G): Show that>. is the largest ordinal, with the property that every proper initial segment of , has cardinality less than >.. By the above exercise, it suffices to prove the following. Claim 12: Every proper initial segment of the strict w.o. (>. x >., <w) has cardinality less than>.. Proof: Let 13" E >., and let 1((13,,)) = {(f3',,') : (13',,') <w (f3,,)} be a proper initial segment of (>. x >., <w). Let a be an ordinal such that 13 +, < a < >.. EXERCISE 20(G): Show that such an a exists. By the choice of a, the inclusion 1((13,,)) ~ {(f3',,') : f3',r' < a} holds. But the right hand side of the above inclusion is equal to a x a, and by the inductive assumption, la x al < >.. This concludes the proof of Claim 12, and simultaneously, of Theorem 11. 0 Cardinal addition and multiplication can also be defined for infinite sets of cardinals. If I is a set, and if {K,i : i E I} is a set of cardinals, then we define
L
=
K,i
iEI
IUKi
X
{i} I·
iEI
In this terminology, Theorem 3.17(b) can be written as follows: Fact 13: If III
= No and if K,i
:::;
No
for every i
LK,i:::; iEI
This fact generalizes as follows:
No.
E
I, then
179
11.2. CARDINAL ARITHMETIC
"'i
Theorem 14: Let", be an infinite cardinal, let III be a cardinal such that "'i ::;",. Then I:iEI"'i ::; ",.
::; ",;
and for each i E I, let
Proof: For each i E I, let fi : "'i ---- ", be a one-to-one function, and let Define a function F : UiEI "'i x {i} ---- ", x ", by letting F(~, i) = (fi(~)' f(i)).
f : J ---- ", also be one-to-one.
EXERCISE 21(G): Convince yourself that F is a one-to-one function. Since I", x
",1 =
0
"', Theorem 14 follows.
Corollary 15: If A is an infinite success~r cardinal, then A is not a union of fewer than A sets, each of which has cardinality less than A. If (f.ti : i E J) is an enumeration of all cardinals that are smaller than a fixed cardinal A, and if", is fixed, then we write ", CON(ZFC "There exists a weakly inaccessible cardinal.") Then ZFC + CON(ZFC) fCON(ZFC + "There exists a weakly inaccessible cardinal"), and since CON(ZFC) is provable in the theory ZFC + "There exists a weakly inaccessible cardinal," we also have ZFC + CON(ZFC) f- CON(ZFC + CON(ZFC)). Godel's Second Incompleteness Theorem, the latter implies that ZFC + CON(ZFC) is an inconsistent theory, contradicting one of our assumptions. 0
+
By
Note that Fact 21 does not imply 'that the existence of weakly inaccessible cardinals is inconsistent with ZFC or ZFC + CON(ZFC). It only implies that if we assume that the existence of a weakly inaccessible cardinal is consistent, then the gentlemen'S agreement of Chapter 6 forces us to be somewhat suspicious of results obtained under such an assumption. In particular, if a theorem has been proved under the assumption that the existence of a weakly inaccessible cardinal is consistent with ZFC, then this assumption has to be stated explicitly in the text of the theorem. Similar remarks apply to other large cardinals. The consistency of the theory ZFC + "There exists a weakly inaccessible cardinal" can be proved in the theory ZFC + "There exists a weakly compact cardinal," and consistency of the latter theory is provable in ZFC + "There exists a measurable cardinal." The larger a cardinal one uses, the more suspicious one should be about the results proved with its help. Let {Ai: i E I} be an indexed family of sets, and for all i E I, denote IAil by K,i· Then the cardinality of the Cartesian product DiEI Ai is denoted by DiE! K,i. Unfortunately, this notation is ambiguous (think about the case where Ai = K,i for all i E I), and one often needs to infer from the context what is meant. The proof of the next theorem provides ample opportunities to practice such context-sensitivity.
Theorem 22 (Konig): Let (K,i)i 2.
Theorem 26: Let>. be an infinite cardinal and assume that 2 ::; '" ::; >.. Then ",A =2A. Proof: By Theorems 11 and 3.13, the following inequalities hold: 2A ::; ",A ::; >.A ::; (2A)A = 2>'->' = 2A. 0 Theorem 27 (Hausdorff): Let "', >. be infinite cardinals. ",A
Then (",+)A
.",+.
Proof: If ",+ ::; >., then by Theorem 26, ",A = (",+)\ and by Theorem 11, = ",A '",+. If >. ::; "', then by regularity of ",+, each function f E A(",+) is bounded in ",+, and thus ",A
By the use of a (instead of >., fL, or v) for the summation index we indicate that the sum is taken over all ordinals < "', not just cardinals. 0 As an application of Theorem 27, let us try to determine the value of N~2 under the assumption that GCR holds. Using Theorem 27 twice, we get: N~2 = N~2 . N4 = N~2 . N3 . N4. Now the combination of Theorem 26, GCR, and Theorem 11 implies that N~2 . N3 . N4 = 2N2 . N3 . N4 = N3 . N3 . N4 = N4. EXERCISE 34( G): Assume that 2Nn = N2n for all nEw. Find N~3 and N~·. Theorem 28 (Tarski): If", is a limit cardinal and 0 ",A = 2: 0 ' ~ 11,. EXERCISE 38(G): Why? Therefore, J-L = La' :::; 10:01>' • 11, = 10:01>' :::; J-L. Hence, J-L = 10:01>' :::; 11,>' :::; (10:01>')>' = 10:01>">' = 10:61 = J-L. (b) Note that in this case cf(J-L) = cf(La cf(",) F. We choose an increasing sequence be : ~ < cf(K,)) of ordinals such that lime->cf("') "fe = 11,. For f E >'11, and ~ < cf(K,), we let H(f)(O be the function g E F such that g({3) = {
~((3)
if f({3) < "fe, otherwise.
