Topologies and Measures
4.3 Closure and Interior Operators in Topological Spaces
Theorem 4.7 implies that for every topological space(S,O)the collectionclosed(O) of closed sets is a closure system on S. For the closure operator attached to this closure system denoted by KS,O, we have the supplementary property:
KS,O(H⊆L)=KS,O(H)⊆KS,O(L) (4.1) for all subsets H,L of S.
Since H,L ∪ H ⊆ L, we have KS,O(H) ∪ KS,O(H ⊆L)and KS,O(L) ∪ KS,O(H⊆L)due to the monotonicity of KS,O. Therefore,
KS,O(H)⊆KS,O(L)∪KS,O(H⊆L).
To prove the reverse inclusion, note that the set KS,O(H)⊆KS,O(L)is a closed set by the third part of Theorem 4.7 and H ⊆L ∪ KS,O(H)⊆KS,O(L). Therefore, the closure of H ⊆L is a subset of KS,O(H)⊆KS,O(L), so KS,O(H ⊆ L) ∪ KS,O(H)⊆KS,O(L), which implies Equality (4.1).
Also, note that KS,O(∈)= ∈because the empty set itself is closed.
If there is no risk of confusion, we denote the closure operator KS,O simply by K.
Note that Equality (4.1) is satisfied for every H,L∈P(S)if and only if the union of two K-closed sets is K-closed. Indeed, suppose that Equality (4.1) is satisfied, and let U and V be two K-closed sets. Since U =K(U)and V =K(V), it follows that U⊆V =K(U)⊆K(V)=K(U⊆V), which shows that U⊆V is K-closed. Conversely, suppose that the union of two K-closed sets is K-closed. Then, K(U)⊆K(V)is K-closed and contains U ⊆V . Therefore, K(U ⊆V)∪ K(U)⊆K(V). The reverse equality follows from the monotonicity of K.
Theorem 4.8 Let S be a set and let K : P(S) −∩ P(S)be a closure operator that satisfies Equality (4.1) for every H,L ∈ P(S)and K(∈)= ∈. The collection OK= {S−U | U∈CK}is a topology on S.
Proof We have K(S)=S, so both∈and S are K-closed sets, which implies∈,S∈ OK.
Suppose that C = {Li | i ∈ I} ∪ OK. Since S−Li ∈ CK, it follows that
⎜{S−Li | i ∈I} =S−
i∈ILi ∈CK. Thus,
i∈ILi ∈OK.
Finally, suppose thatD= {D1, . . . ,Dn}is a finite collection of subsets such that D∪OK. Since S−Di ∈CKwe have S−n
i=1Di =⎜n
i=1(S−Di)∈CK, hence n
i=1Di ∈OK. This proves thatOKis indeed a topology.
Theorem 4.9 Let(S,O)be a topological space and let U and W be two subsets of S. If U is open and U∅W = ∈, then U∅K(W)= ∈.
Proof U ∅W = ∈implies W ∪S−U . Since U is open, the set S−U is closed, so K(W)∪K(S−U)=S−U . Therefore, U∅K(W)= ∈.
152 4 Topologies and Measures Often, we use the contrapositive of this statement: if U is an open set such that U∅K(W)⊕= ∈for some set W , then U∅W ⊕= ∈.
Example 4.10 In the topological space(R,O), every open interval(a,b)with a<b is an open set. Indeed, if x∈(a,b)and|x−u|<σ, whereσ=12min{|x−a|,|x−b|}, then u∈(a,b). A similar argument shows that the half-lines(b,+⊥)and(−⊥,a) are open sets for a,b ∈ R. Therefore, (−⊥,a)⊆(b,+⊥)is an open set which implies that its complement, the interval[a,b], is closed. Also,(−⊥,b]and[a,⊥) are closed sets (as complements of the open sets(b,⊥)and(a,⊥), respectively).
Open sets of the topological space(R,O), whereOis the usual topology on the set of real numbers have the following useful characterization.
Theorem 4.11 A subset U ofRis open in the topological space(R,O)if and only if it equals the union of a countable collection of disjoint open intervals.
Proof Since every open interval (finite or not) is an open set, it follows that the union of a countable collection of disjoint open intervals is open.
To prove the converse, let U be an open set. Note that U can be written as a union of open intervals since for each x ∈U there existsσ>0 such that x∈(x−σ,x+σ)∪U . Define the relationφU on the set U by xφUy if there exist a,b ∈ Rsuch that {x,y} ∪(a,b)∪U , where(a,b)is the open interval determined by a,b. We claim thatφU is an equivalence relation on U .
Since U is open, x ∈ U implies the existence of a positive numberσsuch that {x} ∪(x−σ,x+σ)∪U for every x∈U , soφU is reflexive. The symmetry ofφU is immediate. To prove its transitivity, let x,y,z ∈U be such that xφUz and zφUy.
