computability of a topological poset

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Topology and its Applications 153 (2006) 1132–1140

www.elsevier.com/locate/topo

Computability of a topological poset

W.D. Gillam

Columbia University, Department of Mathematics, New York, NY 10027, USA

Received 22 November 2004; received in revised form 7 March 2005; accepted 9 March 2005

Abstract

For subspacesX andY of Q the notationX �h Y means thatX is homeomorphic to a subspaof Y andX ∼ Y meansX �h Y �h X. The resulting setP(Q)/∼ of equivalence classes�X = {Y ⊆Q: Y ∼ X} is partially-ordered by the relation�X �h

�Y if X �h Y . In a previous paper by the authit was established that this poset is essentially determined by considering only the scatteredX ⊆ Q

of finite Cantor–Bendixson rank. Results from that paper are extended to show that this pcomputable. 2005 Elsevier B.V. All rights reserved.

MSC:54H10; 54H12; 54D35; 06A06

Keywords:Embedding; Partially-ordered; Scattered; Metric space; Computable

1. Introduction

This paper is mainly a continuation of [1]. The notation and terminology frompaper are used here. All topological spaces considered are assumed to be scattereable metric spacesX with finite (Cantor–Bendixson) rank (denotedN(X)). We mayassume each of these is a subspace ofQ. Here we show that the partially-orderedA = {X ⊆ Q: 1< N(X) < ω}/∼ (ordered be embeddability) originally defined in [1, Stion 4] can be generated by a computer program. En route to this conclusion weclassification (or apresentationperhaps) of homeomorphism types of countable me

E-mail addresses:[email protected], [email protected] (W.D. Gillam).

0166-8641/$ – see front matter 2005 Elsevier B.V. All rights reserved.doi:10.1016/j.topol.2005.03.004

W.D. Gillam / Topology and its Applications 153 (2006) 1132–1140 1133

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spaces with finite Cantor–Bendixson rank. For example, in Section 3, we give a comlist of homeomorphism types of spaces of rank� 4, and a presentation of the embeddaity ordering of these spaces. Our focus here is on a closer study of thetypesintroduced inDefinition 20 of the aforementioned paper. There we defined the type of an isolatedto be (∅,∅) and then inductively defined the type of a pointp ∈ X with higher rank tobe (A,B) whereA is the set of typesτ so that every neighborhood ofp in X containsinfinitely many points of typeτ andB is the set of typesσ so that every neighborhood op in X contains a closed infinite set of points of typeσ .

2. The combinatorics of types

In [1, Definition 21] we definedTn to be the set of types arising from points in spaX with N(X) < n. It is easy to see thatTn is finite for eachn ∈ N so we would like todetermine|Tn| as a function ofn. The results presented here will reduce that question(difficult) finite combinatorial problem, which we do not solve. We only give a recuralgorithm (Theorem 1) for computingTn+1 from Tn, but we have no closed form expresion for |Tn|. Another difficulty concerns the functionsfX :TN(X)+1 → ω + 1+ 1 definedin [1, Definition 23]. Recall thatfX(τ) was defined to be the number of points of typeτ

in X with the convention that if there was aclosedinfinite set, we counted its cardinalitas “ω + 1”. Which functionsf :Tn → ω + 1 + 1 are actually of the formf = fX for aspaceX? We answer this in Theorem 3. Also note that [1, Lemma 27] is quite weacause there are certainly scattered spacesX,Y ⊆ Q whereX embeds inY even thoughfX(τ) > fY (τ) for certain typesτ (e.g. takeX := M0, Y := M1, τ = ({(∅,∅)},∅)). Thisweakness is remedied below. Many natural questions are left unanswered. For exwe have no general formula for the size of a maximal anti-chain inTn as a function ofn.