EXERCISE 39(G): Convince yourself that H is one-to-one and maps >'11, into cf(",) F. Conclude that 11,>' :::; J-L cf ("'). 0
Mathographical Remarks Much of the research done on cardinal arithmetic has been inspired by the famous Singular Cardinal Hypothesis, which is usually abbreviated as SCH. This hypothesis states that J-Lcf(J-L) = max{2 cf (J-L), J-L+} whenever J-L is a singular cardinal. If SCH holds, then for every singular cardinal J-L the power 2J-L is the smallest value consistent with conditions (i) and (ii) of Easton's Theorem and with the behavior of the operation 11, f--> 21< on regular cardinals 11,. In the early 1970's, Prikry and Silver showed that if the existence of certain large cardinals is consistent with ZFC, then so is the negation of SCH (see K. Prikry, Changing measurable into accessible cardinals, Diss. Math. 68 (1970), 552, and J. Silver, Reversed Easton iteration and failure of GCH at supercompacts, unpublished notes). In 1977, M. Magidor showed in his paper On the singular cardinals problem I, Israel J. Math. 28 (1) (1977), 1-31, that the SCH may fail at Nw . His construction also uses an assumption that certain large cardinals exist. By a result of Jensen (see K. J. Devlin and R .. B. Jensen, Marginalia to a theorem of Silver, ISILC Logic Conf. Kiel 1974, Lecture Notes in Math. 499, Springer, 1975, 115-142), the use of large cardinals in the construction of models for the failure of SCH is necessary. It is possible to obtain in ZFC some upper bounds for J-Lcf(J-L) for singular cardinals J-L. At the. time of this writing, the best known upper bound for N~o is max (2 No , NN.)' It.was obtained by S. Shelah in 1989. For expositions of this result, see T. J. Jech, Singular Cardinal Problem: Shelah's Theorem on 2Nw , Bull. London Math. Soc. 24 (1992), 127-139, or M. R. Burke and M. Magidor, Shelah's pcf Theory and its Applications, Ann. Pure and Appl. Logic 50 (1990), 207-254.
186
11. THE CARDINALS
Another central problem used to be the question whether the first cardinal where GCH fails can be singular. The models of Prikry and Silver and of Magidor's paper mentioned above violate GCH even below the first cardinal J..L where SCH fails. In 1975, Silver showed in his paper On the singular cardinals problem, Proc. of the International Congress of Math. Vancouver, vol. 1 (1974), 265-268, that the first cardinal where GCH fails cannot be a singular cardinal of uncountable cofinality; and Magidor proved in his paper On the singular cardinal problem II, Ann. of Math. 106 (1977), 517-647, that if the existence of certain large cardinals is consistent with ZFC, then it is possible that Nw is the first cardinal where GCH fails. For a more detailed exposition of cardinal arithmetic we recommend T. J. Jech's book Set Theory, Academic Press 1978, and, at an advanced level, S. Shelah's Cardinal arithmetic, Oxford University Press, 1994.
CHAPTER 12
Pictures of the Universe Throughout this chapter, we will work in the theory ZF. Imagine you are the architect of the universe of sets. Your task is to build up this universe step by step. At each step of the construction, you may build new sets only from elements that have already been constructed at previous steps. How could you proceed? Let us denote the part of the universe built until step a by Va. In the beginning, nothing exists, that is, Va = 0. How can you get started? The only set that does not require preexisting building blocks is the empty set. So, after the first step, your universe will look like this: VI = {0}. In the next step, you have already one building block, and this allows you to form the set {0}. So you get: V2 = {0, {0}}. Next you can form the sets {{0}} and {0, {0}}, which yields: V3 = {0,{0},{{0}},{0,{0}}}. EXERCISE l(G): List all sets that you are allowed to form in the fourth step, and write out the corresponding V4 . Note that Vn+l = P(Vn ) for n < 4. So why not continue like this and define recursively Vn+1 = P(Vn ) for all nEw? This makes sense, since the only sets you are allowed to add at stage n + 1 are sets composed of elements of Vn , that is, subsets of Vn . Moreover, you get an increasing sequence (see Theorem 2(c)). Proceeding like this, after w steps you will have built a collection Vw = UnEw Vn . We used the symbol Vw already in Chapter 7, where we mentioned that (Vw,"E) is a standard model for all the axioms of ZFC, except for the Axiom of Infinity (see Exercise 7.32). Since you want the Axiom of Infinity to hold in the universe V of all sets, you cannot stop at this point. In the next step of the construction, you are allowed to add all subsets of Vw to the universe, then all subsets of subsets of Vw , and so on. After w more steps, you will have constructed a partial universe Vw +w • Unfortunately, this is still not a model of ZFC (see Exercise 24). By now you probably realize that the "steps" involved in the construction of V are the ordinals, and the "set-theoretic timeline" mentioned in the Introduction is the class ON. You have a long way to go before you can relax and contemplate the whole universe V. .. . The construction itself can be described in a remarkably simple way:
Definition 1: Define by recursion over the class ON: Vo = 0; Va+1 = P(Va); Va = U,e 0 is a limit ordinal, then the collection {Vi3 : {3 < o} is a set by the Replacement Axiom Schema (as long as Vi3 is a set for each (3 < 0), and the Union Axiom implies that Va is also a set. Since Vo = 0 is a set, induction over the class ON shows that each Va is a set. Thus the recursion in Definition 1 defines a functional class {(o, Va) : 0 E ON}. This functional class is called the cumulative hierarchy. The next theorem summarizes the basic properties of this hierarchy. Before stating it, let us refresh our memories regarding the notion of a transitive set. EXERCISE 2(G): Recall that a set z is transitive if TC(z) = z (Definition 4.4(c)). Show that a set z is transitive if and only if x ~ z whenever x E z. Theorem 2: Let 0 E ON. Then: (a) For every set x, x ~ Va iff X E Va+ 1. (b) Va is a transitive set. (c) V.o c Va for every {3 < o. (d) Ify ~ x and x E Va, then y EVa' Proof: Point (a) follows immediately from Definition 1. Let us prove (b) by induction over o. Clearly, Vo = 0 is a transitive set. Now suppose 0 > 0 and V.o is transitive for all (3 < o. Let x E Va. We want to show that x ~ Va. If 0 is a limit ordinal, then x E V.o for some (3 < 0, and the inductive assumption implies that x ~ Vi3 ~ Va· If 0 = (3 + 1 for some (3, then x ~ Vi3 by (a), and transitivity of Vi3 implies that every element y of x is an element of Vi3 as well. Invoking transitivity of Vi3 again, we see that every element y of x is also a subset of Vi3 . Now (a) implies that every element of x is also an element of Va, which is another way of saying that x ~ Va. 0 EXERCISE 3(G): Prove points (c) and (d) of Theorem 2. The cumulative hierarchy neatly formalizes the idea of constructing sets step by step. But does every set appear at some step of the hierarchy? Theorem 3: For every set x there exists an
0
E ON such that
x EVa'
Proof: Suppose not, and let x E V\ UaEON Va' Since x ~ TC(x), it follows from Theorem 2(d) that TC(x) ~ U aEON Va. So, without loss of generality, we may assume that x is transitive. An instance of the Comprehension Axiom Schema entitles us to form the set Xo = {y Ex: 30 E ON y E Va}. Define a ,function (! : Xo ---? ON by (!(y) = min{o : y E Va}. By an instance of the Replacement Axiom Schema, rng((!) is a set of ordinals. Let 8 = U rng((!). Then 8 E ON (by Theorem 10.10), and Xo ~ V6+1. By Theorem 2(a), Xo E V6+2' and it follows that x\xo i= 0. By the Foundation Axiom, there exists a Yo E x\xo such that Yo n x\xo = 0. By transitivity of x, this implies that Yo ~ Xo. But then Theorem 2(d) implies Yo E V6+2, hence Yo E xo, contradicting the choice of Yo. 0 Theorem 3 can be restated as follows: V = UaEON Va' But isn't this self-evident? We started this chapter by asking you to play the role of the architect of the universe, and we explained why it is natural for you
189
12. PICTURES OF THE UNIVERSE
to build up the cumulative hierarchy. Now, if you call this process "building the universe V," then, of course, V must turn out to be equal to UaEON Va' And isn't this whole construction process circular anyway? An ordinal a gets constructed only at stage a (see Corollary 6 below). But as the architect of the universe, you must know beforehand that there is such a thing as a step a before you can construct anything at this step. In other words, the class of ordinals would have to be given before you start your construction. What is wrong with the "architect's view" of set theory? Does it perhaps lead to a contradiction similar to Russell's Paradox? As far as we know, nothing is wrong with it. The tale of the architect of the universe is only a metaphor. Like any other metaphor, it can only explain some aspects of reality. You can push this metaphor a little further by imagining that in the beginning the architect is immersed in a kind of primordial soup of sets and strives to arrange this universe in a neat hierarchy. Looked at in this way, Theorem 3 says that the hierarchy built according to Definition 1 comprises the whole universe. 1 But that is about as far as the metaphor of the architect of the universe can take us. It can give us some idea why the axioms of ZFC look the way they do, but it cannot replace the axioms as a kind of First Principle to which ZFC could be reduced. Now let us leave the lofty realm of ontology and show how the cumulative hierarchy can help in studying sets. Theorem 3 allows us to define a rank function on the class of all sets.
