3-Interval irreducible partially ordered sets

Order 11: 97-125, 1994. © 1994 KluwerAcademic Publishers. Printed in the Netherlands.

97

3-Interval Irreducible Partially Ordered Sets* STEFAN

FELSNER

Freie Universitiit Berlin, Fachbereich Mathematik, lnstitut fiir Informatik, Takustr. 9, 14195 Berlin, Germany (E-mail: [email protected]) Communicated by W. I:. Trotter (Received: 6 March i993; accepted: 20 May 1994) Abstract. In this paper we discuss the characterization problem for posets of interval dimension at most 2. We compile the minimal list of forbidden posers for interval dimension 2. Members of this list are called 3-interval irreducible posers. The problem is related to a series of characterization problems which have been solved earlier. These are: The characterization of planar lattices, due to Kelly and Rival [5], the characterization of posets of dimension at most 2 (3-irreducible posers) which has been obtained independently by Trotter and Moore [8] and by Kelly [4] and the characterization of bipartite 3-interval irreducible posers due to Trotter [9]. We show that every 3-interval irreducible poset is a reduced partial stack of some bipartite 3-interval irreducible poset. Moreover, we succeed in classifying the 3-interval irreducible partial stacks of most of the bipartite 3-interval irreducible posers. Our arguments depend on a transformation P ---, B ( P ) , such that I d i m P = dim B(P). This transformation has been introduced in [2]. Mathematics Subject Classifications (1991). 06A07, 05C35. Key words. Partially ordered sets, interval orders, dimension.

1. I n t r o d u c t i o n

and Basics

An extension of a poset P = (X, < p ) is a partial order Q = (X, 10. I f y is an up-made (low-made) element of B(P) and F is a connected subposet of Inc(y), then there exists a low-made (up-made) element x in B(P) so that x ) y (y >t x) and F C_ Inc(x). Proof We prove the theorem when y is up-made. If y is also low-made, then x = y and we are done. Suppose that y is not low-made. Let y' > y be low-made and note that the existence of a low-made element y" > y with y" [[ y' is guaranteed by Theorem 3. Therefore, there are at least two incomparable low-made elements in SUCCB(p)(y). If there is a z E F with x ) z for all low-made elements x > y, then y / > z by Theorem 3. A contradiction. We therefore may assume, that there are elements x, x' E SUCCB(p)(y) and z, z' E F such that z < x and z' < x t but x il z' and x' 1[ z. Now let a l , . . . , an be a fence in F from z' to z. It follows that {x, y, x', a l , . . . , an} contains a crown. A contradiction. [] Let P = (X, < p ) be a 3-interval irreducible poset. Since the dimension of B(P) is 3, B(P) contains a 3-irreducible poset. Suppose first, that B(P) contains a crown A,~ for some n ) 0. Then, it follows from Theorem 5, that P also contains a crown A,~ for some o ~< m ~< n. Since P is irreducible, we conclude, that P = A,~. We may, therefore, assume in the remainder of the argument, that B(P) does not contain a crown An. The remainder of the argument is divided into a sequence of cases. When discussing the case of a 3-irreducible order Q = (Y, e2 and e" > e2 in B(P) with el E Inc(e") but d' > e3 and e3 E Inc(e') but e' > el. This implies that {a, el, e", b3, d, e3} form a copy of D in B(P). The contradiction allows us to assume an element e~. Assume, that bl is low-made, then choose an up-made element b] < bl with b2, c3 C Inc(b]). If b~ < b3, then {c2, be, a, cl, b], 63) form a copy of D d in B(P). We conclude, that b3 E Inc(b]), hence, any U-preimage of b] is b~. Symmetrically, we may assume an element b~. The bipartite order with maximal elements Cl, ~ c2, ~ c3,~b3 and minimal elements a ~, b~, b'~,b~, c~, c~ forms a copy of O2 in P. Hence, P contains a partial stack of 02.