Note that for eachτ ∈ Tn there is a way of associating a canonical scattered sS(τ) ⊆ Q to the typeτ . Sinceτ = (A,B) arises as the type of a pointp in some finite rankspaceX ⊆ Q, it is always possible to choose a neighborhoodU of p in X containing pointsof typeσ only for σ ∈ A and containing closed infinite sets of points of typeσ only forσ ∈ B. Let us call such an open-and-closed neighborhoodstandard. Now takeS(τ) := U .The spaceS(τ) is the unique space (by [1, Theorem 24]) wherefS(τ)(σ ) is ω + 1 whenσ ∈ B, ω whenσ ∈ A \ B, 1 whenσ = τ and 0 otherwise.

Theorem 1. T0 = T1 = ∅, T2 = {(∅,∅)} and for each finiten � 2, Tn+1 \ Tn consists of allelements(A,B) ∈ P(Tn) ×P(Tn) satisfying all of the following:

(i) A ∩ (Tn \ Tn−1) �= ∅,(ii) B ⊆ A,

(iii) (A′,B ′) ∈ A ⇒ A′ ⊆ A andB ′ ⊆ B,(iv) (A′,B ′) ∈ B ⇒ A′ ⊆ B andB ′ ⊆ B.

Proof. We show that this recursive definition is correct using induction onn. Suppose thathe recursive definition ofT1, T2, . . . , Tn given above is correct (certainly it is up ton = 1).Now assumen � 2 and note that the setTn+1 \ Tn is the set of types of(n − 1)-isolated

1134 W.D. Gillam / Topology and its Applications 153 (2006) 1132–1140

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points. If (A,B) ∈ Tn+1 \ Tn then(A,B) arose as the type of some(n − 1)-isolated pointp in a scattered spaceX ⊆ Q and hence it satisfies (ii)–(iv) by [1, Theorem 22] and (isimply the observation that there must be a set of(n − 2)-isolated points converging top,infinitely many of which must have the same type by [1, Theorem 22(i)]. Now fix s(A,B) ∈ P(Tn) × P(Tn) satisfying the four conditions and we will show that it actuaarises as the type of an(n − 1)-isolated point in a scattered spaceX ⊆ Q.

Let A = {τ1, . . . , τm, τm+1, . . . , τl} where (by (ii)) we may assume the elements oA

are listed so thatA \ B = {τ1, . . . , τm} andB = {τm+1, . . . , τl}. For 1� j � m, embed acopy ofS(τj ) into each intervalIn (as described in [1, Section 1]) wheren = j (mod l).Form + 1 � j � l, embed a copy of

⊕ω S(τj ) into each intervalIn wheren = j (mod l).

Add the point 1 to obtain a scattered subspaceX of Q.Note that (by (i)) since some type of an(n−2)-isolated point was inA, it follows that 1

is (n−1)-isolated inX soτ(1;X) = (A∗,B∗) ∈ Tn+1\Tn. Now we verify that(A∗,B∗) =(A,B). The inclusionsA ⊆ A∗,B ⊆ B∗ are immediate. Now take someσ ∈ A∗. It mustbe thatσ is in the support offS(τj ) for somej with 1� j � l. However, for that to happeneitherσ = τj (in which caseσ must have been inA or B henceσ ∈ A by condition (ii)) orσ ∈ A′ whereτj = (A′,B ′) (in which caseσ ∈ A by conditions (ii)–(iv)). This shows thaA∗ = A. Now consider a typeσ ∈ B∗. One (or both) of the following must occur:

Case1: σ is in the support offS(τj ) for somej with m + 1 � j � l. Then if σ = τj ,σ ∈ B is immediate. Otherwise it must be thatσ ∈ A′ whereτj = (A′,B ′) and henceσ ∈ B

by condition (iv).Case2: fS(τj )(σ ) = ω + 1 for somej with 1� j � m. That is to say thatσ ∈ B ′ where

τj = (A′,B ′) and henceσ ∈ B by conditions (ii) and (iii).

