Topology: An Inquiry-Based Approach

Recall that we could characterize a function $f$ from a metric space $(X,d_X)$ to a metric space $(Y,d_Y)$ as continuous at $a\in X$ if $f^{-1}(N)$ is a neighborhood of $a$ in $X$ whenever $N$ is a neighborhood of $f(a)$ in $Y\text{.}$ We have defined neighborhoods in topological spaces, so we can use this characterization as our definition of a continuous function from one topological space to another.

We saw that in metric spaces, a useful characterization of continuity was in terms of open sets. It is not surprising that we have the same characterization in topological spaces. You may assume the result of Theorem 14.2 (the topological space version of Theorem 8.5 for metric spaces) for this activity.

To complete the introduction to this section, we prove Theorem 14.2. We prove one direction now and leave the other for the next activity.

Let $f$ be a function from a topological space $(X,{\tau }_X)$ to a topological space $(Y,{\tau }_Y)\text{.}$ We first assume that $f$ is continuous and show that $f^{-1}(O)$ is an open set in $X$ whenever $O$ is an open set in $Y\text{.}$ Suppose that $O$ is an open set in $Y\text{.}$ To show that $f^{-1}O)$ is open in $X\text{,}$ we will show that $f^{-1}(O)$ is a neighborhood of each of its points. Let $a\in f^{-1}(O)\text{.}$ Then $f(a)\in O\text{.}$ Since $O$ is an open set, $O$ is a neighborhood of $f(a)\text{.}$ The fact that $f$ is continuous means that $f^{-1}(O)$ is a neighborhood of $a\text{.}$ So $f^{-1}(O)$ is a neighborhood of each of its points and $f^{-1}(O)$ is an open set.

The following theorem is the topological analog of Theorem 10.5. The proof is left for Exercise 4.