Case 3. B(P) contains CX3. From Lemma 3 we may assume that there are elements a~, bpb3, ~ ~ c~ and a~,a~,b~. Choose a low-made element b~ > b2 with b3, c E Inc(b~). If b~ > bl, then {a2, bl, b~, a3, e, b3} form a copy of D in B(P). Therefore, we may assume an element b~ in P. If there is no element a~, then there are up-made elements a~ < az and a t < az with a 1 }[ a~ but a~ < a3, and a3 l[ a~ but a~ < a 1. However, it follows, that {b2, al, a~, b3, a~, a3} form a copy of D a. The contradiction shows that we may assume an element a~. The bipartite order corresponding to these elements y~ and y~ as maximals and minimals forms a copy of 270 in P.

Case 4. B(P) contains CX2. From Lemma 3 we obtain elements a~, a~, cz and a~, a u3 ' c" "

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The same argument used in Case 3 allows us to assume an up-made element a~. Next, choose a low-made element b~/> b3 with {bl, b2, al} E lnc(b~). If b~ > c, then B(P) contains CX3. Therefore, we assume an element b~. By symmetry, we may also assume an element b~. Finally, choose a low-made element b~ > b2 with c E Inc(b~). Obviously, bl ~ b~ and e2 ~ b~. Let :c be a preimage of b~ under L. The bipartite ~ c~ and minimal elements a~, a2,a3,c ~ ~ u order with maximal elements a ~1, a l3, b~1, x ,~'3, forms a copy of O~ in P.

Case 5. B(P) contains CXt. If bl is not low-made, then choose a low-made element b] > bl with {a3, bz, c} _C Inc(b]). If b] > b3, then {a2, b3, hi, al, c, be) form a copy of D. Therefore, we may assume that there is an element b~ in P. If there is not element a~, then there are up-made elements a~ < a2 and a~ < a2 with al 11a~ but a~ < a3, and a3 [1 a~ but a~ < al. It tbllows, that {c, al, a~, b3, a~, a3} form a copy of D a. The contradiction shows, that we may assume an element a~ in P. If b2 is not low-made, then choose a low-made element b~ > bz with al, bl C Inc(b~). If b~ > b3, then {al, az, a3, bl, c, b3, b~z} form a copy of CX3. We, therefore, assume an element b~ hi P. From Lemma 3 we obtain elements a], a~, b~ and a~, a~, b~. We also may choose a low-made element d > c with b3 E Inc(d). Let z be a preimage of d under L. The bipartite order with maximal elements a~, at3, ~l,~lb~, b~, z and minimal elements a~, a~'2,a~3, b~'3 forms a copy of Oza in P.

Case 6. B(P) contains EXz. Suppose, that b2 iS not low-made and there is no low-made element b~ > b2 with e, b3 E lnc(b~). Then it follows that there exist points b4, b5 > bz with b3 E Inc(b4) and e, bl E Inc(bs). This implies that b5 > b3. If b4 > e, then {a2, bl, c, a3, bs, b3} generate a copy of D. So we must have b4 [[ e, and thus b4 > bl. However, this implies that {e, bl, b4, b2, b3, a2, a3} generate a copy of CX1 in B(P). The contradiction allows us to assmne an element b~ in P. Since EX2 = EX~ we obtain a~ from the dual argument. If e is not low-made, then choose an element c I > c with a3, b2 E Inc(d). If d > b3, then {d, b3, a2, bz, al, bl} form a copy of D d. We, therefore, may assume an element cz. Dually, we may assume a~. From Lemma 3 we obtain elements a l3, bl, L b3t and a3, ~, bu1, b3. ~, The bipartite order t bl, ~ b~, b~, ce and minimal elements a~, a~, a~, by forms a with maximal elements a3, copy of O3 in P .

Case 7. B(P) contains FX1. If b2 is not low-made, then choose a low-made element b~ > b2 with b~ I1 b3. If b~ > bl, then B(P) contains D. Hence, we assume an element b~ in P.