The finite setsTn can be ordered in a natural way. Declareτ �h σ if and only ifS(τ) �h S(σ ). This makes(Tn,�h) a finite pre-ordered set. Declaringτ ∼ σ if andonly if σ �h τ �h σ and ordering the resulting equivalence classes in the obviousmakes(Tn/∼,�h) into a finite partially-ordered set. The next lemma shows thatpartially-ordered sets(Tn/∼,�h) can be constructed via a recursive algorithm. Note(T0/∼,�h) and(T1/∼,�h) are empty while(T2/∼,�h) is the partially-ordered set witone element.

Lemma 2. Let (A,B) ∈ Tn+1 \ Tn and let(A′,B ′) ∈ Tn+1. Then(A′,B ′) �h (A,B) if andonly if either(i) or (ii) (as below) holds.

(i) For eachσ ∈ A′ there isτ ∈ A such thatσ �h τ and for eachσ ∈ B ′ either(1) there isτ ∈ B such thatσ �h τ , or(2) there is(A′′,B ′′) ∈ A andτ ∈ A′′ such thatσ �h τ ;

(ii) (A′,B ′) ∈ Tn and(A′,B ′) �h σ for someσ ∈ A.

Proof. Clearly condition (ii) is sufficient and it is not too hard to see that condition (sufficient using the “canonical” constructions ofS((A,B)) andS((A′,B ′)) as in the proofof Theorem 1. To show that either (i) or (ii) is necessary, again use the canonical retions ofS((A,B)) andS((A′,B ′)) and suppose there is an embeddingf :S((A′,B ′)) →


ae

fails,f

in

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S((A,B)). The point 1∈ S((A′,B ′)) is the unique point in that space with type(A′,B ′)and similarly for 1∈ S((A,B)).

If f (1) = 1, argue that condition (i) must be satisfied. Indeed, for eachσ ∈ A′, S(A′,B ′)contains infinitely many points of typeσ , so there must be some typeτ ∈ A such thatinfinitely many points of typeσ are taken byf to points of typeτ . Every neighborhoodof a point contains a standard neighborhood of that point, so sincef is an embedding,fmust take a standard neighborhood of a point of typeσ into a standard neighborhood ofpoint of typeτ—this requiresσ �h τ . For eachσ ∈ B ′, S(A′,B ′) contains a closed infinitset of points each of typeσ . Thus some closed infinite set of points all of typeσ must betaken byf to an infinite set of points, all of typeτ , which is closed inf [S((A′,B ′))]. Weclaim thatτ must be as in (1) or (2). Indeed, if this were not the case, then since (2)for each typeτ ′ ∈ A, S(τ ′) contains no points of typeτ if τ �= τ ′ and only one point otype τ whenτ = τ ′. Then since (1) fails, for each intervalIn, In ∩ S((A,B)) contains atmost one point of typeτ . Now sincef (1) = 1 ∈ f [S((A′,B ′))], it follows that no infiniteset of points all of typeτ could possibly be closed inf [S((A′,B ′))].

If f (1) �= 1, then(A′,B ′) cannot be inTn+1. The pointf (1) ∈ S((A,B)) has typeσfor someσ ∈ A and sincef is an embedding,f must take some neighborhood of 1S(A′,B ′) into a standard neighborhood off (1) in S((A,B)). Now (ii) is immediate sinceany neighborhood of 1 inS((A′,B ′)) is homeomorphic toS((A′,B ′)).

The next result classifies the functionsf :Tn → ω+1+1 that arise asfX for a scatteredspaceX ⊆ Q with finite rank.

Theorem 3. A functionf :Tn → ω + 1+ 1 is equal tofX for a spaceX ⊆ Q of finite rankif and only iff satisfies all the following conditions for all(A,B) ∈ Tn:

(i) f ((A,B)) = ω + 1 ⇒ f (σ ) = ω + 1 for all σ ∈ A;(ii) f ((A,B)) > 0⇒ f (σ ) � ω for all σ ∈ A andf (σ ) = ω + 1 for all σ ∈ B;

(iii) f ((A,B)) = ω ⇒ there is(A′,B ′) in the support off with (A,B) ∈ A′.