There are a,b,c,d ∈ Rsuch that{x,z} ∪(a,b)∪U and{z,y} ∪ (c,d) ∪U . Since z∈(a,b)∅(c,d), it follows that(a,b)⊆(c,d)is an interval(e,e∞)such that {x,y} ∪(e,e∞)∪U , which shows that xφUy. Thus,φU is an equivalence on U .
We claim that each equivalence class[x]φU is an open interval or a set of the form (a,+⊥)or a set of the form(−⊥,b). Indeed, suppose that u, v ∈ [x]φU (that is, uφUx andvφUx) and that t ∈(u, v). We now prove that tφUx.
There are two open intervals(a,b)and(c,d)such that{u,x} ∪(a,b)∪U and {x, v} ∪(c,d)∪U . Again,(a,b)⊆(c,d)is an open interval(e,e∞)and we have (u, v)∪(e,e∞)∪U . Thus, if[x]φU contains two numbers u andv, it also contains the interval(u, v)determined by these numbers.
To prove that[x]φU has the desired form, we shall prove that this set has no least element and no greatest element. Suppose that[x]φU has a least elementy. Then, there exist a and b such that a <y <x <b and(a,b)∪U . Sinceyis supposed to be the least element of[x]φU, if a <z<y, we have z ⊕∈ [x]φU. This contradicts yφUz andyφUx. In a similar manner, it is possible to show that[x]φU has no largest element.
Finally, we prove that the partition that corresponds toφU is countable. Select a rational number rx ∈ [x]φU ∅Q. Since the equivalence classes[x]φU are pairwise disjoint, it follows that[x]φU ⊕= [y]φU implies rx ⊕=ry. Thus, we have an injection r : U/φU −∩ Qgiven by r([x]φU) =rx for x ∈ U . By Theorem 1.126, the set U/φU is countable.
Example 4.12 Let S be an infinite set. The family of sets O= {∈} ⊆ {L∈P(S) | S−U is finite} is a topology on S. We refer toOas the cofinite topology on S.
Note that both∈and S belong toO. Further, ifCis a subcollection ofO, then S−
C=⎜
{(S−L) | L∈C}, which is a finite set because it is a subset of every finite set S−L, where L ∈C.
Also, if U,V ∈O, then S−(U∅V)=(S−U)⊆(S−V), which shows that S−(U∅V)is a finite set. Thus, U∅V ∈O.
Example 4.13 Let(S,)be a partially ordered set. A subset T of S is upward closed if x ∈ T and x y impliesy ∈ T . The collection of upwards closed setsO∨is a topology on S.
It is clear that both∈and S belong toO∨. Further, if{Li | i ∈ I}is a family of upwards closed sets, then
{Li | i ∈ I}is also an upwards closed set. Indeed, suppose that x ∈
{Li | i ∈ I}and x y. There exists Li such that x ∈ Li and thereforey∈Li, which impliesy∈
{Li | i ∈I}. Moreover, it is easy to see that any intersection of sets fromO∨belongs toO∨, not just a finite intersection (which would suffice forO∨ to be a topology). This topology is known as the Alexandrov topology on the poset(S,).
Definition 4.14 A topologyOis finer than a topologyO∞or, equivalently,O∞ is a coarser thanO, ifO∞∪O.
Every topology on a set S is finer than the indiscrete topology on S; the discrete topologyP(S)(which has the largest collection of open sets) is finer than any topology on S.
Theorem 4.15 Let(S,O)be a topological space and let T be a subset of S. The collectionOT defined byOT= {L∅T | L ∈O}is a topology on the set T . Proof We leave the proof of this theorem to the reader as an exercise.
Definition 4.16 If U is a subset of S, where(S,O)is a topological space, then we refer to the topological space (U,O U)as a subspace of the topological space (S,O).
To simplify notation, we denote the subspace(U,OU)just by U .
Theorem 4.17 Let(S,O)be a topological space and let(T,OT)be a subspace of this space. Then, a set H is closed in(T,OT)if and only if there exists a closed set H0in(S,O)such that H=T ∅H0.
Proof Suppose that H is closed in(T,OT). Then, the set T −H is open in this space and therefore there exists an open set L0in(S,O)such that T−H =T∅L0. This is equivalent to H =T−(T∅L0)=T∅(S−L0). We define H0as the closed set S−L0.
154 4 Topologies and Measures Conversely, suppose that H = T ∅H0, where H0 is a closed set in S. Since T−H =T ∅(S−H0)and S−H0is an open set in(S,O), it follows that T−H is open in the subspace and therefore H is closed.
Corollary 4.18 Let(S,O)be a topological space and let T ∪S. Denote by KSand KT the closure operators of(S,O)and(T,OT), respectively. For every subset W of T , we have KT(W)=KS(W)∅T .
Proof The set KS(W)is closed in S, so KS(W)∅T is closed in T by Theorem 4.17.