Definition 4: For each set x let rank(x)
= min{a : x
E. Va+d.
Let RANK denote the functional class {(x, rank(x)) : x E V}.
Claim 5: For every transitive set z, the restriction of RANK to z is the rank function of the wellfounded relation E n z x z. EXERCISE 4(PG): Prove Claim 5.
Corollary 6: rank(a) = a for each ordinal a. Proof: Apply Claim 5 to z = a + 1 and note that the rank function on an ordinal is the identity. Corollary 6 can also be proved directly by induction over a. 0 Corollary 7: Let z be a transitive set. If rank(z) = a, then for every (3 there exists an x E z such that rank(x) = (3. Proof: This follows immediately from Exercise lO.15(b).
.) instead of H)..
12. PICTURES OF THE UNIVERSE
191
(i) the domain of ORD is the class PO of all p.o. 's; (ii) "Ix, y E PO (3z ((x, z), (y, z) E ORD) ~ x ~ y). It is instructive to compare Exercise 7 with Exercise 8.16. Note that the class OT constructed in Exercise 8.16 has slightly stronger properties than the class ORD of Exercise 7, but was constructed with the help of a very strong version of the Axiom of Choice (compare Corollary 33 below).
EXERCISE 8(G): Prove that for every formula cp(x,y) of Ls the following holds: 3
(Allcp)
Va ("Ix E a3y cp(x, y)
--+
3z"lx E a3y E z cp(x, y)).
Now let us consider the following question: How could one prove theorems like "CON(ZF) --+ CON(ZFC)" or "CON(ZF) --+ CON(ZFC + GCH)?" In Chapter 6 we mentioned that one might start out with a model of ZF and modify it so as to obtain a model for ZFC or ZFC + GCH. It is also possible to start not with a model (whose universe should be a set) but with the universe of sets V itself (which is a proper class). Now suppose we have decided what kind of structure to start with. Then we have two possibilities: We can try to construct a nonstandard model (X, E) or a standard model (X, E).4 Up to date, nonstandard models of ZF have found only very few applications, and in this chapter we shall concentrate exclusively on standard models. Since in a standard model the interpretation of the symbol E of Ls is always the "real" membership relation E, we shall often identify a standard model with its underlying class. Now we give a formal definition of standard models. Definition 13: (a) Let X be a class, and let cp be a formula of Ls. The relativization of cp to X is the formula cp IX that is obtained by replacing each quantifier "3Vi" occurring in cp by "3Vi E X," and each quantifier ""lvi" occurring in cp by ""IVi EX." (b) If T is a theory in Ls and X is a class such that cp IX holds for every cp in T, then we call X a standard model of T and write X F T. The phrase "cp IX holds" in the above definition is naively interpreted as "cp IX is true in the universe V." As we mentioned in Chapter 8, the use of symbols for proper classes is just a convenience, and can always be eliminated: Suppose X = {x : 1jJ(x)}. Then the relativization cp IX of a formula cp can be obtained by replacing all occurences of "3Vi" in cp by "3Vi(1jJ(Vi) 1\ ... ), and all occurences of ""Iv/' in cp by ""IVi(1jJ(Vi) --+ .•• )." 3The formulas (Allcp) are instances of what is called the Axiom Schema of Collection. Note that the difference between Collection and Replacement is that ill the former we do not require that there exists exactly one y such that cp(x, y). 4The notions of standard and nonstandard model were introduced at the beginning of Chapter 7 for the case where X is a set. If X is a proper class, then, strictly speaking, we cannot form ordered pairs (X, y). We use (X, E) as a suggestive abbreviation, and we hope the reader will forgive us the notational transgression. A rigorous treatment of standard models whose universes are proper classes is given by Definition 13(b). In the remainder of this chapter, we will use the
notation (N, E) only when we want to indicate that N is a set.