Proof. To see that (i) is necessary, take a closed infinite set{x1, x2, . . .} of points all oftype (A,B) in X and fix someσ ∈ A. For eachn, choose a pointan of type σ withd(xn, an) < 1/n and argue that{a1, a2, . . .} is closed inX. Necessity of (ii) is obvious. Tosee that (iii) is necessary, take an infinite set of points inX each of type(A,B) (which isnot closed inX sincef ((A,B)) �= ω + 1) and look at the type of any point in the closuof this set.

Now consider a functionf satisfying (i)–(iii). Letτ1, . . . , τk be the types thatf takes toω + 1 and let(τk+1, n1), . . . , (τk+m,nm) be the pairs with 0< f (τk+i ) = ni < ω. Let τi =(Ai,Bi) for i = 1, . . . , k+m. LetX be the space

⊕ki=1(

⊕ω S(τi))⊕⊕m

j=1(⊕

njS(τk+j ).

We claimf = fX. The types taken byfX to ω + 1 are

(1) τ1 = (A1,B1), . . . , τk = (Ak,Bk),(2) each typeσ ∈ Ai for somei = 1, . . . , k, and(3) each typeσ ∈ Bi for somei = 1, . . . , k + m.


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By definition,f takes the types in (1) toω + 1. Sincef satisfies (i), it also takes thtypes in (2) toω + 1 and sincef satisfies (the second part of) (ii), it also takes the tyin (3) to ω + 1. Thusf (τ) = ω + 1 if and only if fX(τ) = ω + 1 for all τ ∈ Tω. Byconstruction it is also immediate thatf andfX agree on all the types either takes tofinite ordinal. SupposefX(σ ) = ω. Thenσ �= τi for i = 1, . . . , k +m, andσ ∈ Ai for somei ∈ {k + 1, . . . , k + m}. Thusf (σ ) � ω by (the first part of) (ii) andf (σ ) �= ω + 1 becauseσ is not in{τ1, . . . , τk}. Sof (σ ) = ω.

Now supposef (σ ) = ω. By (iii) there is a type (C1,D1) with σ ∈ C1 andf ((C1,D1)) > 0. Notice thatf ((C1,D1)) < ω + 1 by (i). If f ((C1,D1)) = ω, invoke (iii)and (i) again to find(C2,D2) with (C1,D1) ∈ C2 and 0< f ((C2,D2)) < ω + 1. Keep do-ing this until arriving at a type(Ct ,Dt ) with (Ct−1,Dt−1) ∈ Ct and 0< f ((Ct ,Dt )) < ω.Then (Ct ,Dt ) = (Ak+i ,Bk+i ) for somei ∈ {1, . . . ,m} and by [1, Theorem 22(iii)] wehaveAk+i = Ct ⊇ Ct−1 ⊇ · · · ⊇ C1 henceσ ∈ Ak+i and it follows thatfX(σ ) = ω.

Corollary 4. Every countable metric space with finite rank is homeomorphic to a spathe form

⊕mi=1(

⊕αi

S(τi)) for τ1, . . . , τm ∈ Tn andα1, . . . , αm ∈ (ω + 1).

The representation above is generally not unique, but it will be (up to reorderingm

is chosen to be minimal. Our final goal will be to give necessary and sufficient condon fX andfY for X to embed inY (whenX andY have finite rank). Before stating thnext theorem, we need the following notation:

SX := {τ ∈ Tω: fX(τ) �= 0

}(the support offX),

IX := {τ ∈ Tω: fX(τ) = ω + 1

},

FX := {τ ∈ Tω: 0< fX(τ) < ω

},

GX := {τ ∈ Tω: fX(τ) = ω ∧ ∃(A,B), (A′,B ′) ∈ SX with τ ∈ A ∧ (A,B) ∈ A′},

CX := {τ ∈ Tω: fX(τ) = ω ∧ τ is not inGX

}.

