Since flDi⊂fi−1(0) and dimfi−1(0) =n−1 (Proposition 2.1.3(2)), eachDi=
j=i(fj|flDi)−1(R+) is a cell by the inductive assumption. Thus, we have a finite set A = k
i=1D(0)i ⊂ ∂C. Consequently, ∂C = k
i=1Di ⊂ A ⊂ C.
Take any pointv ∈A⊂∂C. For each x∈rintC, v+R+(x−v)⊂C, hence there is a y ∈ ∂C such that x ∈ v, y. Then,x ∈ A because v, y ∈ A. Therefore,C=Ais a cell. ⊓⊔
Later, we will use the following results, which are easily proved.
3.1.9 Additional Results for Cells
(1) For each cellC⊂Eand each flatF⊂EwithC∩F =∅, the intersection C∩F is also a cell.
(2) For every two cellsC, D⊂EwithC∩D=∅, the intersectionC∩Dis also a cell with (C∩D)x=Cx∩Dxfor eachx∈C∩D. If rintC∩rintD=∅, then rint(C∩D) = rintC∩rintD.
(3) Let f :C → E′ be an affine map from a cell C ⊂E into another linear spaceE′. Then,f−1(D) is a cell for every cellD⊂E′ withD∩f(C)=∅, where f−1(D)x = Cx ∩f−1(Df(x)) for each x ∈ f−1(D). When f is injective,f(Cx) =f(C)f(x) for eachx∈C.
Sketch of Proof.For the above three items, apply the characterization 3.1.8 (cf. Proposition 2.2.2 for (3)). The statements about faces in (2) and (3) are the same as 2.2.7(1) and (4), respectively. The statement about the radial interior in (2) is 2.2.7(2).
(4) For every two cells C, D ⊂ E, C×D is also a cell with rintC×D = rintC×rintD and (C×D)(x,y)=Cx×Dy for each (x, y)∈C×D.
Sketch of Proof.Note thatC×D=C(0)×D(0)and see 2.2.7(3).
3.2 Complexes and Subdivisions 143 Sketch of Proof.SinceC D andD C implyC=D, we have (C2′)⇒ (C2′′). To see (C2)⇒(C2′), show thatC∩rintD=∅impliesC∩D=D.
(C2′′)⇒(C2): Assume C∩D =∅and take a point x∈ rint(C∩D).
SinceCx, Dx∈Kby (C1) andx∈rintCx∩rintDx, we haveCx=Dx by (C2′′). It follows from 3.1.9(2) that
C∩D= (C∩D)x=Cx∩Dx=Cx C.
For eachn-cellC, we can define the following cell complexes (cf. Proposi-tion 3.1.6(6)):
F(C) =
DDC
and F(∂C) =
DD < C .
Condition (C1) means that F(C) ⊂ K for each C ∈ K. A cell complex consisting of simplexes is called a simplicial complex. For a simplex σ, F(σ) andF(∂σ) are simplicial complexes.
The next fact follows from (C2) and Proposition 3.1.6(3):
Fact For eachC, D∈K,DC⇔D(0) ⊂C(0).
LetK be a cell (simplicial) complex. We callK(0) =
C∈KC(0) theset of vertices. It is said that K is finite, infinite, or countable according to cardK (equivalently cardK(0)). If cardK is infinite, we have cardK = cardK(0).
Indeed,K∋C→C(0)∈Fin(K(0)) is an injection by Proposition 3.1.2 (or the above Fact). Then, cardK(0)is also infinite, hence it follows that
cardK card Fin(K(0)) = cardK(0) cardK.
The dimension of K is dimK = supC∈KdimC. If dimK = ∞, K is said to be infinite-dimensional (abbrev. i.d.). When dimK < ∞, K is finite-dimensional (abbrev. f.d.). It is said that K is n-dimensional if dimK=n. Note that every cell complexK with dimK1 is simplicial.
Thepolyhedron|K|ofK is defined as follows:
|K|=
K=
C∈K
C=
C∈K
rintC (⊂E).
