Topologies and Measures
Theorem 4.81 The image of a connected topological space through a continuous function is a connected set
Proof Let(S1,O1)and(S2,O2)be two topological spaces and let f : S1 −∩ S2
be a continuous function, where S1is connected. If f(S1)were not connected, we would have two nonempty open subsets L and L∞of f(S1)that are complementary.
Then, f−1(L) and f−1(L∞) would be two nonempty, open sets in S1 which are complementary, which contradicts the fact that S1is connected.
A characterization of connected spaces is given next.
Theorem 4.82 Let (S,O)be a topological space and let ({0,1}),P({0,1})be a two-element topological space equipped with the non-discrete topology. Then, S is connected if and only if every continuous application f :S−∩ {0,1}is constant.
Proof Suppose that S is connected. Both f−1(0)and f−1(1)are clopen sets in S because both{0}and{1}are clopen in the discrete topology. Thus, we have either f−1(0)= ∈and f−1(1)=S, or f−1(0)=S and f−1(1)= ∈. In the first case, f is the constant function f(x)=1; in the second, it is the constant function f(x)=0.
Conversely, suppose that the condition is satisfied for every continuous function f : S −∩ {0,1}and suppose(S,O)is not connected. Then, there exist two non-empty disjoint open subsets L and L∞ that are complementary. Let f =1L be the indicator function of L, which is continuous because both L and L∞are open. Thus, f is constant and this implies either L= ∈and L∞ =S or L =S and L∞ = ∈, so S is connected.
Example 4.83 Theorem 4.82 allows us to prove that the connected subsets ofRare exactly the intervals.
Suppose that T is a connected subset of S but is not an interval. Then, there are three numbers x,y,z such that x<y<z, x,z∈ T buty⊕∈T . Define the function
170 4 Topologies and Measures f : T −∩ {0,1}by f(u) = 0 if u < y and f(u) = 1 ify < u. Clearly, f is continuous but is not constant, and this contradicts Theorem 4.82. Thus, T must be an interval.
Suppose now that T is an open interval ofR. We saw that T is homeomorphic toR(see Example 4.68), so T is indeed connected. If T is an arbitrary interval, its interior I(T)is an open interval and, since I(T)∪T ∪K(I(T)), it follows that T is connected.
Definition 4.84 A topological space(S,O)is totally disconnected if, for every x∈ S, the connected component of x is Kx = {x}.
Example 4.85 Any topological space equipped with the discrete topology is totally disconnected.
Theorem 4.86 Let(S,O)be a topological space and let T be a subset of S.
If for every pair of distinct points x,y∈T there exist two disjoint closed sets Hx and Hysuch that T ∪Hx⊆Hy, x∈ Hx, andy∈Hy, then T is totally disconnected.
Proof Let Kxbe the connected component of x, and suppose thaty∈ Kxandy⊕=x, that is, Kx = Ky = K . Then, K ∅Hx and K ∅Hyare nonempty disjoint closed sets and K =(K ∅Hx)⊆(K ∅Hy), which contradicts the connectedness of K . Therefore, Kx = {x}for every x∈T and T is totally disconnected.
4.8 Separation Hierarchy of Topological Spaces
We introduce a hierarchy of topological spaces that is based on separation properties of these spaces.
Definition 4.87 Let(S,O)be a topological space and let x andybe two arbitrary, distinct elements of S. This topological space is:
(i) a T0space if there exists U ∈Osuch that one member of the set{x,y}belongs to U and the other to S−U ;
(ii) a T1space if there exist U,V ∈Osuch that x∈U−V andy∈V −U ; (iii) a T2space or a Hausdorff space if there exist U,V ∈ Osuch that x ∈U and
y∈V and U∅V = ∈;
(iv) a T3space if for every closed set H and x ∈S−H there exist U,V ∈Osuch that x∈U and H ∪V and U∅V = ∈;
(v) a T4space if for all disjoint closed sets H,L there exist U,V ∈ Osuch that H∪U , L∪V , and U∅V = ∈.
Theorem 4.88 A topological space(S,O)is a T1space if and only if every singleton {x}is a closed set.
Proof Suppose that(S,O)is a T1, space and for everyy ∈ S− {x}let Uyand Vy be two open sets such as x ∈Uy−Vyandy∈ Vy−Uy. Then, x ∈
y⊕=xUyand x ⊕∈
y⊕=xVy, soy∈
y⊕=xVy∪ S− {x}. Thus, S− {x}is an open set, so{x}is closed.
Conversely, suppose that each singleton{u}is closed. Let x,y∈ S be two distinct elements of S. Note that the sets S− {x}and S− {y}are open and x ∈(S− {y})− (S− {x})andy∈(S− {x})−(S− {y}), which shows that(S,O)is a T1-space.
Theorem 4.89 Let(S,O)be a T4-separated topological space. If H is a closed set and L is an open set such that H ∪ L, then there exists an open set U such that H ∪U∪K(U)∪L.