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12. PICTURES OF THE UNIVERSE
EXERCISE 9(R): Show that if X is a set, then the notion of X F T established by Definition 13(b) coincides with the notion (X, E) F T defined in Chapter 5. Hint: We did something similar in the proof of Theorem 7.1. EXERCISE lO(R): Define a notion of nonstandard models of Ls which are (possibly proper) classes. Make sure that if your models happen to be sets, then you get the satisfaction relation (X, E) F T defined in Chapter 5. If we get the same concept as in Chapter 5, why do we have a separate definition for class models? And why do we bother to look at class models in the first place? One can prove in ZF that if a theory has a set model, then it is consistent. But by G6del's Second Incompleteness Theorem it is impossible to prove in ZF that ZF is a consistent theory. Therefore, if we want to work with set models, then we have to transcend the theory ZF by making the unprovable assumption that there is such a model. Class models are different in this respect. Let {'Po, . .. ,'Pn, ... } be a list of all axioms of ZF. Given nEw, the symbols "V F {'Po, ... ,'Pn}" and "'Po /\ ... /\ 'Pn" stand for exactly the same thing. Thus, for every nEw, ZF f- "V F {'Po, ... ,'Pn}." Now common sense would dictate that V F ZF, but the latter is not a theorem of ZF! How do we know? Well, the symbol "V F ZF" would stand for something like "'Po /\ ... /\ 'Pn ... ," and this is not even a well-formed expression of Ls, let alone a theorem of ZF. In other words, it is not provable that V is a model of ZF, and no conflict with G6del's Second Incompleteness Theorem arises. So what is the meaning of Definition 13(b)? Shouldn't the use of "X F T" be outlawed when T is infinite? Maybe so. But consider the following situation: Suppose X is a class such that for all nEw,
ZF f- (AC)/x/\ GCH/ x /\ 'Po/x /\ ... /\ 'Pn/X' Moreover, it is conceivable that the following is a theorem of ZF: (a) . "For every nEw, there exists a formal proof p of the statement (AC)/x/\ GCH/ x /\ 'Po/x /\ ... /\ 'Pn/X such that the only axioms used in p are from the list {'Pn : nEw}." Note that such a theorem says nothing about the consistency of the list {'Pn : n E w} itself and hence does not contradict G6del's Second Incompleteness Theorem. In order to prove a theorem like (a) on the purely syntactical level, one would have to show how to construct formal proofs of GCH/ x , 'Po/x, etc. straight from the axioms. But human minds are not designed to shuffle meaningless symbols. Fortunately, Definition 13(b) allows us to express (a) in the crisper form: ((3) "If V F ZF then X F ZFC + GCH." It is much easier for us to understand what is going on if we imagine that we can start with a model V of ZF, and, by throwing away a few sets if need be, construct another class X that is a model of ZFC + GCH. In this sense, ((3) is more user-friendly than (a). Once we have proved theorem ((3) in this way, it is not hard to convince ourselves that ((3) has a formal ZF-counterpart (a), and that it would be "in principle" possible to translate the proof of ((3) into a ZF-proof of (a). Most of the standard models considered in the set-theoretic literature are transitive. The reason is that transitive models are easier to work with than non-transitive ones. Moreover, every interesting consistency result that can be obtained by means
12 .. PICTURES OF THE UNIVERSE
193
of a standard model can also be obtained by means of a standard transitive model. More precisely, the following holds. Theorem 14: Let T be a theory in L8 that contains the Axiom of Extensionality. If T has a standard model, then T has a standard transitive model. Proof: We prove the case where T has a standard set model and leave the case of class models as an exercise. As in Chapter 7, we abbreviate the Axiom of Extensionality by (AI).
Claim 15: Let N be a set. Then (N, 'E) onN.
P (AI)
iff 'E is an extensional relation
EXERCISE l1(G): Recall Definition 4.38 (of an extensional relation) and prove Claim 15. Now suppose T is a theory in L8 such that (AI) E T, and assume that (N, 'E) P T. By Theorem 4.39, there exist a transitive set M (the Mostowski collapse of N) and a bijection F : N -+ M such that
(M)
Vx,y E N (x'Ey
+->
F(x)'EF(y)).
Condition (M) asserts that F is an isomorphism of the models (N, 'E) and (M, 'E). Theoff~m 6.4, (M, 'E) p T. All notions used in this proof can be generalized to classes. In particular, we call a proper class X transitive if Vx E XVy E X (y E x -+ Y EX).
By
EXERCISE 12(R): Prove that if T is as in the assumptions of Theorem 14, and if a class X is a standard model of T, then there exists a transitive class Y such thatypT. 0 Now suppose you are immersed in the universe of sets V, you are being told that the axioms of ZF hold, and you want to find a model for ZFC or even ZFC + GCR. By what was said above, it would be wise to look for a standard, transitive model. Now you can look either for a set model or for a proper class. Let us first try the former. Earlier in this chapter we defined the Va's and the H).,'s. These are transitive sets; are some of them perhaps models of ZF? Exercise 7.32 settles this problem for Vw = HNo.5 If you did this exercise, you already know that it is rather tedious to verify all the axioms of ZF one by one. Fortunately, there are some shortcuts. Lemma 16: Let X be a transitive class. Then X p (AI) 1\ (A2) 1\ (A3). Proof: Theorem 7.1 asserts that (AI) holds in all standard transitive (set) models; and by Exercise 7.6(b), the Empty Set Axiom holds in all standard (set) models. 5Recall that all axioms except the Axiom of Infinity hold in this model.
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12. PICTURES OF THE UNIVERSE
EXERCISE 13(PG): (a) Generalize the results of Theorem 7.1 and Exercise 7.6(b) to class models. (b) Prove that X F (A3) holds for every class X. 0 We say that a class X is closed under the power set operation if P(x) E X for every x E X. EXERCISE 14(G): Convince yourself that Va is closed under the power set operation if and only if a E LIM. Lemma 17: Let N be a transitive set closed under the power set operation. Then (N, E') satisfies the Power Set Axiom and the Comprehension Axiom Schema.
Proof: This may look utterly trivial, since the Power Set Axiom apparently "says" that N is closed under the power set operation. But not so fast. If x, YEN, is it really true that (E)
y
= P(x) iff (N, E') F y = P(x)?
Recall that expressions like "P(x)" are abbreviations. To be on the safe side, let us unravel the meanings of both sides of (E). The expression "y = P(x)" stands for:
I;jz(z
(Ihs) whereas "(N, E') (rhs)
Fy=
~
x ...... z E y),
P(x)" abbreviates: I;jz E N (z
~
x ...... z E y).
There is no reason to expect that (lhs) should always be equivalent to (rhs). For example, if N is not transitive, there could be some z in y which is not a subset of x (and hence witnesses the failure of (Ihs)) , but is not an element of N either (and hence does not invalidate (rhs)). But wait a minute. What do we mean by "z is not a subset of x?" Do we want to interpret this as z rz. x or as (N, E') F z rz. x? To get a handle on this confusing issue, we need a new concept.
Definition 18: Let cp(vo, ... ,Vk) be a formula of Ls with free variables Vo, ... ,Vk, and let X be a class. We say that cP is absolute for X if cp(xo, .. · ,Xk) ...... cP jX(xo, ... ,Xk) for all Xo, ... ,Xk E X. We say that cP is absolute for transitive models if cp is absolute for every transitive class X. A formula cp is absolute if it is absolute for every class X. Let us list some formulas whose absoluteness for transitive models will be used in this chapter.