Recall that each cellC∈Kis given the unique topology, as mentioned in the previous section, and if dimC=nthenCwith this topology is homeomorphic to the unit closedn-ballBn(Proposition 2.5.8). The topology for|K|is defined as follows:
U ⊂ |K| is open in|K| ⇔ ∀C∈K, U∩C is open inC equiv.A⊂ |K| is closed in |K| ⇔ ∀C∈K, A∩C is closed inC
. This topology is called theWhitehead(orweak)topology. Then,K(0) is discrete in|K|. EachC∈K is a closed subspace of|K| becauseC∩DD for anyD∈K withC∩D=∅. The following fact is used very often:
Fact For an arbitrary spaceX, eachf :|K| →X is continuous if and only iff|C is continuous for everyC∈K.
Remark 1.IfV is a neighborhood ofx∈ |K|, thenV∩C is a neighborhoodx in C for everyC∈K[x]. However, the converse does not hold. For example, letKbe the 2-dimensional simplicial complex in RNf defined as follows:
K=
0,ei,0,ei,e1,ei+1,0,e1,ei+1i∈N . We define V =
i∈N0,2−ie1,2−iei+1 ⊂ |K|. For each simplex σ ∈ K[0], V ∩σ is a neighborhood of0in σ. Nevertheless,V is not a neighborhood of 0in|K|. Indeed, for eachi∈N,
(int|K|V)∩ 0,e1,ei+1 ⊂ 0,2−ie1,2−iei+1. Hence, (int|K|V)∩0,e1 ⊂
n∈N0,2−ie1={0}, which implies (int|K|V)∩ 0,e1=∅. Thus,V is not a neighborhood of0in|K|.
Each x∈ |K| is contained in the interior of the unique cell cK(x)∈ K, which is called thecarrier of xin K. In other words,cK(x) is the smallest cell ofKcontainingx. Ifx∈C∈KthenCx=cK(x). A cellC∈Kis said to beprincipalifCis not a proper face of any cell ofK, that is, it is a maximal cell of K. A cellC ∈ K is principal if and only if int|K|C =∅. In general, int|K|C = rintC even if C∈K is principal. If dimK =n, then everyn-cell ofK is principal.
A cell complex L is called a subcomplex of a cell complex K if L ⊂ K. A subcollection L ⊂ K is a subcomplex of K if and only if L satisfies condition (C1). Evidently, unions and intersections of subcomplexes of K are also subcomplexes of K. Every subcomplex of a simplicial complex is a simplicial complex. Then-skeletonofK is the subcomplex:
K(n)=
C∈KdimCn
⊂K.
The 0-skeleton is the set of vertices. For each cell C ∈K,F(C) and F(∂C) are subcomplexes ofK andF(∂C) =F(C)(n−1) ifn= dimC.
3.2.1 Proposition For every subcomplexL of a cell complex ofK,|L|is a closed subspace of|K|.
Proof. As is easily observed,A∩ |L|is closed in|L|for each closed set A in
|K|. Then, it suffices to show that every closed set Ain |L|is closed in |K|. For eachC∈K,
A∩C=A∩C∩ |L|=
D∈L
A∩C∩D=
D∈L∩F(C)
A∩D.
For eachD ∈L∩F(C), D is a closed subspace ofC andA∩D is closed in D, hence it is closed in C. BecauseL∩F(C) is finite, A∩C is closed inC.
Therefore,Ais closed in|K|. ⊓⊔
3.2 Complexes and Subdivisions 145 3.2.2 Proposition The polyhedron|K|is perfectly normal.
Proof. By definition, it is obvious that|K|is T1. For any disjoint closed sets A, B ⊂ |K|, it suffices to find a mapf :|K| → Isuch that f−1(0) =A and f−1(1) =B. We will inductively construct maps fn :|K(n)| →I, n∈N, so that fn−1(0) =A∩ |K(n)|and fn−1(1) =B∩ |K(n)|. Then, the mapf can be defined byf||K(n)|=fn for eachn∈N.