Proof Observe that H and S−L are two disjoint closed sets under the assumptions of the theorem. Since(S,O)is a T4-separated topological space, there exist U,V ∈O such that H ∪U , S−L ∪V and U∅V = ∈. This implies U ∪S−V ∪L. Since S−V is closed, we have
H∪U∪K(U)∪K(S−V)=S−V ∪L, which proves that U satisfies the conditions of the theorem.
The next theorem is in some sense a reciprocal result of Theorem 4.61, which holds in the realm of Hausdorff spaces.
Theorem 4.90 Each compact subset of a Hausdorff space(S,O)is closed.
Proof Let H be a compact subset of(S,O)and letybe an element of the set S−H . It suffices to show that the set S −H is open. For every x ∈ H , we have two open subsets Uxand Vx such that x∈Ux,y∈Vx and Ux∅Vx = ∈. The collection {Ux | x∈H}is an open cover of H and the compactness of H implies the existence of a finite subcover Ux1, . . . ,Uxn of H . Consider the open set V =⎜n
i=1Vxi, which is disjoint from each of the sets Ux1, . . . ,Uxn and, therefore, it is disjoint from H . Thus, for everyy ∈ S−H there exists an open set V such thaty ∈ V ∪S−H , which implies that S−H is open.
Corollary 4.91 In a Hausdorff space(S,O), each finite subset is a closed set.
Proof Since every finite subset of S is compact, the statement follows immediately from Theorem 4.90.
It is clear that every T2 space is a T1 space and each T1 space is a T0 space.
However, this hierarchy does not hold beyond T2. This requires the introduction of two further classes of topological spaces.
Definition 4.92 A topological space(S,O)is regular if it is both a T1 and a T3
space;(S,O)is normal if it is both a T1and a T4space.
172 4 Topologies and Measures Theorem 4.93 Every regular topological space is a T2 space and every normal topological space is a regular one.
Proof Let(S,O)be a topological space that is regular and let x andybe two distinct points in S. By Theorem 4.88, the singleton{y}is a closed set. Since(S,O)is a T3, space, two open sets U and V exist such that x ∈U ,{y} ∪V , and U∅V = ∈, so (S,O)is a T2space. We leave the second part of the theorem to the reader.
4.9 Products of Topological Spaces
Theorem 4.94 Let{(Si,Oi) | i ∈ I}be a family of topological spaces indexed by the set I . Define on the set S=
i∈I Sithe collection of setsB= {⎜
j∈J p−j1(Lj) | Lj ∈Ojand J finite}. Then,Bis a basis.
Proof Note that every set⎜
j∈J p−j1(Lj)has the form
i∈I−J×
j∈J Lj. We need to observe only that a finite intersection of sets inBis again a set inB. Therefore, Bis a basis.
Definition 4.95 The topology TOP(B)generated on the set S byBis called the product of the topologiesOi and is denoted by
i∈IOi.
The topological space {(Si,Oi) | i ∈ I}is the product of the collection of topological spaces{(Si,Oi) | i ∈I}.
The product of the topologies{Oi | i ∈ I}can be generated starting from the subbasisSthat consists of sets of the form Dj,L = {t | t ∈
i∈I | t(j) ∈ L}, where j ∈ I and L is an open set in(Sj,Oj). It is easy to see that any set in the basis Bis a finite intersection of sets of the form Dj,L.
Example 4.96 LetRn = R× · · · ×R, where the product involves n copies ofR and n 1. In Example 4.45, we saw that the collection of open intervals{(a,b) | a,b ∈Rand a<b}is a basis for the topological space(R,O). Therefore, a basis of the topological space (Rn,O× · · ·O)consists of parallelepipeds of the form (a1,b1)× · · · ×(an,bn), where ai <bi for 1i n.
Theorem 4.97 Let{(Si,Oi) | i ∈ I}be a collection of topological spaces. Each projection pν :
i∈I Si −∩ Sν is a continuous function forν ∈ I . Moreover, the product topology is the coarsest topology on S such that projections are continuous.
Proof Let L be an open set in(Sν,Oν). We have
pν−1(L)=
t ∈
i∈I
Si | t(ν)∈ L
,
which has the form
The proof of the second part of the theorem is left to the reader.
A preliminary result to a theorem that refers to the compactness of products of topological spaces is shown next.
Lemma 4.98 LetCbe a collection of subsets of S =
i∈ISi such thatChas the f.i.p.andCis maximal with this property.
We have⎜
D∈Cfor every finite subcollectionDofC. Furthermore, if T∅C⊕= ∈ for every C ∈C, then T ∈C.
Proof LetD= {D1, . . . ,Dn}be a finite subcollection ofCand let D=⎜ D⊕= ∈. Note that the intersection of every finite subcollection ofC⊆ {D}is also nonempty.
The maximality ofCimplies D∈C, which proves the first part of the lemma.
For the second part of the lemma, observe that the intersection of any finite subcollection ofD⊆ {T}is not empty. Therefore, as above, T ∈C. topological spaces(Si,Oi)is compact because each projection piis continuous.
Conversely, suppose that each of the topological spaces(Si,Oi)is compact.
