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:

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

• By the way, in more general topological spaces, as you’ll probably see sooner rather than later, this property (the preimage of an open set is always open) is used as the definition of continuity. Mar 23, 2019 at 5:33

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

• Since the $\delta$-neighborhood of $p$ is only a subset of $f^{-1}(V)$, how do we know that the whole set $f^{-1}(V)$ is open? Mar 23, 2019 at 5:31
• Because $p$ is an arbitrary point in $f^{-1}(V)$. Mar 23, 2019 at 5:35
• "$\delta$-neighborhood of $p$ is a subset of $f^{-1}(V)$" shows that $p$ is an interior point of $f^{-1}(V)$. Since our $p$ is arbitrary taken from $f^{-1}(V)$, every point of $f^{-1}(V)$ is an interior point. Thus $f^{-1}(V)$ is open by definition. Mar 23, 2019 at 5:35
• Ohhhhhhh ok! Still goofing up my logic in these proofs - "You only showed it for one $p$!" - "Yeah? Want me to show it for every other $p$?" xD Mar 23, 2019 at 5:40
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.