SinceK(0) is discrete, f0 can be easily constructed. Assume that f(n−1) has been constructed. For eachn-cellC∈K, we apply the Tietze Extension Theorem 1.2.2 to obtain a map gC : C → I such that gC|∂C = fn−1|∂C, gC(A∩C) ={0}, and gC(B∩C) ={1}. On the other hand, becauseC∈K is metrizable (so perfectly normal), there is a maphC :C→Iwith
h−C1(0) = (A∩C)∪(B∩C)∪∂C.
We define a mapfC:C →Iby
fC(x) = (1−hC(x))gC(x) +12hC(x).
Then,fC|∂C =fn−1|∂C,fC−1(0) =A∩CandfC−1(1) =B∩C. Hence,fn can be defined byfn||K(n−1)|=fn−1 andfn|C=fC for everyn-cell ofK. ⊓⊔
A full complex (or full simplicial complex) is a simplicial complex K such thatK(0) is affinely independent andv1, . . . , vn ∈K for all finitely many distinct verticesv1, . . . , vn∈K(0). Every simplex is the polyhedron of a finite full complex. For each affinely independent setAinE, let∆(A) denote the full complex withAthe set of vertices (i.e.,∆(A)(0)=A). In the case when A is infinite,|∆(A)| might be considered as an infinite-dimensional simplex.
In fact, |∆(σ(0))|=σ for a simplex σ, where note that∆(σ(0)) =F(σ). An infinite full complex has no principal simplexes. For a simplicial complexK, if K(0) is affinely independent, then K is a subcomplex of the full complex
∆(K(0)).
On the other hand, it is said that a subcomplexLof a simplicial complex K is full in K or a full subcomplex of K if σ ∈ L for any σ ∈ K with σ(0) ⊂ L(0), that is, L is a maximal subcomplex of K such that the set of vertices is L(0).5 The n-skeleton K(n) is not full in K unless K(n) = K. In general, a full subcomplex of a simplicial complex is not a full complex, but a full subcomplex of a full complex is always a full complex.
The following subcomplex ofK is called thestaratC∈K:
St(C, K) =
D∃D′ ∈K such that CD′, DD′ .
Evidently, St(C, K) =F(C) (i.e.,|St(C, K)|=C) if and only ifCis principal in K. We must not confuse St(C, K) with st(C, K).6 Note that|St(C, K)|=
5It should be noted that although the same wordfullis used,full subcomplexand full complexare different concepts. The former is used in the relative sense, but the latter is in the absolute sense.
6For anyA⊂ |K|, we denoteK[A] ={C∈K|C∩A=∅}and st(A, K) =
Ë
K[A].
WhenA={x},K[x] ={C∈K|x∈C}and st(x, K) =ËK[x]. See§1.2.
st(rintC, K) for eachC∈K but, in general,|St(C, K)|st(C, K). Observe that
st(x, K) =|St(cK(x), K)| for eachx∈ |K|.
IfK is a simplicial complex, thelinkofσ∈Kcan be defined as follows:
Lk(σ, K) = St(σ, K)\K[σ] =
τ∈St(σ, K)τ∩σ=∅
=
τ ∈Kτ σ∈K .
Note that Lk(σ, K) = ∅ if and only if σ is principal in K. For each non-principal simplexσ∈K, we have|St(σ, K)|=
τ∈Lk(σ,K)στ.
We define theopen staratx∈ |K|(with respect toK) as follows:
OK(x) =|K| \ |K\K[x]|=
C∈K[x]
rintC,
whereK\K[x] is a subcomplex ofK, henceOK(x) is an open neighborhood of xin|K|. SinceOK(x)⊂st(x, K), it follows that st(x, K) (=|St(cK(x), K)|) is a closed neighborhood ofxin |K|. Note the following equivalences:
y∈OK(x)⇔cK(y)∈K[x]⇔cK(x)cK(y)⇔cK(x)(0)⊂cK(y)(0)
⇔ ∀v∈cK(x)(0), cK(y)∈K[v]⇔ ∀v∈cK(x)(0), y∈OK(v).
