# Converse of Rudin Theorem 4.8

I am not able to fully follow the converse of Theorem 4.8 in Baby Rudin. The theorem states:

A mapping $$f$$ of a metric space $$X$$ into a metric space $$Y$$ is continuous on $$X$$ if and only if $$f^{-1} (V)$$ is open in $$X$$ for every open set $$V$$ in $$Y$$.

Rudin's proof (of the converse) is:

Conversely, suppose $$f^{-1} (V)$$ is open in $$X$$ for every open set $$V$$ in $$Y$$. Fix $$p \in X$$ and $$\epsilon > 0$$, let $$V$$ be the set of all $$y \in Y$$ such that $$d_Y (y, f(p)) < \epsilon$$. Then $$V$$ is open; hence $$f^{-1} (V)$$ is open; hence there exists $$\delta > 0$$ such that $$x \in f^{-1} (V)$$ as soon as $$d_X (p,x) < \delta$$. But if $$x \in f^{-1} (V)$$, then $$f(x) \in V$$, so that $d_Y (f(x), f(p)) < \epsilon. The step that confuses me is "hence there exists $$\delta > 0$$ such that $$x \in f^{-1} (V)$$ as soon as $$d_X (p,x) < \delta$$. The sequencing of the proof confuses me mostly. $$f^{-1} (V)$$ is open, but it isn't necessarily a neighborhood (which, by Rudin's definition, is an open ball.) The way I would be inclined to write this is: $$V = N_{\epsilon} (f(p))$$ by definition. By continuity, for this $$\epsilon$$, there exists $$\delta > 0$$ so that $$d_X (p,x) < \delta$$ implies $$d_Y (f(x), f(p)) < \epsilon$$, so $$f(x) \in V$$, so $$x \in f^{-1} (V)$$. Rudin works this argument backwards, and I don't understand how. Can anyone shed some light on how Rudin did this? •$p \in f^{-1}(V)$and because$f^{-1}(V)$is open, it contains an open ball around$p\$. Commented Jun 6, 2021 at 18:21

The set $$\{y\in Y\mid d_X(y,f(p))<\delta\}$$ is an open set (since it is an open ball), and therefore $$f^{-1}\bigl(\{y\in Y\mid d_X(y,f(p))<\delta\}\bigr)$$ is an open set, to which $$p$$ belongs. Therefore, by the definition of open set, there is an open ball with center $$p$$ and radius $$\delta$$ contained in it. But what this means is that if $$d_X(x,p)<\delta$$, then$$x\in f^{-1}\bigl(\{y\in Y\mid d_X(y,f(p))<\delta\}\bigr)\subset f^{-1}(V).$$