Theorem 5. LetX andY be countable metric spaces with finite rank. ThenX �h Y if andonly if there are functionsΦ : IX → SY \ FY and Ψ :FX → ωSY satisfying the followingfive conditions:

(i) τ �h Φ(τ) for all τ ∈ IX;(ii) For all τ ∈ FX and for allσ ∈ SX, Ψ (τ)(σ ) > 0 impliesτ �h σ ;(iii)

∑σ∈SY

Ψ (τ)(σ ) = fX(τ) for all τ ∈ FX;(iv)

∑τ∈FX

Ψ (τ)(σ ) � fY (σ ) for all σ ∈ SY ;(v) For all σ ′ ∈ Φ[IX] ∩CY there is some(A,B) ∈ SY with σ ′ ∈ A such that the inequal

ity in (iv) is strict whenσ = (A,B).

Proof. To prove these conditions are necessary, assume we have an embeddingf :X → Y .For eachτ in IX, X contains a (there may be many such—choose one) closed infinitepoints of typeτ all mapped byf to points of the same typeΦ(τ) in Y . For eachσ ∈ FY ,Y only has finitely many points of typeσ , so Φ(τ) cannot be inFY . For eachτ ∈ FX

and eachσ ∈ SY , let Ψ (τ)(σ ) be the number of points of typeτ taken byf to points of


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typeσ . Sincef is an embedding, for each pointx ∈ X there is a standard neighborhoof x taken byf into a standard neighborhood off (x). ThusΨ andΦ as defined abovmust satisfy conditions (i) and (ii). Clearly (iii) is satisfied sincef is injective andfX(τ)

is the number of points of typeτ in X and each such point is taken byf to a point inY

with sometype. Condition (iv) is also trivially verified. Let us now verify the most difficcondition: number (v). Fix someσ ′ ∈ Φ[IX] ∩ CY , so there is a typeτ in IX and a closedinfinite setC of points ofX all with typeτ that are taken byf to points of typeσ ′. Lookclosely at the wayCY is defined to see that any infinite set of points of typeσ ′ must havea cluster point (choose one and call itp) whose type (call it(A,B)) is in FY . Thenσ ′ ∈ A

and the inequality in (iv) must be strict because if not, then every point of type(A,B) in Y

would be inf [X]. However, this cannot be the case becausep cannot be inf [X] becausef is an embedding sof [C] should be closed inf [X] andp is in the closure off [C].

Now we prove that these conditions are sufficient. Assume we are givenX, Y andΨ ,Φ satisfying all the conditions. As in the proof of the previous theorem,X can be writtenin the form

⊕ki=1(

⊕ω S(τi)) ⊕ ⊕m

j=1(⊕

njS(τk+j ) whereIX = {τ1, . . . , τk} andFX :=

{τk+1, . . . , τk+m}. There are only finitely many points inY of type σ for each typeσ =(A,B) ∈ FY , so choose standard neighborhoods of each such point which are paidisjoint. By possibly making these neighborhoods a bit smaller, we may assume thcomplement of the union of all of these neighborhoods contains a closed infinitepoints of typeσ ′′ for eachσ ′′ ∈ B for eachσ ∈ FY and at least one point of typeσ ′′ ∈ A

for eachσ ∈ FY .For eachσ ∈ SY and eachj ∈ {1, . . . ,m}, embedΨ (τk+j )(σ )-many of theS(τk+j )

summands into pairwise-disjoint standard neighborhoods of points of typeσ (using thepreviously-chosen neighborhoods whenσ ∈ FY and ensuring disjointness from all the prviously defined neighborhoods whenσ is not in FY ) in Y . Furthermore, by making thstandard neighborhoods small enough, we may assume that the complement of thsure of the) range of the union of these embeddings (call this mapf ′) still contains a closedinfinite set of points of typeσ ′′ for eachσ ′′ ∈ B for eachσ ∈ FY and at least one point otype σ ′′ ∈ A for eachσ ∈ FY . All of this can be done since conditions (ii) and (iii) asatisfied.