Therefore, we have
OK(x) =
v∈cK(x)(0)
OK(v).
Then,|K|has the following open and closed covers:
OK =
OK(v)v∈K(0)
; SK =
|St(v, K)|v∈K(0) , whereOKcl =SK. IfK is a simplicial complex,
OK(v) =|St(v, K)| \ |Lk(v, K)| for eachv∈K(0).
3.2.3 Proposition Let K be a simplicial complex and v1, . . . , vn ∈ K(0). Then,v1, . . . , vn ∈Kif and only if n
i=1OK(vi)=∅. Proof. Ifσ=v1, . . . , vn ∈K, thenn
i=1OK(vi)⊃rintσ=∅. Thus, we have the “only if” part. To prove the “if” part, assume thatn
i=1OK(vi) contains a point x. Then,v1, . . . , vn∈cK(x)(0). Hence,v1, . . . , vn ∈K. ⊓⊔
3.2.4 Proposition Let K be a simplicial complex and L a subcomplex of K. Then,OK||L|=OL.
Proof. Since OL(v) = OK(v)∩ |L|for each v∈L(0), we haveOL ⊂ OK||L|. To proveOK||L| ⊂ OL, letv∈K(0) withOK(v)∩ |L| =∅. We have a simplex σ ∈ L such that OK(v)∩σ =∅, which means that v ∈ σ(0) ⊂ L(0). Then, OK(v)∩ |L|=OL(v). Therefore,OK||L|=OL. ⊓⊔
3.2 Complexes and Subdivisions 147 A cell (or simplicial) complex K is said to be locally finite, locally countable, or locally finite-dimensional (abbrev. l.f.d.) according to whether the star St(v, K) at every v ∈ K(0) is finite, countable, or finite-dimensional, respectively. Every locally finite cell complex is l.f.d. Note that K[OK(v)] =K[v] and St(v, K) =
C∈K[v]F(C) for everyv∈K(0). Then, we have the following:
3.2.5 Proposition A cell complexKis locally finite (or locally countable) if and only ifK is locally finite (or locally countable) as a collection of subsets in the space|K|. ⊓⊔
For compact sets in|K|, we have the following:
3.2.6 Proposition Let K be a cell complex. Every compact set A ⊂ |K| is contained in |L| for some finite subcomplex L ⊂K. Consequently,|K| is compact if and only ifK is finite.
Proof. It suffices to show thatKA ={D ∈K|A∩rintD =∅}is finite. For each D ∈KA, take xD ∈A∩rintD. SinceC∩ {xD | D ∈KA} is finite for each C∈ K by (C2′), any subset of {xD |D ∈ KA} is closed in |K|, hence {xD |D ∈KA} is discrete in |K|. Since Ais compact, it follows that KA is finite. ⊓⊔
When two cell complexes have the same polyhedron, the following propo-sition holds:
3.2.7 Proposition For each pair of cell complexesK1 and K2 with|K1|=
|K2|as sets,|K1|=|K2|as spaces if and only if each cell ofKi is covered by finitely many cells of K3−i fori= 1,2.
Proof. The intersection of a cell ofK1and a cell ofK2is also a cell by 3.1.9(2).
This intersection has the unique topology mentioned in the previous section, hence it is a subspace of both spaces|K1|and|K2|. If each cell ofKiis covered by finitely many cells of K3−i, then every closed set in |Ki| is also closed in
|K3−i| for i = 1,2. Thus, we have proved the “if” part. The “only if” part follows from Proposition 3.2.6. ⊓⊔
It is said that a cell complexK′ is asubdivisionof a cell complexK, or K′ subdividesK, if the following conditions are satisfied:
(S1) Each cell ofKis covered by finitely many cells ofK′; (S2) K′≺K (i.e., eachC′∈K′ is contained in someC∈K).
Due to Proposition 3.2.7,
• For every subdivisionK′ of a cell complexK,|K′|=|K|as spaces.
Evidently, if K′ is a subdivision of K and K′′ is a subdivision of K′, then K′′ is a subdivision ofK. A simplicial complexK′ subdividingK is called a simplicial subdivisionand denoted as follows:
K′⊳K or K⊲K′.
