Q: Baby Rudin Theorem 4.8 ($f^{-1}(V)$ is open for every open set V iff $f$ is continuous) 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 $x\in f^{-1}(V)$ 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:


*

*"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 $f^{-1}(V)$

*"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 $x\in f^{-1}(V)$ 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 $p\in X$ 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 $d_Y(f(x),f(p))<\epsilon$. 

 A: 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 $f^{-1}(V)$)
A: *

*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 $B_{\delta}(p) \subset f^{-1}(V).$

*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.