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s TEFAN FELSNER

If there is no element a~, then there are up-made elements a] < al and a I!i < al with a2, b3 E Inc(a]) but a] < a3, and a3 E Inc(a~) but a~ < b3. It follows, that {c, b3, a~~, bl, a], a3} form a copy of D d. The contradiction shows, that we may assume an element a~ in P. If there is no element a~, then there are up-made elements a~ < a2 and a~ < a2 with ai 11a~ but a~ < az and a3 ]l a~ but a~ < ai. It follows that {b2, al, a~, b3, a~, a3} form a copy of D a. The contradiction shows that we may assume an element a~. From Lemma 3 we obtain low-made elements a3,t bi,t b~, e~ and up-made elements a3,u bl,u b3.u The bipartite order with maximal elements a3,t bl,l b~, b~3,cl and minimal elements ai, a2, a~, by, b~ forms a copy of ~'0 in P.

Case 8. B(P) contains EX1. Choose a low-made element b[ > b2 with a3, b3 E Inc(b~). If b~ > bl and b[ > b4, then {hi, hi, ax, b3, a2, b4} generate D d. If b[ > bl and b[ 1[ b4, then replacing b2 by b[ we obtain EX2. If b[ > b4 and b~ [1 bi, then the same elements also generate EX2. Hence, we may assume an element b~ in P. Choose a low-made element b~ > b3, such that b2 E Inc(b~). If b~ > hi, then {ai, a2, a3, ba, b2, b3, b~} forms a copy of FX1 in B(P). If b~ > b4, then {al, a2, a3, b2, b3, b~, b4} generates FXI. We may, therefore, assume that there is an element b~ in P. If there is no element a~, then choose an up-made element a t < al with a2, b4 E Inc(a]). If a] < a3, then B(P) contains a copy of FX~. Thus, we may assume a3 I[ at, i.e., we may assume an element a~ in P. By symmetry, we may also assume an element a~ in P. From Lemma 3 we obtain elements a3, bl, b~ and a~, b~, b~. The bipartite order (X ~' × XZ)n < v is a copy of go in P.

Case 9. B(P) contains FX2. If there is no low-made element d > c with bi, b3 E Inc(d), then choose Cl > c and c2 > c with b3 [[ el but bi < % and bl [[ c2 but b3 < c2. It follows, that B(P) contains a copy of D. We, therefore, may assume an element cl. Dually, we also assume an element a~ in P. Now, choose a low-made element b~ ~> b3 with c, b2 E Inc(b~). If b~ > bi, then B(P) contains D. We, therefore, assume an element b~ in P. Dually, we also assume an element a~. From Lemma 3 we may assume elements a3, b l , l~ b~ and a3,bi,b It '~ ~' 2 in P. The bipartite order generated by these elements is a copy of 5% h / P . In cases 10 to 13, the next lemma will find repeated applications.

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L E M M A 4. Let P be a poset such that B ( P ) contains no D, CX3, nor their duals

nor a crown An. Let aa, bb az, bz, a3,...,an, bn be a fence in B ( P ) and x > bi for some i E { 1 , . . . , n } . (1) I f j l is the minimal, J2 the maximal integer with z > ai~ and z > aj2 , then

z > aj when j l bj2 , then x > bj w h e n j l bj+l, then B ( P ) contains D d. Finally, suppose that z > bj. and x > bj. and j . + 2 < j* but x If bj for all j . < j < j*. Then from the first part we know that z > a j . + l , a j . + 2 and the elements of {aj., aj,+l, aj.+2, bj., bjo+l, bjo+2, x} generate a copy of CX3 in B(P). [] Note that the dual o f this lemma is also true, i.e., if z is below some element ai, then the elements of the fence comparable with x are intervals of a's and b's.

