Basic Analysis – Gently Done Topological Vector Spaces

4 Basic Analysis
Proof The statements (ii) and (iii) are contrapositives. We shall show that (i)
and (ii) are equivalent. The proof rests on the observation that if {Uα } is any
collection of sets, then K ⊆ �
αUα if and only if K ∩�α
(X \Uα ) = ∅. We first
show that (i) implies (ii). Suppose that K is compact and let {Fα } be a given
family of closed sets such that K ∩ �
αFα = ∅. Put Uα = X \Fα . Then each Uα is open, and, by the above observation, K ⊆ �
αUα . But then there is a finite set
I such that K ⊆ �
α∈I Uα , and so K ∩�α∈I
Fα = ∅, which proves (ii).
Now suppose that (ii) holds, and let {Uα } be an open cover of K. Then each
X\U α is closed and K∩ �
α (X\U α ) = ∅. By (ii), there is a finite set I such that
K ∩ �
α∈I (X \U �
α ) = ∅. This is equivalent to the statement that K ⊆ α∈I Uα .
Hence K is compact.
Remark 1.12 We say that a family {A α } α∈J has the finite intersection property
if �
α∈I A α
�= ∅ for each finite subset I in J. Thus, we can say that a topological
space (X,T) is compact if and only if any family of closed sets {F α } α∈J in X
having the finite intersection property is such that �
α∈J F α
�= ∅.
Definition 1.13 A set N is a neighbourhood of a point x in a topological space
(X,T) if and only if there is U ∈ T such that x ∈ U and U ⊆ N.
Note that N need not itself be open. For example, in any metric space (X,d),
the closed sets {x ∈ X : d(a,x) ≤ r}, for r > 0, are neighbourhoods of the point a.
We note also that a set U belongs to T if and only if U is a neighbourhood of
each of its points. (Indeed, to say that U is a neighbourhood of x is to say that x
is an interior point of U. We have already observed that a set is open if and only
if each of its points is an interior point and so this is the same as saying that it is
a neighbourhood of each of its points.)
Definition 1.14 A topological space (X,T) is said to be a Hausdorff topological
space if and only if for any pair of distinct points x,y ∈ X, (x �= y), there exist
sets U,V ∈ T such that x ∈ U, y ∈ V and U ∩V = ∅.
We can paraphrase the Hausdorff property by saying that any pair of distinct
points can be separated by disjoint open sets. Example 3 above is an example of
a non-Hausdorff topological space—take {x,y} to be {1,2}.
Proposition 1.15 A non-empty subset A of the topological space (X,T) is
compact if and only if A is compact with respect to the induced topology, that is,
if and only if (A,T A ) is compact. If (X,T) is Hausdorff then so is (A,T A ).
Proof Suppose first that A is compact in (X,T), and let {G α } be an open cover
of A in (A,T A ). Then each G α has the form G α = A ∩U α for some U α ∈ T. It
follows that {U α } is an open cover of A in (X,T). By hypothesis, there is a finite
Department of Mathematics King’s College, London