Claim 19: The following formulas are all absolute for transitive models: (a) CPo(x,y): Y = TC(x); (b) CP1(X): x EON; (c) CP2 (x, y): x ~ y; (d) CP3(X,y): y = Ux; (e) CP4(X,y,Z): z = {x,y}; (f) CP5(X): x = w; (g) CP6(X): x E w; (h) CP7(X,y,Z): x,y,z E ON 1\ x +y = z; (i) CPs (x).' X is a function; (j) cPg(x, y, z): z is a bijection from x onto y.
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12. PICTURES OF THE UNIVERSE
It is worth noting that none of the above formulas is absolute. This is perhaps most surprising for 'P4; so let us demonstrate nonabsoluteness of this formula. Consider the model (N, E), where N = {0, {0}, {0, {0}, {{0}}}}. Of course,
{0,{0},{{0}}}
i {0,{0}},
but nevertheless
(N, E)
F 'P4(0, {0}, {0, {0}, {{0}}})
EXERCISE 15(PG): Show that none of the formulas 'PO-'P9 is absolute. Proof of Claim 19: We give the proof only for 'P2 and leave the other formulas as an exercise. Suppose x, yEN and N is transitive. Then the formula "x ~ y" is an abbreviation for
(A)
Vz (z
and "(N, E)
Fx
~
E x
-+
z E y),
y" stands for
(B)
Vz
E N
(z
E
x
-+
z
E
y).
It is clear that (A) implies (B), even if N is not transitive. But if N is transitive, then z E x already implies that zEN. In other words, the implication z E x -+ z E Y is vacuously true for all z tJ. N (because the antecedent fails), so that the truth or falsity of (A) only depends on the z's in N, which is another way of saying that (B) implies (A).
EXERCISE 16(PG): Prove the remaining parts of Claim 19. Note that the formula "y = P(x)" does not appear in Claim 19. For good reason: It is not absolute for transitive models. EXERCISE 17(G): Let a E ON. Show that ON Luckily, the formula y power set operation.
F P(a) = a + 1.
= P(x) is absolute for transitive models closed under the
EXERCISE 18(G): (a) Assume that N is transitive, x, yEN, and y Show that (N, E) F "y = P(x)." (b) Deduce Lemma 17.
= P(x). 0
EXERCISE 19(PG): Show that for every ordinal a and every infinite cardinal ). the models (Va, E) and (H).., E) satisfy the Union Axiom. EXERCISE 20(G): Show that if a> w and ). is an uncountable cardinal, then (Va, E) and (H).., E) satisfy the Axiom of Infinity. EXERCISE 21(G): Show that if a is a limit ordinal and ). is an infinite cardinal, then (Va, E) and (H).., E) satisfy the Pairing Axiom. EXERCISE 22(G): Let a E ON, and let). be an infinite cardinal. Show that if (AC) holds in V , then (Val "E) and (HAl "E) satisfy the Axiom of Choice.
196
12. PICTURES OF THE UNIVERSE
EXERCISE 23(0): Show that if A is a regular infinite cardinal, then (H>., E") satisfies the Axiom Schema of Replacement. For which 0: does Va satisfy the Axiom Schema of Replacement? This is a delicate question. Consider the following formula:
(1)
cp(n, y) : nEw 1\ y = w + n.
By Claim 19, this formula is absolute for transitive models. Therefore,
(2)
(Vw+w,"E) pVnEw3!ycp(n,y). EXERCISE 24(G): (a) Convince yourself that
(3)
(Vw+w,"E)
P -,3zVn E w3y E zcp(n,y).
Hint: Use Corollary 6. (b) Conclude that (Vw+w, "E) is not a model of the Replacement Axiom Schema. EXERCISE 25(PG): Show that neither (Vw.w,"E) nor any model of the form (Va +w . 2 , "E) satisfies the Replacement Axiom Schema. EXERCISE 26(R): Show that (VWl,"E) does not satisfy the Replacement Axiom Schema.
Remark 20: Somewhat surprisingly, it is not possible to show that if cf(o:) < 0:, then (Va,"E) does not satisfy Replacement. In Exercise 25 we used the fact that the ordinals w . wand 0: + w . 2 are definable (the latter ordinal is definable with parameter 0: E Va +w . 2 )' We will return to this issue in Chapter 24. Remark 21: Note that we have shown that the Replacement Axiom Schema is not a consequence of the remaining axioms of ZF (all of them, and possibly even the Axiom of Choice, are satisfied in (Vw+w, "E)). By deleting the Replacement Axiom Schema from the axioms of ZF, one obtains a theory called Zermelo's Set Theory and abbreviated by Z. Adding the Axiom of Choice one obtains the Theory ZC. Since the most extensively studied mathematical objects (like, for example, the real line) can be coded as elements of Vw +w , an overwhelmingly large portion of mathematics can be formalized in ZC. On the other hand, ZC does not even imply the existence of the ordinal w + w. Since getting a model of the form (Va, "E) for all axioms of ZF seems a bit tricky, let us try our luck with the H>.'s. These models are much more inclined to satisfy Replacement, but there is a little problem with the Power Set Axiom. Let us investigate which H>. 's are closed under the power set operation. An initial ordinal >. is called a strong limit cardinal if 21< < >. for every'" < >.. Note that this definition makes sense even in ZF. 6 Moreover, note that No is a strong limit cardinal, and if GCH holds, every limit cardinal is strong limit. EXERCISE 27(PG): Show that H>. is closed under the power set operation if and only if>. is a strong limit cardinal. Combining Lemmas 16 and 17 with Exercises 19-22, we obtain the following. 6If A is a strong limit cardinal, then P(K,) can 'be wellordered for every K, < A. Thus, in ZF alone one cannot prove the existence of an uncountable strong limit cardinal. But it is also relatively consistent with ZF that an uncountable strong limit cardinal exists while the full Axiom of Choice fails.
12. PICTURES OF THE UNIVERSE
197
Theorem 22: Suppose A is an uncountable strong limit cardinal. Then H).. is a model of ZF. EXERCISE 28(PG): Show that if A is a strong limit cardinal, then (H)..,"E) (AC), even if the Axiom of Choice fails in V.
F
Apparently we have found a set model of ZF. But wait a minute. If A = K+ for some K, then 2K :;::: A, and hence A is not a strong limit cardinal. In other words, every strong limit cardinal is a limit cardinal. Therefore, if A satisfies the assumptions of Theorem 22, then A is weakly inaccessible. Even more is true. Let us call an uncountable strong limit cardinal strongly inaccessible. If GCH holds, then the notions of strongly inaccessible and weakly inaccessible cardinals coincide, but in general, it need not be the case that every weakly inaccessible cardinal is strongly inaccessible. Theorem 22 and Exercise 28 imply the following:
Corollary 23: ZF
+ ":3
a strongly inaccessible cardinal"
r-
CON(ZFC).
