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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 $p_n\in f(V)$.

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  • $\begingroup$ @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? $\endgroup$ Apr 25, 2023 at 16:33
  • $\begingroup$ This second argument looks better and correct. $\endgroup$ Apr 27, 2023 at 2:52

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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.

You can construct the argument based on this.

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.

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  • $\begingroup$ 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. $\endgroup$ Apr 26, 2023 at 16:14
  • $\begingroup$ 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? $\endgroup$ Apr 26, 2023 at 17:17

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