LetEbe a family of sets in S=
i∈ISi that has thef.i.p.and let(C,∪)be the partially ordered set whose elements are collections of subsets of S that have the f.i.p.and contain the familyE.
Let{Ci | i∈ I}be a chain in(C,∪). It is easy to verify that
{Ci | i ∈I}has the f.i.p., so every chain in(C,∪)has an upper bound. Therefore, by Zorn’s Lemma (see Theorem 2.81), the poset(C,∪)contains a maximal collectionCthat has thef.i.p.and containsE. We aim to find an element t∈
i∈ISithat belongs to⎜
{K(C) | C ∈C}
because, in this case, the same element belongs to⎜
{K(C) | C ∈E}and this would imply, by Theorem 4.54, that(S,O)is compact.
LetCi be the collection of closed subsets of Si defined by Ci = {Ki(pi(C)) | C∈C}
for i ∈ I , where Ki is the closure of the topological space(Si,Oi).
It is clear that each collectionCihas thef.i.p.in Si. Indeed, sinceChas thef.i.p., if{C1, . . . ,Cn} ∪Cand x ∈⎜n
k=1Ck, then pi(x)∈ ⎜n
k=1K(pi(Ck)), soCi has thef.i.p.Since(Si,Oi)is compact, we have⎜
Ci ⊕= ∈, by Part (iii) of Theorem 4.53.
174 4 Topologies and Measures Let ti ∈⎜
Ci =⎜
{Ki(pi(C)) | C ∈C}and let t ∈ S be defined by t(i)=ti for i ∈ I .
Let Dj,L = {u | u ∈
i∈I | u(j)∈ L}, a set of the subbasis of the product topology that contains t , defined earlier, where L is an open set in(Sj,Oj). Since g(j)∈ L, the set L has a nonempty intersection with every set Ki(pi(C)), where C ∈ C. On the other hand, since pi(Dj,L) = Si for i ⊕= j , it follows that for every i ∈ I we have pi(Dj,L)∅⎜
C∈CKi(pi(C)) ⊕= ∈. Therefore, pi(Dj,L)has a nonempty intersection with every set of the form Ki(pi(C)), where C ∈ C. By the contrapositive of Theorem 4.9, this means that pi(Dj,L)⊆pi(C)⊕= ∈for every i ∈ I and C ∈ C. This in turn means that Dj,L ⊆C ⊕= ∈for every C ∈ C. By Lemma 4.98, it follows that Dj,L ∈ C. Since every set that belongs to the basis of the product topology is a finite intersection of sets of the form Dj,L, it follows that any member of the basis has a nonempty intersection with every set ofC. This implies that g belongs to
{K(C) | C ∈ C}, which implies the compactness of (
i∈ISi,
i∈IOi).
Example 4.100 In Example 4.57, we have shown that every closed interval[x,y] of Rwhere x < y is compact. By Theorem 4.99, any subset ofRn of the form [x1,y1] × · · · × [xn,yn]is compact.
4.10 Fields of Sets
In this section, we introduce collections of sets that play an important role in measure and probability theory.
Definition 4.101 Let S be a set. A field of sets on S is a family of subsetsEof S that satisfies the following conditions:
(i) S∈E;
(ii) if U∈E, thenU¯ =S−U ∈E; (iii) if U0, . . . ,Un−1belong toE, then
0in−1Ui belongs toE.
Aπ-field of sets on S is a family of subsetsEof S that satisfies conditions (i) and (ii) and, in addition, satisfies the following condition:
(iii∞) if{Ui |i∈N}is a countable family of sets included inE, then
i∈NUibelongs toE. Clearly, everyπ-field is also a field on S.
IfEis aπ-field of sets on S, we refer to the pair(S,E)as a measurable space.
Example 4.102 The collection E0 = {∈,S}is aπ-field on S; moreover, for every π-fieldEon S, we haveE0∪E.
The setP(S)of all subsets of a set S is aπ-field on S.
If T is a subset of S, then the collection{∈,T,S−T,S}is aπ-field on S.
Theorem 4.103 The class of all fields (π-fields) of sets on S is a closure system on
A similar argument proves that the class of allπ-fields of sets is also a closure system onP(S).
Example 4.104 Let A be a subset of the set S. Theπ-field generated by the collection {A}is{∈,A,A¯,S}.
Definition 4.105 Let(S,O)be a topological space. A subset T of S is said to be a Borel set if it belongs to theπ-field generated by the topologyO.
Theπ-field of Borel sets of(S,O)is denoted byBO.
It is clear that all open sets are Borel sets. Also, every closed set, as a complement of an open set, is a Borel set.
Example 4.106 We identify several families of Borel subsets of the topological space (R,O).
It is clear that every open interval(a,b)and every set(a,⊥)or(−⊥,a)is a Borel set for a,b∈ Rbecause they are open sets. The closed intervals of the form [a,b]are Borel sets because they are closed sets in the topological space.
Since[a,b)=(−⊥,b)−(−⊥,a), it follows that the half-open intervals of this