Case 10. B ( P ) contains Fn. We also assume that B ( P ) does not contain Fro, Era or Edm when 0 ~< m < n. Now suppose, that 2 ~< i ~< n + 2. Choose a low-made element x with z >1 bi and

Hd. Suppose, that z < e, it follows that x 11bl. Let Jl be the minimal, J2 the maximal integer with z > ajl and x > aj2. If J2 = j l + 1, then there is an element b~ in P. Otherwise, { a l , . . . , a j ~ , a j 2 , . . . ,an+e} U {bl,...,bj~,x, bj2+l . . . . ,bn+2} U {c,d,e} generate Fra for some m < n. Now suppose, that z II e. If z II a +2 then

{aj2,... ,an+2} U {x, bj2+l,... ,bn+2} U {c, d,e}. generates Fra for some m < n. We, henceforth, may assume that x > an+2. If z > bi, then {e, d, e, am+z, z, bl} form a copy of D. Now assume, that z II bl and Jl > 1, then

{al,...,aj~} U {bl,...,bj~,z} U {d,e} generate Era for some m < n. Finally, if bl, e E Inc(z) and x > al, x > an+2, then {al, a,~+2, bl, c, d, e, z} form a copy of F X d. We may, therefore, assume an element b~ in P . By symmetry and duality we may also assume elements a.~ when l 1 then { a j o - b ajo,.. ., at+l} L.J {b/o , b j o + l , . . . , bi+l) k.J {a t ,c) forms a copy of E~m where m < n. If J0 = 1 then {al, a2, . . . , at+l} U {bl, b 2 , . . . , bi+l} U {a', c, d} forms a copy of F d where m < n. The contradiction allows us to conclude a' II c. Let Jl (J2) be the minimal (maximal) index j with a' < bj. If J2 • Jl -4- 1, then {al, . . . , ajl_l, a', ajz+l,..., an+2} LJ { b l , . . . , bjl , bj2,... , bn+3} U {c, d} forms a copy of E ~ where m < n. We may, therefore, assume elements a ~ , . . . , a,~+lu and by symmetry also a~+ 2. Next choose an integer i with 2 ~< i ~< n + 2 and a low-made element x >>,bi with x 1[ c. Suppose, that x II d. Suppose x > al and x > an+2. If bl, bn+3 E Inc(x), a copy of EX1 is generated by {bl, z, bn+3, d, al, an+2, e}. The same point set generates EX2 if only one of bl and b~+3 is incomparable with z. It contains D a if bl, bn+3 < x. We now assume, without loss of generality, that x [I a,~+2. Let Jo be the maximal index j with x > aio, it follows that { a j o , . . . , an+2} U {x, b j 0 + a , . . . , b n + 3 } L.J {c, d} generates Earn where m < n. Now suppose that x < d. Thus bl, bn+3 E Inc(x). Note that if z > aj for some j ~ i - l, i, then B(P) contains Edm for some m < n. £ b3,..., £ l 2 in P. Therefore, we may assume the existence of elements b2, bn+ z c~,d~ and bl, u bn+3, ~ c~. The Finally, from Lemma 3 we obtain elements bl, bn+3, bipartite order corresponding these elements forms a copy of j r in P.

Case 12. B ( P ) contains B. Suppose that a is not up-made. If there is an up-made element a ~ < a with ISucc(a') n {bl, b2, b3}l = 1 then B ( P ) contains Eg. Otherwise, there are up-made elements al, a2, a3 c3, then B ( P ) contains C a. So we assume c] II c3. If > b3, then B ( P ) contains CX1, so we may assume that c] 11b3. This gives an element c~ in P. By symmetry, we may also assume elements c~2 and ~3. From Lemma 3 we obtain elements ~1,~2,~3~l ~ ~ and bl,b2,b3 . u~ ~ The bipartite order corresponding to these elements forms a copy of O~ in P.

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Case 13. B(P) contains In. We also assume that B(P) does not contain a copy of Im when m < n. Choose an integer i with 2 ~< i ~< n + 2 and a low-made element x ) bi with x I[ c. Let Jl (3"2) be the minimal (maximal) integer with x > aj~ (aj2) and suppose that J2 5k Jl + 1. Assume, that ~: < d l and x < d2. It follows that bl, bn+3 E Inc(x) and B(P) contains a copy of Im for m < n. If x II d2 and Jz < n + 2, then