3.2.8 Lemma LetK′ be a subdivision of a cell complexK. For eachC∈K′, letD∈K be the smallest cell containingC. Then,rintC⊂rintD.
Proof. Take any x∈rintC. Then, C ⊂ Dx by the definition ofDx. By the minimality, Dx=D, which meansx∈rintD. ⊓⊔
It should be noted that condition (S1) can be strengthened as follows:
3.2.9 Proposition Let K′ be a subdivision of a cell complex K. For each C∈K, there are finitely manyD1, . . . , Dk ∈K′such thatC=k
i=1rintDi= k
i=1Di.
Proof. Because of condition (S1), we can find finitely manyD1, . . . , Dk∈K′ such that C ⊂ k
i=1rintDi ⊂ k
i=1Di, where it can be assumed that each rintDi meets C. By Lemma 3.2.8, rintDi ⊂rintCi for someCi ∈K. Then, C∩rintCi = ∅, which means Ci C by (C2′). Thus, we have k
i=1Di ⊂ k
i=1Ci⊂C. ⊓⊔
With regard to subdivisions, we have the following:
3.2.10 Theorem Every cell complexK has a simplicial subdivisionLwith the same vertices, i.e.,K(0)=L(0).
Proof. Give an order onK(0) so thatC(0) has the maximumvC = maxC(0) for eachC ∈K (e.g., a total order). Let L0=K(0) andL1=K(1). Suppose that a simplicial subdivisionLn⊳K(n)has been defined so that
(1) L(0)n =K(0) andLn−1⊂Ln; (2) vcK(ˆσ)∈σ(0) for eachσ∈Ln,
where cK(ˆσ) is the carrier of the barycenter ˆσ ofσin K (note thatcK(ˆσ)∈ K(n) because ˆσ ∈ |K(n)|). Let C ∈ K be an (n+ 1)-cell. For each σ ∈ Ln
with σ⊂∂C, we have cK(ˆσ)C,σ⊂cK(ˆσ) by Lemma 3.2.8 and vcK(ˆσ)∈ σ(0) by the assumption. If vC ∈ cK(ˆσ)(0) then vC = vcK(ˆσ) ∈ σ(0). When vC ∈cK(ˆσ)(0), sincecK(ˆσ)(0) =C(0)∩flcK(ˆσ) (cf. 3.1.6(3)), it follows that vC ∈flcK(ˆσ), sovC∈flσ. Therefore,vCis joinable toσ, that is, we have the simplex vCσinC withσ < vCσ. Now, we define
Ln+1=Ln∪
vCσC∈K(n+1)\K(n), σ∈Ln
withσ⊂∂CandvC ∈cK(ˆσ)(0) .
3.2 Complexes and Subdivisions 149 It is easy to verify thatLn+1 is a simplicial complex andLn+1⊳K(n+1). By definition,Ln+1 satisfies conditions (1) and (2).
By induction, we have simplicial subdivisionsLn ⊳K(n),n∈N, such that L(0)n =K(0) and Ln−1⊂Ln. Then,L=
n∈NLn is a simplicial subdivision ofK withL(0)=K(0). ⊓⊔
In the above proof of Theorem 3.2.10, we give an order on the setK(0) of vertices such thatC(0) has the maximum for each cellC∈K. A cell complex K with such an order onK(0) is called an ordered cell complex. If K is a simplicial complex, σ(0) has the maximum for each simplex σ∈ K if and only ifσ(0) is totally ordered for eachσ∈K. Thus, anordered simplicial complexis a simplicial complex K with an order on K(0) such thatσ(0) is totally ordered for eachσ∈K.
3.2.11 Theorem LetK1andK2be cell complexes such that|K1|=|K2|as spaces. Then,K1andK2have a common subdivisionK. In addition, ifK0is a subcomplex of bothK1 andK2, thenK0 is also a subcomplex ofK.
