Kenneth Kuttler

168 CHAPTER 7. METRIC SPACES AND TOPOLOGICAL SPACES

compact subset of X . Then there exist sets from C , U1, · · · ,Ur which cover K. Therefore,a finite subcover of X̃ is U1, · · · ,Ur,U∞.

To see the last claim, suppose U contains ∞ since otherwise there is nothing to show.Notice that if C is a compact set, then X \C is an open set. Therefore, if x ∈U \{∞} , andif X̃ \C is a basic open set contained in U containing ∞, then if x is in this basic open setof X̃ , it is also in the open set X \C ⊆U \{∞} . If x is not in any basic open set of the formX̃ \C then x is contained in an open set of τ which is contained in U \{∞}. Thus U \{∞}is indeed open in τ .

Definition 7.12.21 If every finite subset of a collection of sets has nonempty intersection,the collection has the finite intersection property.

Theorem 7.12.22 Let K be a set whose elements are compact subsets of a Hausdorfftopological space, (X ,τ). Suppose K has the finite intersection property. Then /0 ̸= ∩K .

Proof: Suppose to the contrary that /0 = ∩K . Then consider

C ≡{

KC : K ∈K}.

It follows C is an open cover of K0 where K0 is any particular element of K . But thenthere are finitely many K ∈K , K1, · · · ,Kr such that K0⊆∪r

i=1KCi implying that∩r

i=0Ki = /0,contradicting the finite intersection property.

Lemma 7.12.23 Let (X ,τ) be a topological space and let B be a basis for τ . Then K iscompact if and only if every open cover of basic open sets admits a finite subcover.

Proof: Suppose first that X is compact. Then if C is an open cover consisting of basicopen sets, it follows it admits a finite subcover because these are open sets in C .

Next suppose that every basic open cover admits a finite subcover and let C be an opencover of X . Then define C̃ to be the collection of basic open sets which are contained insome set of C . It follows C̃ is a basic open cover of X and so it admits a finite subcover,{

U1, · · · ,Up}

. Now each Ui is contained in an open set of C . Let Oi be a set of C whichcontains Ui. Then

{O1, · · · ,Op

}is an open cover of X . This proves the lemma.

In fact, much more can be said than Lemma 7.12.23. However, this is all which I willpresent here.

7.13 Connected SetsStated informally, connected sets are those which are in one piece. More precisely,

Definition 7.13.1 A set, S in a general topological space is separated if there exist sets,A,B such that

S = A∪B, A,B ̸= /0, and A∩B = B∩A = /0.

In this case, the sets A and B are said to separate S. A set is connected if it is not separated.

One of the most important theorems about connected sets is the following.

168 CHAPTER 7. METRIC SPACES AND TOPOLOGICAL SPACEScompact subset of X. Then there exist sets from @, U),--- ,U, which cover K. Therefore,a finite subcover of X is Uj,--- ,U;,Us.To see the last claim, suppose U contains © since otherwise there is nothing to show.Notice that if C is a compact set, then X \C is an open set. Therefore, if x € U \ {>}, andif X \C is a basic open set contained in U containing ©, then if x is in this basic open setof X, it is also in the open set X \C CU \ {o>}. If x is not in any basic open set of the formX \C then x is contained in an open set of t which is contained in U \ {co}. Thus U \ {co}is indeed open in T.Definition 7.12.21 If every finite subset of a collection of sets has nonempty intersection,the collection has the finite intersection property.Theorem 7.12.22 Let % be a set whose elements are compact subsets of a Hausdorfftopological space, (X,7). Suppose H has the finite intersection property. Then®#NX.Proof: Suppose to the contrary that @=.%. Then considerC={K°: KE XH}.It follows @ is an open cover of Ko where Ko is any particular element of “%. But thenthere are finitely many K € .#, Ki,--- ,K, such that Ko C U!_, K€ implying that N_)Ki = 9,contradicting the finite intersection property.Lemma 7.12.23 Let (X,7) be a topological space and let B be a basis for t. Then K iscompact if and only if every open cover of basic open sets admits a finite subcover.Proof: Suppose first that X is compact. Then if @ is an open cover consisting of basicopen sets, it follows it admits a finite subcover because these are open sets in @.Next suppose that every basic open cover admits a finite subcover and let @ be an opencover of X. Then define @ to be the collection of basic open sets which are contained insome set of @. It follows @ is a basic open cover of X and so it admits a finite subcover,{U on Up}. Now each U; is contained in an open set of @. Let O; be a set of @ whichcontains U;. Then {O1 oo Op} is an open cover of X. This proves the lemma.In fact, much more can be said than Lemma 7.12.23. However, this is all which I willpresent here.7.13 Connected SetsStated informally, connected sets are those which are in one piece. More precisely,Definition 7.13.1 A set, S in a general topological space is separated if there exist sets,A,B such that _S=AUB, A,B £40, andANB=BNA=0.In this case, the sets A and B are said to separate S. A set is connected if it is not separated.One of the most important theorems about connected sets is the following.