Since W ∪KS(W)∅T , it follows that KT(W)∪KS(W)∅T .
To prove the converse inclusion, observe that we can write KT(W) = T ∅H , where H is a closed set in S because KT(W)is a closed set in T . Since W ∪H , it follows that KS(W)∪H , so KS(W)∅T ∪H∅T =KT(W).
Corollary 4.19 Let(S,O)be a topological space and let T ∪ S. If U ∪ S, then KT(U∅T)∪KS(U)∅T .
Proof By applying Corollary 4.18 to the subset U∅T of T we have KT(U∅T)= KS(U∅T)∅T . The needed inclusion follows from the monotonicity of KS. Definition 4.20 A set U is dense in a topological space (S,O)if K(U) = S. A topological space is separable if there exists a countable set U that is dense in (S,O).
Theorem 4.21 If T is a subspace of a separable topological space(S,O), then T itself is separable.
Proof Since S is separable, there exists a countable set U such that KS(U)=S. On the other hand, KT(U∅T)=KS(U∅T)∅T ∪KS(U)∅T =S∅T =T , which implies that the countable set U∅T is dense in T . Thus, T is separable.
Theorem 4.22 If T is a separable subspace of a topological space(S,O), then so is KS(T).
Proof Let U be a countable subset of T that is dense in T , that is, KT(U)=T . We need to prove that KKS(T)(U)=KS(T)to prove that U is dense in KS(T)also.
By Corollary 4.18, we have
KKS(T)(U)=KS(U)∅KS(T)=KS(U) due to the monotonicity of KS.
Note that T =KT(U)=KS(U)∅T , so T ∪KS(U), which implies KS(T)∪ KS(U). Since KSis monotonic, we have the reverse inclusion KS(U)∪KS(T), so KS(U) = KS(T). This allows us to conclude that KKS(T)(U)= KS(T), so U is dense in KS(T).
Theorem 4.23 Let(S,O)be a topological space. The set U is dense in(S,O)if and only if U ∅L⊕= ∈for every non-empty open set L.
Proof Suppose that U is dense, so K(U)=S. Since K(U∅L)=K(U)∅K(L)= S ∅K(L) = K(L), U ∅ L = ∈would imply K(L) = K(∈) = ∈, which is a contradiction because∈ ⊕=L∪K(L).
Conversely, suppose that U has a non-empty intersection with every non-empty open set L. Since K(U)is closed, S−K(U)is open. Observe that U∅(S−K(U))= ∈, so the open set S−K(U)must be empty. Therefore, we have K(U)=S.
Theorem 4.24 The following statements hold for any topological space(S,O)and x∈S:
(i) if U,V ∈neighx(O), then U∅V ∈neighx(O);
(ii) if U∈neighx(O)and U∪W ∪S, then W ∈neighx(O); (iii) a set L is open if and only if L is a neighborhood of all its points.
Proof The first two parts follow immediately from Definition 4.7. We discuss here only the third statement.
If L is open, it is immediate that L is a neighborhood of all its points. Conversely, suppose that L is a neighborhood of all its members. Then, for each x ∈ L there exists Wx ∈Osuch that x ∈Wx ∪L. Therefore,
L=
x∈L
{x} ∪
x∈L
Wx ∪L,
which implies L =
x∈LWx. This in turn implies L∈O.
In Chap.2, we discussed the notion of an interior system of sets on a set S and the notion of an interior operator. Since∈is an open set in any topological space(S,O) and any union of open sets is an open set, it follows that the topology itself is an interior system on S. In addition, an interior system of open sets is closed to finite intersection. Definition 4.25 which follows is a restatement of the definition of the interior operator associated to an interior system contained by Theorem 1.185.
Definition 4.25 Let(S,O)be a topological space. The interior of a set U , U ∪S, is the set I(U)=
{L ∈O | L ∪U}.
The interior I(U)of a set U is the largest open set included in U , because the union of any collection of open sets is an open set. Furthermore, a set is open in a topological space if and only if it equals its interior.
Theorem 4.26 Let(S,O)be a topological space and let U be a subset of S. The closure K(S−U)of the set S−U equals S−I(U).
Proof Since I(U)is an open set, the set S−I(U)is closed. Note that S−U ∪ S−I(U). Therefore, K(S−U)∪S−I(U).
Conversely, the inclusion S−U ∪K(S−U)implies S−K(S−U)∪U . Since S−K(S−U)is an open set included in U and I(U)is the largest such set, it follows that S−K(S−U)∪I(U), which implies S−I(U)∪K(S−U).
156 4 Topologies and Measures Corollary 4.27 For every subset U of a topological space(S,O), we have I(U)= S−K(S−U)and K(U)=S−I(S−U).
Proof The first equality is immediate; the second follows from Theorem 4.26 by replacing U by S−U .
Theorem 4.28 The following statements are equivalent for a topological space