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Topology: An Inquiry-Based Approach

Steven Schlicker

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Section Continuity and Open Sets

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Recall that we showed that a function f from a metric space (X,dX) to a metric space (Y,dY) is continuous if and only if fβˆ’1(N) is a neighborhood of a∈X whenever N is a neighborhood of f(a) in Y. We can now provide another characterization of continuous functions in terms of open sets. This is the characterization that will serve as our definition of continuity in topological spaces.
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Proof.

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Let (X,dX) and (Y,dY) be metric spaces, and let f:Xβ†’Y be a function. To prove this biconditional statement we need to prove both implications. First assume that f is a continuous function. We must show that fβˆ’1(O) is an open set in X for every open set O in Y. So let O be an open set in Y. To demonstrate that fβˆ’1(O) is open in X, we will show that fβˆ’1(O) is a neighborhood of each of its points. Let a∈fβˆ’1(O). Then f(a)∈O. Now O is an open set, so there is an open ball B(f(a),Ο΅) around f(a) that is entirely contained in O. Since B(f(a),Ο΅) is a neighborhood of f(a), we know that fβˆ’1(B(f(a),Ο΅)) is a neighborhood of a. Thus, there exists Ξ΄>0 so that B(a,Ξ΄)βŠ†fβˆ’1(B(f(a),Ο΅)). Now f(B(a,Ξ΄))βŠ†B(f(a),Ο΅)βŠ†O, and so B(a,Ξ΄)βŠ†fβˆ’1(O). We conclude that fβˆ’1(O) is a neighborhood of each of its points and is therefore an open set in X.
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The proof of the reverse implication is left for the next activity.
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Activity 8.5.

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Let f be a function from a metric space (X,dX) to a metric space (Y,dY).
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(a)

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What assumption do we make to prove the remaining implication of Theorem 8.5? What do we need to demonstrate to prove the conclusion?
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(b)

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Let a∈X, and let N be a neighborhood of f(a) in Y. Why does there exist an Ο΅>0 so that B(f(a),Ο΅)βŠ†N.
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(c)

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What does our hypothesis tell us about fβˆ’1(B(f(a),Ο΅)) in X?
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(d)

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Why is fβˆ’1(N) a neighborhood of a? How does this show that f is a continuous function?
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Recall that every open set is a union of open balls, so we can simplify proofs of continuous functions in metric spaces by working only with open balls instead of arbitrary open sets. The next activity provides the details.
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Activity 8.6.

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In this activity we prove the following corollary to Theorem 8.5.
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To set up the proof, let (X,dX) and (Y,dY) be metric spaces, and let f:X→Y be a function.
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(a)

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Since the corollary is a biconditional statement, we need to prove both implications. First, assume that f is continuous. Use Theorem 8.5 to explain why fβˆ’1(B) is open in X whenever B is an open ball in Y.
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(b)

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For the remaining implication, assume that fβˆ’1(B) is an open set in X for any open ball B in Y. To show that f is a continuous function, we will use Theorem 8.5 and show that fβˆ’1(O) is open in X whenever O is an open set in Y. So let O be an open set in Y.
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(ii)
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Recall that Lemma 2.11 tells us that if {BΞ²} is a collection of subsets of Y for Ξ² in some indexing set J, then
fβˆ’1(β‹ƒΞ²βˆˆJBΞ²)=β‹ƒΞ²βˆˆJfβˆ’1(BΞ²).
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Use Lemma 2.11 to show that fβˆ’1(O) is open in X and conclude that f is a continuous function.
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Example 8.7.

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As an example of Corollary 8.6, we prove that the square function from R to R is a continuous function. Let X=R with the Euclidean metric dE, and let f:Xβ†’X be defined by f(x)=x2. We will show that f is a continuous function by verifying that fβˆ’1(B) is open in X for every open ball B in X. Let B=B(b,Ξ²)=(bβˆ’Ξ²,b+Ξ²) be an open ball in X. Let Bβ€²=B(b,Ξ²)∩(R+βˆͺ{0}). We consider cases.
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    Suppose that Bβ€²=βˆ…. Then fβˆ’1(B)=βˆ… and fβˆ’1(B) is open in X.
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    Suppose that Bβ€²=[0,b+Ξ²). Then fβˆ’1(B)=(βˆ’b+Ξ²,b+Ξ²) and fβˆ’1(B) is open in X.
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    The final case is Bβ€²=(bβˆ’Ξ²,b+Ξ²). Then
    fβˆ’1(B)=(βˆ’b+Ξ²,βˆ’bβˆ’Ξ²)βˆͺ(bβˆ’Ξ²,b+Ξ²)
    and fβˆ’1(B) is open in X.
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Since the inverse image of every open ball is an open set, we conclude that f is a continuous function.
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