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Topology: An Inquiry-Based Approach

Steven Schlicker

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Section Continuity and Neighborhoods

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We can define continuity now in terms of neighborhoods instead of using metrics. The advantage here is that this idea does not explicitly depend on the existence of a metric, so we will be able to adopt this concept of continuity for arbitrary topological spaces.
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Recall that a function f from a metric space (X,dX) to a metric space (Y,dY) is continuous at a∈X if, for any Ο΅>0 there exists Ξ΄>0 so that dX(x,a)<Ξ΄ implies dY(f(x),f(a))<Ο΅. We can interpret this definition of continuity to say that for every Ο΅>0, the inverse image under f of the open ball B(f(a),Ο΅) contains the open ball B(a,Ξ΄) for some Ξ΄>0. It is not unreasonable to wonder if the set fβˆ’1(B(f(a),Ο΅)) itself is an open ball. We investigate this question in the following activity.
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Activity 7.3.

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Let f be a function from a metric space (X,dX) to a metric space (Y,dY) that is continuous at a∈X. Using the notation from the paragraph above, in this activity we determine if fβˆ’1(B(f(a),Ο΅)) must equal B(a,Ξ΄) for some Ξ΄.
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Define f:R→R by
f(x)=x2,
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where we use the Euclidean metric dE throughout. Assume that f is a continuous function. Then f is continuous at x=2.
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(c)

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Is fβˆ’1(B(f(2),1)) an open ball centered at 2? Explain.
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The conclusion to be drawn from Activity 7.3 is that if f is continuous, we can only conclude that the inverse image of B(f(a),Ο΅) contains an open ball centered at a. By definition of continuity, if for every Ο΅>0 there exists a Ξ΄>0 so that the open ball fβˆ’1(B(f(a),Ο΅)) contains B(a,Ξ΄), then f is continuous at a. We summarize this in the next theorem.
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We can extend this idea of continuity to describe continuity in terms of neighborhoods. This condition will allow us to later consider continuous functions even if there are no metrics on our 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 continuous at some point a∈X. We will show that for any neighborhood N of f(a) in Y, its inverse image fβˆ’1(N) in X is a neighborhood of a in X. Let N be a neighborhood of f(a) in Y. To demonstrate that fβˆ’1(N) is a neighborhood of a in X, we need to find an open ball around a that is contained in fβˆ’1(N). Since N is a neighborhood of f(a), by definition there exists Ο΅>0 so that B(f(a),Ο΅)βŠ†N. Since f is continuous at a, there exists Ξ΄>0 such that B(a,Ξ΄)βŠ†fβˆ’1(B(f(a),Ο΅)). So if x∈B(a,Ξ΄), then f(x)∈B(f(a),Ο΅)βŠ†N. So B(a,Ξ΄)βŠ†fβˆ’1(N), and fβˆ’1(N) is a neighborhood of a in X.
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The proof of the reverse implication is left for the next activity.
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Activity 7.4.

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Let (X,dX) and (Y,dY) be metric spaces, and let f:Xβ†’Y be a function. Let a∈X. In this activity we prove that if the inverse image of every neighborhood of f(a) is a neighborhood of a, then f is continuous at a.
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(a)

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What does Theorem 7.5 tell us that we need to do to show that f is continuous at a?
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Suppose Ο΅ is greater than 0, why is B(f(a),Ο΅) a neighborhood of f(a) in Y?
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(b)

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What does our hypothesis tell us about fβˆ’1(B(f(a),Ο΅))?
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We conclude this section with some important facts about neighborhoods. Assume that (X,d) is a metric space and a∈X.
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    There is a neighborhood that contains a.
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    If N is a neighborhood of a and NβŠ†M, then M is a neighborhood of a.
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    If M and N are neighborhoods of a, then so is M∩N.
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The proofs are straightforward and left for Exercise 8.
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