19.1. GENERAL TOPOLOGICAL SPACES 503
Regular
pU
C V
Definition 19.1.10 The topological space, (X ,τ) is said to be normal if wheneverC and K are disjoint closed sets, there exist disjoint open sets U, V with C ⊆U, K ⊆ V .Thus any two disjoint closed sets can be separated with open sets.
Normal
CU
K V
Definition 19.1.11 Let E be a subset of X. E is defined to be the smallest closedset containing E.
Lemma 19.1.12 The above definition is well defined.
Proof: Let C denote all the closed sets which contain E. Then C is nonempty becauseX ∈ C .
(∩{A : A ∈ C })C = ∪{
AC : A ∈ C},
an open set which shows that ∩C is a closed set and is the smallest closed set whichcontains E. ■
Theorem 19.1.13 E = E ∪{limit points of E}.
Proof: Let x∈ E and suppose that x /∈ E. If x is not a limit point either, then there existsan open set U , containing x which does not intersect E. But then UC is a closed set whichcontains E which does not contain x, contrary to the definition that E is the intersection ofall closed sets containing E. Therefore, x must be a limit point of E after all.
Now E ⊆ E so suppose x is a limit point of E. Is x ∈ E? If H is a closed set containingE, which does not contain x, then HC is an open set containing x which contains no pointsof E other than x negating the assumption that x is a limit point of E. ■
The following is the definition of continuity in terms of general topological spaces. Itis really just a generalization of the ε − δ definition of continuity given in calculus.
Definition 19.1.14 Let (X ,τ) and (Y,η) be two topological spaces and let f : X→Y . f is continuous at x ∈ X if whenever V is an open set of Y containing f (x), there existsan open set U ∈ τ such that x ∈U and f (U)⊆V . f is continuous if f−1(V ) ∈ τ wheneverV ∈ η .
Then the following comes from the definition.
Proposition 19.1.15 In the situation of Definition 19.1.14 f is continuous if and onlyif f is continuous at every point of X.
Proof: ⇒ Suppose f is continuous and let f (x) ∈ V an open set in Y . Then x ∈f−1 (V )≡U ∈ τ .⇐ Next suppose f is continuous at every point. Then if V ∈ η , and x ∈ f−1 (V ) ,
continuity at x implies there is open Ux ⊆ f−1 (V ). Thus f−1 (V ) = ∪x∈ f−1(V )Ux and sof−1 (V ) is open. ■