This follows from Proposition 3.2.7 and the next proposition:
3.2.12 Proposition For each pair of cell complexesK1andK2, the following K is a cell complex with|K|=|K1| ∩ |K2| (as sets):
K=
C∩DC∈K1, D∈K2 such that C∩D=∅ .
If K0 is a subcomplex of bothK1 and K2, then K0 is also a subcomplex of K. Moreover, if each cell ofKi is covered by only finitely many cells ofK3−i
fori= 1,2, then|K|is a closed subspace of both|K1|and |K2|.
Proof. For each pairC∈K1andD∈K2withC∩D=∅,C∩D is a cell and (C∩D)x=Cx∩Dxfor eachx∈C∩Dby 3.1.9(2). Thus,K satisfies (C1).
We will show thatK satisfies (C2′′), that is, for C, C′ ∈K1 andD, D′ ∈ K2, the following implication holds:
rint(C∩D)∩rint(C′∩D′)=∅ ⇒C∩D=C′∩D′.
Letx∈rint(C∩D)∩rint(C′∩D′). Then, (C∩D)x=C∩D and (C′∩D′)x= C′∩D′. On the other hand, by Proposition 2.2.5(8),
x∈rintCx∩rintCx′ ∩rintDx∩rintDx′.
Then, it follows from (C2′′) that Cx =Cx′ and Dx =D′x. Therefore, we can apply 3.1.9(2) to obtain the following equality:
C∩D= (C∩D)x=Cx∩Dx=Cx′ ∩D′x= (C′∩D′)x=C′∩D′. By the definition of K, if K0 is a subcomplex of both K1 and K2 then K0 is also a subcomplex of K. Moreover, it is evident that the inclusions
|K| ⊂ |Ki|,i= 1,2, are continuous. When each cell ofKi is covered by only finitely many cells ofK3−i fori= 1,2, it is easy to see that each closed set in
|K|is closed in|K1|and |K2|. ⊓⊔
Combining Theorems 3.2.11 and 3.2.10, we have the following:
3.2.13 Corollary LetK1and K2be simplicial complexes such that|K1|=
|K2| as spaces. Then, K1 and K2 have a common simplicial subdivision K.
Additionally, if K0 is a subcomplex of both K1 and K2, then K0 is also a subcomplex ofK. ⊓⊔
3.2.14 Proposition LetKbe a cell complex andK′be a subdivision ofK.
Then, every subcomplexLofKis subdivided by the subcomplex L′={C′∈ K′|C′ ⊂ |L|}ofK′.
Proof. Obviously,L′ is a subcomplex ofK′. For each C′∈L′, let C∈K be the smallest cell containingC′. On the other hand, we haveD∈Lcontaining some x ∈ rintC′. Since rintC′ ⊂ rintC by Lemma 3.2.8, it follows that D∩rintC =∅, which implies CD by (C2′), henceC ∈L. Thus, we have L′ ≺L.
For each C ∈ L, we have finitely manyD′1, . . . , Dk′ ∈ K′ such that C = k
i=1D′i by Proposition 3.2.9. SinceD′i ⊂ |L|, it follows thatD′i∈L′. Thus, C is covered by finitely many cells ofL′. ⊓⊔
3.2.15 Proposition Let K′ be a subdivision of a cell complex K. Then, OK′(v′)⊂OK(v)for eachv′ ∈K′(0)andv∈cK(v′)(0). Consequently,OK′ ≺ OK.
Proof. For each C ∈ K′ with v′ ∈ C(0), there is some D ∈ K such that rintC ⊂ rintD by Lemma 3.2.8. Then, D∩rintcK(v′) = ∅, which implies cK(v′) D by (C2′), hence v ∈ cK(v′)(0) ⊂ D(0). Thus, rintC ⊂ rintD ⊂ OK(v). Therefore,OK′(v′)⊂OK(v). ⊓⊔
3.2.16 Some Topological Properties of Polyhedra LetK be a cell complex.
(1) |K|is separable if and only ifKis countable.
Sketch of Proof. EachC ∈ K has a countable dense set DC. IfK is countable, thenD=ËC∈KDC is a countable dense set in|K|.
