50 CHAPTER 2. BASIC TOPOLOGY AND ALGEBRA

Definition 2.5.18 Let f : X → Y be a function where (X ,d) and (Y,ρ) are metricspaces. Then f is continuous at x ∈ X if and only if the following condition holds. Forevery ε > 0, there exists δ > 0 such that if d (x̂,x) < δ , then ρ ( f (x̂) , f (x)) < ε . If f iscontinuous at every x ∈ X we say that f is continuous on X. The notation f−1 (S) means{x ∈ X : f (x) ∈ S} . It is called the inverse image of S.

For example, you could have a real valued function f (x) defined on an interval [0,1] . Inthis case you would have X = [0,1] and Y =R with the distance given by d (x,y) = |x− y|.Then the following theorem is the main result. Recall that if (X ,d) is a metric space andS⊆ X is a nonempty subset, then (S,d) is also a metric space so this latter case is includedin what follows.

Theorem 2.5.19 Let f : X → Y where (X ,d) and (Y,ρ) are metric spaces. Thenthe following two are equivalent.

a f is continuous at x ∈ D( f ) .

b Whenever xn→ x ∈ D( f ) , each xn ∈ D( f ) , it follows that f (xn)→ f (x) .

Also, the following are equivalent.

c f is continuous on X .

d Whenever V is open in Y, it follows that f−1 (V )≡ {x ∈ X : f (x) ∈V} is open in X .

e Whenever H is closed in Y, it follows that f−1 (H) is closed in X.

Proof: a⇒ b: Let f be continuous at x and suppose xn→ x. Then let ε > 0 be given.By continuity, there exists δ > 0 such that if d (x̂,x) < δ , then ρ ( f (x̂) , f (x)) < ε. Sincexn→ x, it follows that there exists N such that if n≥ N, then d (xn,x)< δ and so, if n≥ N,it follows that ρ ( f (xn) , f (x))< ε. Since ε > 0 is arbitrary, it follows that f (xn)→ f (x).

b ⇒ a: Suppose b holds but f fails to be continuous at x. Then there exists ε > 0such that for all δ > 0, there exists x̂ such that d (x̂,x)< δ but ρ ( f (x̂) , f (x))≥ ε . Lettingδ = 1/n, there exists xn such that d (xn,x) < 1/n but ρ ( f (xn) , f (x)) ≥ ε . Now this is acontradiction because by assumption, the fact that xn → x implies that f (xn)→ f (x). Inparticular, for large enough n, ρ ( f (xn) , f (x))< ε contrary to the construction.

c⇒d: Let V be open in Y . Let x∈ f−1 (V ) so that f (x)∈V. Since V is open, there existsε > 0 such that B( f (x) ,ε)⊆V . Since f is continuous at x, it follows that there exists δ > 0such that if x̂ ∈ B(x,δ ) , then f (x̂) ∈ B( f (x) ,ε) ⊆ V.( f (B(x,δ ))⊆ B( f (x) ,ε)) In otherwords, B(x,δ ) ⊆ f−1 (B( f (x) ,ε)) ⊆ f−1 (V ) which shows that, since x was an arbitrarypoint of f−1 (V ) , every point of f−1 (V ) is an interior point which implies f−1 (V ) is open.

d⇒ e: Let H be closed in Y . Then f−1 (H)C = f−1(HC)

which is open by assumption.Hence f−1 (H) is closed because its complement is open.

e⇒ d: Let V be open in Y. Then f−1 (V )C = f−1(VC)

which is assumed to be closed.This is because the complement of an open set is a closed set. Thus f−1 (V ) is open becauseits complement is closed.

d ⇒ c: Let x ∈ X be arbitrary. Is it the case that f is continuous at x? Let ε > 0be given. Then B( f (x) ,ε) is an open set in V (Recall that open balls are open.) and sox ∈ f−1 (B( f (x) ,ε)) which is given to be open. Hence there exists δ > 0 such that x ∈B(x,δ )⊆ f−1 (B( f (x) ,ε)) . Thus, f (B(x,δ ))⊆ B( f (x) ,ε) so if d (x, x̂)< δ meaning that