For eachj ∈ {1, . . . , k}, consider the points of typeΦ(τj ) in X. If Φ(τj ) is not inCY ,then there are infinitely many points of typeΦ(τj ) in the complement of the closure of thrange off ′. Choose an infinite set of such points with at most one cluster point (not iclosure of the range of the union of the previously-defined embeddings) and inducchoose pairwise-disjoint standard neighborhoods of these points, and use conditiofind an embedding of the

⊕ω S(τj ) summand into an infinite and co-infinite union

these neighborhoods. This method ensures that the same process can be conducfor any j ′ ∈ {1, . . . , k} with Φ(τj ) not in CY . If Φ(τj ) ∈ CY , use condition (v) to find apoint p in Y of type (A,B) with Φ(τj ) ∈ A that is not in the closure of the range of aof the previously-defined embeddings. Then there is an infinite sequence of pointstypeΦ(τj ) converging top (and to no other point). Inductively choose pairwise-disjostandard neighborhoods of these points and embed (using (i)) the

⊕ω S(τj ) summand

into an infinite and co-infinite union of these neighborhoods. Taking the union of aembeddings defined in this manner gives an embeddingf :X → Y .


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Notice that since condition (iii) must be satisfied, to determine whether there aretions Φ andΨ satisfying conditions (i)–(v), it is not necessary to consider all functΨ :FX → ωSY . It is enough to consider only thoseΨ with Ψ (τ)(σ ) � max{fX(τ ′): τ ′ ∈FX} for all σ ∈ SY for all τ ∈ FX. Thus only finitely many possibilities forΨ andΦ needto be considered, and checking whether conditions (i)–(v) are satisfied can be carralgorithmically in a finite amount of time using Theorem 1 and Lemma 2. Using tresults and Theorem 5 proves thatA is acomputablepartially-ordered set in the followinsense.

Theorem 6. There is a decidable subsetD of 2<ω (:= the set of all finite length binarsequences) and a computable functionf : 2<ω × 2<ω → {0,1} such that(D, {(d1, d2) ∈2<ω × 2<ω: f (d1, d2) = 1}) is order-isomorphic toA.

We close this section with the following question. LetT be the first-order theory o(P(Q)/∼,�h) in the language of partially-ordered sets (i.e., the language with one brelation symbol “�”). Is T decidable? Recall that not every computable partially-ordset has decidable theory.

3. Spaces with rank at most 4

The setT5 can be computed from Lemma 1. We avoid using the cumbersome,definition of typeand simply list the elements ofT5 in the form in wherein ∈ Ti+1 \ Ti

and we list the types starting fromn = 0 in arbitrary order (e.g., 10 := (∅,∅) is the uniqueelement ofT2 \ T1 = T2 while 20 := ({(∅,∅)},∅) and 21 := ({(∅,∅)}, {(∅,∅)}) are theelements ofT3 \ T2). Carrying on in this manner we can compute the elements ofT4 \ T3by hand. They are:

30 := ({10,20},∅), 31 := ({10,20}, {10}

),

32 := ({10,20}, {10,20}), 33 := ({10,21}, {10}

),

34 := ({10,21}, {10,21}), 35 := ({10,20,21}, {10}

),

36 := ({10,20,21}, {10,20}), 37 := ({10,20,21}, {10,21}

),

38 := ({10,20,21}, {10,20,21}).