Proof: "CON(ZFC)" can be interpreted as "there exists a model for ZFC." 0 Corollary 24: CON(ZF) cardinals. ")
r
CON(ZF
+
"there are no strongly inaccessible
Proof: This follows immediately from Corollary 23 and Godel's Second Incompleteness Theorem. However, the proof of Godel's Theorem is beyond the scope of this book. Therefore, let us give an alternative, entirely self-contained argument. EXERCISE 29(PG): Show that if A is strongly inaccessible, then H).. =
V)...7
Now suppose all axioms of ZF hold in the universe V. If there are no strongly inaccessible cardinals in V, then there is nothing to prove. So suppose there are strongly inaccessible cardinals. By wellfoundedness of ON there exists a smallest one. The following lemma concludes the proof of Corollary 24.
F
Lemma 25: Let A be the smallest strongly inaccessible cardinal. Then (V)..,"E) "There are no strongly inaccessible cardinals."
Proof: Let ), be as in the assumptions, and let !PlO(K) be the formula: "K is a strongly inaccessible cardinal." We want to show that if K E V).., then (V)..,"E) F -'!PlO(K). Of course, by the choice of A, if K E V).., then K is not a strongly inaccessible cardinal in V, so !PlO(K) fails (in V) for every K E V)... But the formula !P1O is not absolute for transitive models in general. 8 Fortunately, the following holds. Sublemma 26: Let 8 be a limit ordinal. Then!plO is absolute for V6 · Proof: Let 8 be as in the assumptions, and let K E V6. Note that !PlO(K) can be written as !P1(~) 1\ !Pll(K) 1\ !P12(K), where: 7S 0 , after all, we have found a model of ZF (even ZFC) of the form (V"" E)! 8In order to show this, one needs more sophisticated tools than the ones presented here.
198
12. PICTURES OF THE UNIVERSE
CPl(K): K EON; CPll (K): "K is a regular cardinal;" CPI2(K): "K is strong limit." The idea of the proof is to show that failure of anyone of the formulas CPl (K), CPll (K), CPI2(K) in V implies its failure in (VO, 'E). The least troublesome of the three formulas is CPI, since Claim 19 says that it is absolute for transitive models. The case of CPI2(K) is also relatively easy. Note that CPI2(K) can be written as: Vci. E KCPI3(Ci.,K), where
CPI3(Ci., K): ~f~p~(3 E K(CP9(p, (3, J) /\ P = P(ci.)). By Claim 19, CP9 is absolute for transitive models. Moreover, by Exercise 18, the formula y = P(x) is absolute for models closed under the power set operation (in particular, for Vo). Therefore, repeated application of the following claim shows that if CP12 fails in V, then it fails in (VO, 'E).
Claim 27: Let X be a transitive class, let y E X, and let cp(x) be a formula that is absolute for X. Then the formulas ~x E y cp(x) and Vx E y cp(x) are absolute for
X. Moreover, if the formula ~x cp(x) fails in V, then X if the formula Vx cp(x) holds in V, then X F Vx cp(x).
F= -dxcp(x). Analogously,
EXERCISE 30(PG): (a) Prove Claim 27. (b) Infer that if CPI2(K) fails in V, then CPI3(Ci., K) fails in (VO, 'E) for all ci. E K. (c) Conclude that if CPI2(K) fails in V, then CPI2(K) fails in (VO, 'E). Now consider the formula CPll (K). It can be written as: Vf CPI4(J, K), where
CPI4(J, K): "CP8(J) /\ dom(J) E K /\ rng(J) ~ K --> ~(3 E K (rng(J) ~ (3)." EXERCISE 31(PG): Prove that the formula CPI4(J, K) is absolute for transitive models. From Claim 27 and Exercise 31 we can infer the following. Corollary 28: Let X be a transitive class and assume in V, then X F= CPll(K).
K
E X. If CPu (K) holds
This corollary will come in handy at the end of this chapter, but for the present purpose it is useless, since we need its converse. The following observation helps out.
Claim 29: Let Ci.,(3 E ON, assume f maps ci. into (3, and let 'Y = max{Ci.,(3}. Then f E VI'+3' EXERCISE 32(G): Prove Claim 29. By Claim 29, CPll(K) is equivalent to the formula Vf E V"'+3 CPI4(J,K). Since V"'+3 E Va' we can infer from the first part of Claim 27 that if CPU(K) fails in V, then (VO, 'E) F= -'CPll(K). This concludes the proof of Sublemma 26, and hence of Lemma 25. D
12. PICTURES OF THE UNIVERSE
199
Have we perhaps overlooked some interesting set models? In other words, are there transitive set models that are not of the form V,,? Perhaps yes, but such models cannot be closed under the power set operation. EXERCISE 33(R): (a) Show that if X is a transitive class, then either XnON = ON or XnON EON. (b) Prove that the formula a such that for every z E x: (L{3,"E)
F cp[z, ao, ...
