# Rudin's Theorem 4.17 by a contradiction

I have recently started self-studying Rudin's well-known blue book by proving every theorem before looking at a given proof. Here is the statement of Theorem 4.17 in Rudin's Principles of Mathematical Analysis:

Suppose $$f$$ is a continuous 1-1 mapping of a compact metric space $$X$$ onto a metric space $$Y$$. Then the inverse mapping $$f^{−1}$$ defined on $$Y$$ by $$f^{−1}(f(x))=x$$ $$(x \in X)$$ is a continuous mapping of $$Y$$ onto $$X$$.

The proof in the textbook is pretty straightforward and understandable, but my proof, being a proof by a contradiction, is therefore different from Rudin's. I used Theorem 4.8:

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

Now, here is my attempt at proving 4.17:

Let $$V\subseteq X$$ be open and suppose $$f(V)$$ is not open. We then have a point $$q\in f(V)$$ that is not an interior point. For every $$\frac{1}{n}$$, choose $$p_n$$ such that $$d_Y(p_n,q)<\frac{1}{n}$$ and $$p_n \notin f(V)$$. Clearly, $$p_n\rightarrow q$$. For each such point, by bijective property, there is exactly one unique $$x_{p_n}\in X$$ with $$f(x_{p_n})=p_n$$. For the point $$q$$ we also have $$x_q\in X$$ with the same property. Clearly, $$x_q\in V$$, so we can have a neighborhood $$N$$ with $$N\subseteq V$$. Note that $$f(N)\subseteq f(V)$$. Now, by compactness, there is a subsequence $$x_{p_{n_i}}$$ that converges to some $$\alpha \in X$$. By continuity, $$\lim_{x \to x_{p_{n_i}}}f(x)=f(x_{p_{n_i}})=p_{n_i}.$$ Taking $$i\rightarrow \infty$$, we have $$\lim_{x \to \alpha}f(x)=f(\alpha)=q$$. Hence, by stated uniquiness, $$\alpha=x_q$$. But then, for large $$i$$, we must have $$p_{n_i}\in f(V)$$.

My questions:

• Are there any serious flaws, lack of rigour, or blatant errors? Is it even correct?
• Is there anything that must be improved to look more formal?

(EDIT) According to the lemma given by User1865345:

Let $$V\subseteq X$$ be open and suppose $$f(V)$$ is not open. We then have a point $$q\in f(V)$$ that is not an interior point. For every $$\frac{1}{n}$$, choose $$p_n$$ such that $$d_Y(p_n,q)<\frac{1}{n}$$ and $$p_n \notin f(V)$$. For each such point, by bijective property, there is exactly one unique $$x_{p_n}\in X$$ with $$f(x_{p_n})=p_n$$. For the point $$q$$ we also have $$x_q\in X$$ with the same property. Clearly, $$x_q\in V$$, so we can have a neighborhood $$N$$ with $$N\subseteq V$$. Note that $$f(N)\subseteq f(V)$$. Now, we have that $$f(x_{p_n})\to f(x_q)$$, since $$p_n\to q$$. By lemma (see answer @User1865345), it implies that $$x_{p_n}\to x_q$$, so that, for large $$n$$, we have $$p_n\in f(V)$$.

• @User1865345, the last step particularly concerns me; is it justified to take $i\to\infty$ and conclude what I did conclude, or is it some sort of notation abuse? Apr 25, 2023 at 16:33
• This second argument looks better and correct. Apr 27, 2023 at 2:52

One concern to me is you didn't explicitly make it clear how $$f(\alpha) = q.$$
You could have shown if $$f:K\to Y$$ where $$K\subseteq X$$ is a compact subset and $$f$$ is continuous and one-one on $$K,$$ then $$\langle x_{p_n}\rangle\in K, ~x_q\in K,$$ and $$f(x_{p_n})\to f(x_q)\implies x_{p_n}\to x_q.$$
This is easy to see: if $$f(x_n) \to f(x)$$ and $$x_n\nrightarrow x,$$ by compactness one can find a subsequence $$x_{n_i}\to y\in K~\wedge~y\ne x.$$ However, we know, since $$f$$ is continuous, $$f(x_{n_i}) \to f(y) \implies f(y) =f(x) \overset{\textrm{one-one}}{\implies} x= y,$$ leading to contradiction.
As for your comment, you could have argued for a given $$\varepsilon> 0,~\exists \delta> 0~ \wedge ~\exists N(\delta)\in \mathbb N~:\forall n_i\geq N(\delta)~d(x_{p_{n_i}},\alpha)<\delta/2\implies d'(f(x_{p_{n_i}}),f(\alpha)) <\varepsilon/2.$$ Similarly $$\exists N'(\delta)\in \mathbb N~:\forall n_i\geq N'(\delta)~d(x_{p_{n_i}},x)<\delta/2\implies d'(f(x_{p_{n_i}}),f(x)) <\varepsilon/2.$$ Define $$N'':=\sup\{N(\delta),N'(\delta)\}$$ and employ triangle inequality to deduce your desired result.
• Seems like this lemma was crucial to make the argument clear. See my edit, is the proof finished now? As for vagueness of $f(\alpha)=q$, I suppose the argument given by you in first-order symbolics clarifies it, doesn't it (since my proposed result was that after taking $i\to\infty$ the expression comes from the limit equality given prior, and your argument was exactly about it)? If not, please explain how I could show this explicitly, as I, probably, have no idea then. Apr 26, 2023 at 16:14
• Uh, no, it doesn't. Just shows how one could justify my last step about taking to infinity. But then this argument is useless, as one could just invoke continuity at $\alpha$. I suppose the equality can be shown as follows: as $p_{n_i}\to q$ and $f(x_{p_{n_i}})\to f(\alpha)$, given $\epsilon$ and invoking triangle inequality with $d(f(\alpha),q)$, and from the fact that $f(x_{p_{n_i}})=p_{n_i}$, one sees that $f(\alpha)=q$. Am I right here? Apr 26, 2023 at 17:17