So we have|T4| = 12= 9+ 2+ 1. A simple computer program (about 75 lines of cocan be run to list all the types inT5. We find that|T5| = 20106= 20094+ 9 + 2 + 1,though beyond this there is no hope of listing all the types since there are probably105000 types inT6. Let us take stock of the situation. We know that all sufficiently smneighborhoods of a fixed point (in one of the spaces we are studying) are homeomFurthermore, restricting our attention to spaces with rank at most 4, we see that e20106 homeomorphism types arise in this manner. Next one can check by hand ttypes inT4 fit into 8 embeddability equivalence classes:

{10} {20} {21} {30,31} {32} {33,35} {36} {34,37,38}.It is actually possible to calculate all the embeddability equivalence classes ofT5 and theordering of those classes by hand. There are the 8 classes above, plus 31 additiona


ach ofses.es incommas

re

s

The author’s hand calculation was verified by a computer program to check that ethe 20094 classes inT5 \ T4 is embeddability-equivalent to exactly one of these 31 clasWe list these classes below, numbered arbitrarily, together with the number of typeach equivalence class, and typical representatives of each class. Set brackets andbetween elements are dropped so that, for example, the type({10,20,32}, {10,20}) appearsas(102032,1020) and “T4” abbreviates the list of all types inT4.

No. Types Representatives0 13 (102030,∅) ∼ (1020213031,1020)

1 8 (102032,1020) ∼ (102021303132,1020)

2 25 (102033,1020) ∼ (102021303133,1020)

3 12 (10203233,1020) ∼ (1020213031323335,1020)

4 32 (102036,1020) ∼ (102021303132333536,1020)

5 6 (10202130,102021) ∼ (1020213031,102021)

6 4 (10202132,102021) ∼ (102021303132,102021)

7 25 (10202133,102021) ∼ (10202130313335,102021)

8 12 (1020213233,102021) ∼ (1020213031323335,102021)

9 32 (10202136,102021) ∼ (102021303132333536,102021)

10 498 (10202134,102021) ∼ (T4,102021)

11 15 (102030,102030) ∼ (1020213031,1020213031)

12 15 (10203032,102030) ∼ (102021303132,1020213031)

13 30 (1020213033,102030) ∼ (10202130313335,1020213031)

14 30 (102021303233,102030) ∼ (1020213031323335,1020213031)

15 80 (1020213036,102030) ∼ (102021303132333536,1020213031)

16 560 (1020213034,102030) ∼ (T4,1020213031)

17 27 (102032,102032) ∼ (102021303132,102021303132)

18 54 (1020213233,102032) ∼ (1020213031323335,102021303132)

19 72 (1020213236,102032) ∼ (102021303132333536,102021303132)

20 504 (1020213234,102032) ∼ (T4,102021303132)

21 54 (10202133,10202133) ∼ (10202130313335,10202130313335)

22 45 (1020213233,10202133) ∼ (1020213031323335,10202130313335)

23 90 (1020213336,10202133) ∼ (102021303132333536,10202130313335)

24 1285 (1020213334,10202133) ∼ (T4,10202130313335)

25 45 (1020213233) ∼ (1020213031323335,1020213031323335)

26 45 (102021323336) ∼ (102021303132333536,1020213031323335)

27 630 (102021323334,1020213233) ∼ (T4,1020213031323335)

28 243 (10202136,10202136) ∼ (102021303132333536,102021303132333536)

29 1701 (1020213436,10202136) ∼ (T4,102021303132333536)

30 13902 (102134,102134) ∼ (T4, T4)

A pictorial representation of the poset(T5/∼,�h) appears in Fig. 1. In this picture theis a downhill line fromσ to τ only whenσ < τ and there is no classρ with σ < ρ < τ .Notice that some classes in(T5 \ T4)/∼ are incomparable to classes inT4/∼. There arealso some fairly large antichains (e.g.,{3,6,7,11}).

We close by answering a question of Rehana Patel: IsA (equivalently(P(Q)/∼,�h))

a lattice? Consider the embeddability classes of the spacesS(32) andS(33). These classe


82.

Fig. 1. Nearest neighbor representation of(T5/∼,�h).

have no least upper bound since (the classes of)S(32)⊕S(33) andS(36) are incomparableclasses which are both minimal among all classes of spaces into which bothS(32) andS(33) embed.

References

[1] W.D. Gillam, Embeddability properties of countable metric spaces, J. Topology Appl. 148 (2005) 63–

computability of a topological poset

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