,ak-l]
iff L
F cp[z, ao, ... , ak-l]
Lemma 35 is an instance of a more general theorem, called the Reflection Principle. The Reflection Principle will be discussed in Chapter 24. EXERCISE 36(R): Prove that L satisfies the remaining axioms of ZF. How could one construct a wellorder of L? The very definition of L suggests that one should do this by recursion. To begin with, let (CPn)nEw be an enumeration of all formulas of Ls that are not sentences. Now suppose LOI. has been wellordered by a relation ::;01.' This generates a lexicographic order relation ::;~ on (LOI.)k. Note that ::;~ is also a wellorder relation. Now suppose CPn has one free variable x and parameters ao, ... ,ak(n)-I' Given ao, ... ,ak(n)-1 E LOI., let
Yn(ao, ... , ak(n)-I) = {x E LOI. : (LOI.,"E)
F CPn[X, ao,···
,ak(n)-d}·
For Y E LOI.+1 \LOI.' we let n(y) = min{n : y = Yn(ao, ... , ak(n)-I)}, and denote by a(y) the ::;~(n(Y))-minimal k(n(y»-tuple (ao, ... , ak(n)-I) such that y = yn(y) (ao, . .. , ak(n(y))-d· Now we define a relation ::;01.+1 on LOI.+1 as follows: y ::;01.+1
z iff (i) y, z E LOI. /\ Y ::;01. z or (ii) y E LOI. /\ z ~ LOI. or (iii) y, z E LOI.+1 \LOI. /\ n(y) < n(z) or (iv) y, z E LOI.+1 \LOI. /\ n(y) = n(z) /\ a(y) ::;~ a(z). It is a routine, but somewhat tedious exercise to check that ::;01.+1 is a wellorder relation on LOI.+1 that extends ::;01.' Now we can inductively define ::;{3 for every /3. At limit stages, we simply take ::;')'= UOI. "" such that y E L",. Since the theory T of Claim 38 is finite, we can form the conjuction 'fl of all sentences in T. By Lemma 35, there exists a f3 ~ a such that (L(3, E") 1= 'fl, or equivalently, (L(3, E") 1= T. By a result that will be proved in Chapter 24, there exists a set N ~ L(3 such that: (1) "" ~ N and yEN; (2) INI = ""; (3) for every k E w, for every formulaW(ao, ... ,ak-I) of Ls with k free variables, and for every k-tuple (ao, ... ,ak-I) of elements of N we have (N, E") 1= W[ao, ... ,ak-I] iff (L(3, E") 1= W[ao, ... , ak_d· lO In particular, by (3), (N, E") 1= T. However, N is not in general transitive. Now consider the Mostowski collapse M of N. This M is transitive. Moreover, the collapsing function f is an isomorphism of the models (N,"E) and (M, E"). So, in particular, (M,E") 1= T. By Claim 38, M = L, for some I E ON. By (2), I < ",,+. EXERCISE 38(PG): Let M, N be as above, and let f : N -> M be the collapsing function. Show that f(a) = a for every a E "", and conclude that f(y) = y. Hint: Use (1). 10 An
N that satisfies (3) is called an elementary submodel of L{3.
12. PICTURES OF THE UNIVERSE
203
Exercise 38 implies that Y E L-y. Thus, by Lemma 31(v), y E LI. be weakly inaccessible. Then I>. E L, and by Corollary 28, L 1= "I>. is a regular cardinal." The following exercise shows that we also have L 1= "I>. is a limit cardinal." EXERCISE 39(PG): Show that for every>.
.+
1>.,
if L
1= ">' is a cardinal,"
then
< 1>..
Therefore, L 1= "I>. is a weakly inaccessible cardinal." Since L satisfies the GCH, 0 we can conclude that L F "/1, is a strongly inaccessible cardinal."
204
12. PICTURES OF THE UNIVERSE
Mathographical Remarks Kurt Godel first announced Theorem 32 in two short articles that appeared in 1938 and 1939. The complete proof was published in 1940. The monograph Constructibility, by Keith J. Devlin, Springer, 1984, gives a thorough presentation of the constructible universe and its various properties. Devlin's book is also a good starting point for those readers who wish to learn more about inner model theory.
Subject Index absolute for transitive models, 194 Aczel, P., 120 aleph, 32 almost disjoint, 137 Anti-Foundation Axiom, 120 architect's view of set theory, 5 Aristotle, 4 arity, 71 automorphism, 82 axiom, 78 Axiom of Choice, 118 Axiom of Determinacy, 148 Axiom of Extensionality, 11, 107 Axiom of Infinity, 119 Axiom Schema of Collection, 191 axiom schemata, 95 axiom system complete, 98 effective, 98 sound, 98 axiomatic set theory, 5 axiomatizable, 92
cardinal exponentiation, 176 cardinal multiplication, 36, 176 cardinal number, 173 cardinal successor, 174 cardinality, 30 Cartesian product, 12, 13 categorical, 83 chain, 133 maximal, 133 choice function, 131 Church's Thesis, 89 class, 23, 121 domain, 124 functional, 123 injective, 124 proper, 23, 121 relational, 123 transitive, 193 wellfounded, 126 wellorder, 125 closed under countable intersections, 137 closed under finite intersections, 134 closed under supersets, 134 closed under the power set operation, 194 closed under unions, 133 closure, downward, 60 cofinal, 159, 179 cofinality, 179 cofinite, 135 Cohen, P. J., 41, 99 collapsing function, 65, 155 Compactness Theorem, 79 comparable, 18 complete linear order, 54 Comprehension Axiom Schema, 112 consequence, 78 consistency strength, 203 consistent, 76 consistent with, 79 constant symbol, 71 constructible hierarchy, 200 constructible set, 200 constructible universe, 200 continuum, 40 Continuum Hypothesis, 40
back and forth construction, 53 Banach-Tarski Paradox, 143, 153 Bernstein set, 145 bijection, 13 Borel set, 41 bound, upper, 134 Bukovsky-Hechler Theorem, 184 Cantor's Normal Form Theorem, 168 Cantor's set definition, 1 Cantor's Theorem, 32 Cantor, G., 1, 50 Cantor-Bernstein Theorem, 42 Cantor-Schrader-Bernstein Theorem, 31, 46 cardinal large, 180 regular, 179 singular, 179 strong limit, 196 strongly inaccessible, 197 weakly inaccessible, 180 cardinal addition, 36, 176 205
206
SUBJECT INDEX
countable, 32 hereditarily, 190 Countable Axiom of Choice, 147 countably complete, 137 cumulative hierarchy, 188 Dedekind complete linear order, 54 Dedekind cut, 53 Dedekind-finite, 146 DefinabiIity by Recursion over a Wellfounded Relation, 115 definition (of new functions and relations), 102 dense, 22 dense subset, 54 denumerable, 32 Descriptive Set Theory, 42 diagonalization, Cantor's, 33 domain, 13 Easton's Theorem, 182 elementarily equivalent, 83 elementary submodel, 202 embedding, 82 Empty Set Axiom, 108 end-extension, 21 enumeration, 32 one-to-one, 32 equiconsistency, 203 equipotent, 29 equivalence, 23 equivalence class, 14 equivalence relation, 14 equivalent, 130 equivalent in T, 130 expansion in superbase, 169 expression, 73 family indexed, 13 mad, 137 Fermat's Theorem, 95 filter, 134 cocountable, 137 cofinite, 135 proper, 134 finite, hereditarily, 190 First Foundational Thesis, 88 first-order definitions, 86 first-order language, 76 first-order property, 86 first-order theory, 76 forcing method, 99 formal language, 71 formula, 73 absolute, 194 provable, 76 Fraenkel, A., 5, 107