{aj~_,..., an+2} U {x, bjz+l,... , bn+3} U {c, d2} is a copy of Era for some m < n. We, therefore, assume that J2 "- n + l . I f x < dl, then {a~+2, b~+3, de, c, dl, x} generates a copy of D in B(P). Otherwise, if x II dl, then, by symmetry, we may also assume z > al. If x ~ bl, then {a,,+2, al, bl, c, dl, d2, x} generates a copy of FX2 in B(P). Again, by symmetry, we now assume x > b 1 and x > bn+3, but then {bl, dl, c, d2, bn+3, x} generates a crown A3 in B(P). Hence, we may assume elements by for 2 ~< i ~< n + 2 in P. Now, let i be an integer with 1 ~< i ~< n + 2. Choose an up-made element a~ a~, then {a~, b,~+3, d2, c, dl, bl) generates a copy of D in B(P). So we may assume, without loss of generality, that a~ II b +3. It follows easily, that B(P) contains Im where 0 ~ m < n, whenever there exists an integer j with bj > a~ and j ~ i, i + 1. From Lemma 3 we obtain elements b~, b,~+3, ~ e~ and bl,b,~+3, ~ u c",d~,d ~ in P. The bipartite order corresponding to these elements forms a copy of 77,~+1 in P.

Case 14. B(P) contains Gn. We also assume that B(P) does not contain a copy of Gin, Jm or I-Im when m < n. We first show that we may assume an element a~ when 2 ~< i ~< n + 2. Choose a low-made element x ) ai with hi-l, bi E Inc(x). Note that from z ]] bi-i it follows t h a t z ~ a j a n d x ~ b j for a l l i < j ~ < n + 3 . Assume that x > c and note that this implies x [I aj when i < j ~< n + 3 and x II bj when i - 1 ~< j < n + 3. Let i 7~ n + 2. If x < bn+3, then

{ai, a ~ + l , . . . , a~+3} U {bi, bi+l,..., bn+3} U {x} forms a copy of Gm for some m < n. Otherwise, if x ]l b,z+3, then the same set together with c generates Jm for some m < n. Now, let i = n + 2. If x < bn+3, then {bn+3, bn+2, bn+b x, a,~+2, a,~+3} forms a copy of D d. Otherwise, if z II bn+3, then the same set together with c generates CX d. Therefore, we may assume, that x II c and we have found an element ai i n P . We now show, that we may assume an element b~ when 2 ~< i ~< n + 2. Choose a low-made element x/> bi with ai, ai+l E Inc(x).

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STEFAN FELSNER

Suppose, that z > c. If i > 2, then {al,...,ai}

U {bl,...,bi-l,X}

U {c}

forms a copy of Gm for s o m e m < n. Otherwise, if i = 2, then {al, a2, a3, bx, z, c} generates D. Therefore, x [] c and we have found an element b~ in P . We have s h o w n that we m a y assume elements a li and b~i for all 2 ~< i ~< n + 2. By duality, we m a y as well assume elements a i a n d b ~ ' i n P for a l l 2 ~ < i ~ < n + 2 . L ~ bn+3, ~ u Finally, f r o m L e m m a 3 we obtain elements an+3, bl, c~ and a~, an+3, b~, c u. The bipartite order corresponding to these elements forms a copy of Gn in P.

Case 15. B ( P ) contains J , . We also assume, that B ( P ) does not contain a copy of Gin, Jm or I-Ira w h e n m < n. In complete analogy with Case 14, we m a y assume elements a~L i n P w h e n 2 ~
2 and x > d, then

{al,...,ai}U

{bl,...,bi_l,X

)U {c}

forms a copy of Gm for s o m e m < n. If x 1[ d, then the same set together with d generates J m for s o m e rn < n. If i = 2 and z > d, then {al, a2, a3, bl, z, d} generate D. If z II d, then {al, a2, a3, bl, x, c, d) generates CX3. The contradiction shows that

]lc. We have shown that we m a y assume elements a lt a n d bi for all 2 ~< i ~< n + 2. By duality, we m a y as well assume elements a~' and b~ for all 2 e. If i >/3, then {al,...,ai-l,x}

U { b l , . . . , b i ) U {c}

forms a copy of Gm for s o m e m < n. Otherwise, if i = 1, 2, then {bx, bz, b3, al, ~, c} forms a copy of D.

