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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?

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    $\begingroup$ $p \in f^{-1}(V)$ and because $f^{-1}(V)$ is open, it contains an open ball around $p$. $\endgroup$
    – user295959
    Commented Jun 6, 2021 at 18:21

1 Answer 1

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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$$x\in f^{-1}\bigl(\{y\in Y\mid d_X(y,f(p))<\delta\}\bigr)\subset f^{-1}(V).$$

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