For a countable subsetA⊂ |K|, the following is countable:
{C∈K| ∃D∈K such thatC D, A∩rintD=∅}.
IfKis uncountable, we can find a cellC0∈Ksuch thatA∩rintC=∅ ifC ∈K and C0 C. Let x∈rintC0. Then, OK(x)∩A=∅, which means thatAis not dense in|K|.
(2) The following are equivalent:
3.2 Complexes and Subdivisions 151 (a) K is locally finite;
(b) |K|is locally compact;
(c) |K|is metrizable;
(d) |K|is first-countable.
Sketch of Proof.Because of Proposition 3.2.6, we have (a)⇒(b). The implication (c)⇒(d) is obvious. Since a space is metrizable if it is a locally finite union of metrizable closed subspaces (1.4.5(2)), the impli-cation (a)⇒(c) follows from Proposition 3.2.5.7
(b) ⇒ (a): Let V be a compact neighborhood of v ∈ K(0). Due to Proposition 3.2.6,V ⊂ |L|for some finite subcomplexL⊂K. IfC∈K andv∈C(0)thenV∩Cis a neighborhood ofvinC. SinceV∩rintC=
∅, it follows that rintCmeets someD∈L, which impliesC D. Hence, St(v, K)⊂L.
(d)⇒(a): Assume that St(v, K) is infinite for somev∈K(0). Then,v is a vertex of distinct 1-cells (1-simplexes)An,n∈. Let{Un|n∈} be a neighborhood basis at v in |K|. For each n ∈ , choose an ∈ rintAn∩Un. Then,|K| \ {an|n∈}is an open neighborhood ofvin
|K|but it does not contain anyUn, which is a contradiction.
(3) The following are equivalent:
(a) K is countable and locally finite;
(b) |K|is second-countable;
(c) |K|is separable and metrizable.
Sketch of Proof.Combine (1) and (2) above.
(4) In general,w(|K|)= dens|K|(i.e.,w(|K|)>dens|K|).
Sketch of Proof.LetK(0)={vi|i∈ω}and defineK=K(0)∪{v0, vi | i ∈ }. Then, dens|K| =ℵ0 by (1) above. However, |K| is not first countable by (2), hence it is not second countable, that is,w(|K|)>ℵ0. (5) Letf :X→ |K|be a map of a metrizable (more generally, first countable)
spaceX. Then, eachx∈Xhas a neighborhoodUxinXsuch thatf(Ux)⊂
|Kx|for some finite subcomplexKxofK.
Sketch of Proof. If x ∈ X does not have such a neighborhood, then we can find a sequence (xi)i∈in X such that limi→∞xi = x and cK(xi) = cK(xj) if i = j. Because {f(x), f(xi) | i ∈ } ⊂ |K| is compact, this contradicts Proposition 3.2.6.
Apolyhedron(or a topological polyhedron) is defined as a spaceP such that P =|K| (or P ≈ |K|) for some cell complexK. A subspace Qof a polyhedron (or a topological polyhedron)P is called asubpolyhedronof P if there exists a pair (K, L) of a cell complex and a subcomplex such that P =|K| and Q=|L| (or (P, Q)≈ (|K|,|L|)). Every subpolyhedron ofP is closed in P according to Proposition 3.2.1. It follows from 3.2.16(2) that a (topological) polyhedron is metrizable if and only if it is locally compact. In
7This can be shown as follows: IfKis a simplicial complex, (a)⇒(c) will be proved in Theorem 3.5.6. Due to Theorem 3.2.10, every cell complex has a simplicial subdivision. Evidently, every subdivision of a locally finite cell complex is also locally finite. Thus, (a)⇒(c) is also valid for every cell complex.
general, for a (topological) polyhedron P, w(P) = densP (cf. 3.2.16(4)). A triangulationof a polyhedron (or a topological polyhedron)P is a simplicial complexK such that|K| =P (or |K| ≈P). Then, it is also said that P is triangulated by K or K triangulates P. According to Theorem 3.2.10, every (topological) polyhedron has a triangulation.