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Assume that x > d. Let i # n. If z < bn+3, then

{ a i , . . . , a n + 2 } U {bi+l,...,bn+3} U {x} forms a copy of G,~ for some m < n. Otherwise, if z II bn+3, then the same set together with c generates J,~ for some m < n. If i = n + 1 and z < bn+3, then {bn+3, bn+2, bn+l, x, an+l, an+2} forms a copy of D d. Otherwise, if z 11 bn+3, then the same set together with d generates CX~. If i = n + 2, then {bn+l, bn+2, bn+3, d, x, an+2} forms a copy of D. Therefore, we may assume c, d E Inc(z) and have thus found an element a i. We now show that we may assume an element b~ when 2 ~< i ~< n + 2. Choose a low-made element z >i bi with ai-1, ai E Inc(z). Suppose that z > c and z > d, then {bl, a', b,~+3, d, z, e} forms a copy of D where a t is one of a i_ 1 or a{. Suppose x > c and x tl d; it follows that x I1 aj and z l[ bj when j > i. I f i < n + 2 then

( a i , . . . , an+2} U { h i , . . . , bn+3} [-.J{:g, d} generates Hm for some m < n. If i = n + 2, then {a,~+l, an+2, b,~+l, bn+2, bn+3, d, z ) generates C X d. Finally, let az > d and z

11e. If

i > 2, then

(al . . . . , a i - 1 ) U {bl, . . ., bi-1, x} U {c, d} generates tim for some m < n. If i = 2, then depending on z < bn+ 3 or z II bn+3, a copy of either CX1 or CX2 is formed by (al, an+2, bl, bn+3, e, d, x}. We, therefore, may assume c, d E Inc@) and have found elements b~ for 2 ~< i ~< n+2. !

If bn+3 is not low-made, then choose a low-made element bn+ 3 > bn+ 3 with an+2 [1 bn+ ' 3. If bn+ ' 3 > c, then {al, a,~+2, bl, bn+3, c, d, z} generate a copy of FX1. Therefore, we may assume an element bn+ 3 ill P. We have shown that we may assume elements a~ for 1 ~< / ~< n + 2 and be for 2 ~< i ~< n + 3. Dually, we may assume elements a~ for 1 ~< i ~< n + 2 and b~ in P for 1 ~< i ~< n + 2. Finally, from Lemma 3 we obtain elements c~, de and c ~', d u. The bipartite order corresponding to these elements forms a copy of 7-/,~ in P. 2.2. 34NTERVAL IRREDUCIBLE PARTIAL STACKS We begin this part with an analysis of the relation between the interval dimension of a bipartite order P and the interval dimension of partial stacks of P.

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LEMMA 5. Let P = (X, I d i m P . Proof Let I~,..., I~ be a interval-realizer of P'. Now tran_sfonn each of the interval orders I~ as follows. Extend the intervals of elements in Min(P) to the left to the leftmost endpoint of an interval in I~; symmetrically, extend the intervals of elements in Max(P) to the fight to the rightmost endpoint of an interval in I[. From this transformation we obtain a new family I 1 , . . . , I~ of interval orders. This family is an interval-realizer of P. Hence, Idim p / / > Idim P as claimed. [] As a consequence we can now sharpen the result of Theorem 4. THEOREM 7. If P = (X, < p ) is a 3-interval irreducible poset then P is a reduced

partial stack of some bipartite 3-interval irreducible poset. Proof From Theorem 4 we know that P contains a reduced partial stack of some bipartite 3-interval irreducible poset R = (Y, L(b) for all b E B. The intervals (L'(z), U'(z)) then define an ordering P' = (X, b~ but c~ 1t a~'. These partial stacks are not representable on B(O2). Therefore we require a more formal description of partial stacks. Let Succsp(x) (Predsp(z)) denote the successor sets (predecessor sets) of x in Stack(P). To describe a partial stack Q = (X,