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I think I understand each line of this proof, but don't see how it actually proves the theorem.

4.8 Theorem $ $ A mapping f of a metric space X into a metric space Y is continuous on X if an only if $f^-1(V)$ is open in X for every open set V in Y

Proof $ $ Suppose $f$ is continuous on X and V is an open set in Y. We have to show that every point of $f^-1(V)$ is an interior point of $f^-1(V)$. So, suppose $p in X$ and $f(p) in V$. Since $V$ is open, there exists $epsilon>0$ such that $y in V$ if $d_Y(f(p),y)<epsilon$; and since $f$ is continuous at $p$, there exists $delta>0$ such that $d_Y(f(x),f(p))<epsilon$ if $d_X(x,p)<delta$. Thus $xin f^-1(V)$ as soon as $d_x(x,p)<delta$.

That's the first part (if $V$ is open and $f$ continuous then $f^-1(V)$ is open). I have two problems with this:

1. "Since $V$ is open, there exists $epsilon>0$ such that $y in V$ if $d_Y(f(p),y)<epsilon$" - seems like we're only considering a small subset of $V$, so I don't see how we can conclude anything about the whole $f^-1(V)$
   


2. "since $f$ is continuous at $p$, there exists $delta>0$ such that $d_Y(f(x),f(p))<epsilon$ if $d_X(x,p)<delta$. Thus $xin f^-1(V)$ as soon as $d_x(x,p)<delta$." - this just says that for every $x$ in the $delta$-neighborhood of $p$, $f(x)$ is in the $epsilon$-neighborhood of $f(p)$. But there can still be other points $y$ in $V$ (even other points $y$ in the $epsilon$-neighborhood of $f(p)$) such that $f^-1(y)$ is not in the $delta$-neighborhood of $p$.

So I don't see how this shows that $f^-1(V)$ is open.

The second part of the proof is the converse (if $f^-1(V)$ is open for every open $V$ then $f$ is continuous). I include it here for completeness.

Conversely, suppose $f^-1(V)$ is open in $X$ for every open set $V$ in $Y$. Fix $pin X$ and $epsilon>0$, let $V$ be the set of all $yin Y$ 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 $xin f^-1(V)$ as soon as $d_X(p,x)<delta$. But if $xin f^-1(V)$, then $f(x)in V$, so that $d_Y(f(x),f(p))<epsilon$.

In the first problem, the purpose of the statement about $V$ is to give us a useful $epsilon$. We'll use the $epsilon$ to construct our $delta$-neighborhood of $p$.

About the second problem, what you concern about is
$$d_Y(f(x),f(p))<epsilon nRightarrow d_X(x,p) <delta.$$
But we only need a $delta>0$ such that
$$d_Y(f(x),f(p))<epsilon Leftarrow d_X(x,p) <delta. (*)$$
Such $delta$ exists since $f$ is continuous (as Rudin states).

Then for every $x$ in $delta$-neighborhood of $p$, $f(x)$ is in $epsilon$-neighborhood of $f(p)$ by $(*)$. This shows that $f(x) in V$ by our choice of $epsilon$. Then our proof is complete. ($delta$-neighborhood of $p$ is a subset of $f^-1(V)$)

1. The point is that to show $f^-1(V)$ is open means there is a ball of finite radius contained in it. Since $B_epsilon(f(p)) subset V$, by continuity of $f$, there exists a number $delta > 0$ such that $B_delta(p) subset f^-1(V).$




2. I think you mean $y in V - B_epsilon(f(p))$, if it is outside, nothing is guaranteed (by continuity), so yes it could be missed by the $delta-$ball in the domain for that particular $epsilon$ you chose.