.'. function, 13 characteristic, 33 functional symbol, 71 Godel's Completeness Theorem, 79 Godel's First Incompleteness Theorem, 93, 97 Godel's Second Incompleteness Theorem, 97 Godel's Theorem, 200 Godel, K, 204 Gale, D., 150 game, 148 determined, 148 outcome, 148 gap, 53 Generalized Continuum Hypothesis, 183 Goodstein's Theorem, 170 Hausdorff's Maximal Principle, 133 Hausdorff's Theorem, 184 Hausdorff, F., 151 hereditarily, 190 hereditarily finite, 46 Hessenberg's Theorem, 40, 177 higher-order properties, 87 homomorphism, 81 image, 13 inverse, 13 immediate successor, 158 implicitly established, 94 inconsistent, 76 independence result, 3 independent, 79 induction, 44 induction step, 43 inductive assumption, 43 initial ordinal, 173 initial segment, 56, 60 proper, 60 injection, 13 inner model, 41, 203 instance of an axiom schema, 112 interpretation, 76 invariant, 86 irreducible representations, 151 isomorphic, 82 isomorphism, 82 Jensen, R. B., 185 Konig's Theorem, 181 Kuratowski, K, 134 Lowenheim-Skolem Theorem, 85 language, 71 countable, 71 language of group theory, 73 language of set theory, 72
SUBJECT INDEX
Lebesgue measure, 144 regular, 145 limit cardinal, 174 limit- ordinal, 159 linear order, 18 logical connective, 71 Luzin set, 100 m-categorical, 85 Magidor, M., 185, 186 Martin, D. A., 150 mathematical induction, 43 maximal,18 maximum, 18 measurable, 144 measure, 144 invariant, 144 vanishing on the points 144 membership relation, 1 ' metalanguage, 89, 112 minimal,18 minimum, 18 model,76 model theory, 105 modus ponens, 75 Monster group, 80 Morse-Kelley set theory, 127 Mostowski collapse, 65 Mycielski, J., 148 naive set theory, 5 nonconstructive axioms, 114 nonstandard model, 107 number algebraic, 39 tran~cendental, 40 nUmeral,45
207
ordinal multiplication, 165 ordinal subtraction, 164 pair ordered, 12 unordered, 12 parameter, 112 partial pre-order, 26 partition, 15 Peano Arithmetic, 96 permutation, 48 Plato, 4 Platonism, 121 Polish space, 175 power set, 17, 114Power Set Axiom, 114 Priciple of Generalized Recursive Definitions, 162 Prikry, K., 185 PrinCiple of Dependent Choices, 147 PrinCiple of Generalized Recursive Constructions, 118 Principle of Induction over a Wellfounded Set, 56 product antilexicographic, 25 lexicographic, 25 simple, 25 product of an infinite set of cardinals, 181 proof,75 properly extends, 134 property group-, 86 order-,86 topological, 86 quantifier, 71
Ockham's razor, 108 one-to-one, 13 orbit, 152 order antilexicographic, 25 lexicographic, 25 partial, 17 strict partial, 17 order invariant, 22 order property, 22 order sum, 24 order type, 23, 156 order types associativity and distributivity laws order-embedded, 21 ' 25 order-isomorphic, 21 . order isomorphism, 21 order-preserving, 21 ordinal, 23, 155 ordinal addition, 163 ordinal exponentiation, 167
range, 13 range of a quantifier, 74 rank function, 21, 62, 160 recursion, 44 recursion over the length of formula.:., 77 recursive, 92 recursive construction, 48 recursively enumerable, 92 reduct of a model, 104 Reflection Principle, 201 relation 3-ary, 13 antisymmetric, 14 asymmetric, 14 binary, 12 extensional, 64 irreflexi v, 14 reflexive, 14 symmetric, 14 transitive, 14 relational symbol, 71
208
relatively consistent, 99 relativization, 191 Replacement Axiom, 113 Replacement Axiom Schema, 113 representation, 122 representation theorem, 22 representative, 14 rules of inference, 75 complete, 75 effective, 75 sound, 75 Russell's Paradox, 4 a-algebra, 144 a-field, 144 satisfaction reliition, 77 satisfies, 77 Schroder-Bernstein Theorem, 42 Second Foundational Thesis, 98 second-order definition, 86 second-order property, 87 selector, 131 sentence, 74 Separation Axiom Schema, 112 set empty, 11 finite, 47 Set Existence Axiom, 112 set-like, 125 set-theoretic timeline, 187 Shelah, S., 185 Silver, J., 185, 186 Singular Cardinal Hypothesis, 185 Soundness Theorem, 79 standard model, 107, 191 Steinhaus, H., 148 Stewart, F. M., 150 strategy, 148 strict w.o., 19 strictly wellfounded, 19 structure, 76 subclass, 123 subformula, 74 submodel, 82 substitution, admissible, 74 substructure, 82 successor cardinal, 174 successor ordinal, 159 sum of an infinite set of cardinals, 178 surjection, 13
SUBJECT INDEX
symbol logical, 71 nonlogical, 71 Tarski's Theorem, 184 Tarski, A., 146 term, 73 theory, 77 complete, 79 of a model, 91 of a structure, 91 transfinite induction, 161 transfinite recursion, 161, 162 transfinite sequences, 162 transitive closure, 45 transitive set, 46 translation-invariant, 144 transversal, 131 Tychonoff's Theorem, 141 ultrafilter, 135 fixed, 135 free, 135 principal, 135 uncountable, 32 uniformization, 131 Union Axiom, 111 universe, 76 urelement, 107 valuation, 77 variable bound, 74 free, 74 variable symbol, 71 Vitali set, 144 von Neumann-Bernays-Godel set theory, 127 W-minimal, 56 wellfounded, 19 wellorder, 19 definable, 126 Wellorder Principle, 132 Wiles, A., 95 winning strategy, 148 Zermelo's Set Theory, 196 Zermelo's Theorem, 132 Zermelo, E., 5, 107 ZFC,79 Zorn's Lemma, 134
Index of Notation Greek letters (and ~) are listed under the first letter of their English name. Special symbols (0, =, etc.) and operations (01 + (3, x;+, etc.) are grouped separately at the end. H A ,190 HMP, 133
AC, 131 (AC)NO' 147 (AC)+, 142 (AD),148 (AG),142 ARI(t),94 2(=
(II),43 (12), 43 (13), 43 (IC), 142 I(x),56 Iw(x),56 Indw(x), 56
(A,(R.;)iEI,(Fj)jEJ,(CkhEK), 76
~o,
32 ~o" 176 ck,71
L,71 L, 115, 199 LA,92 L"o 200 La, 72 Ls,72 L., 24
CAR,125 CARD(x), 123 Cf(Oi), 179 CH,40 CON(T),97 c,40 Xx' 33 (DC),147 DC(Y),60 Def (x), 200
DIS, 133
ml,144
DLONE,84
m3,144
MK,127
EQ,123 167
eO,
n VJI , 102 NBG,127
T/,24
Fa, 135
01,123 ON,156 OT,126 ot(X), 23 ot«(X, -
