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(Exercise 7.5.7 Introduction to Real Analysis by Jiri Lebl): A continuous function $f: X \to Y$ for metric space $(X, d_X)$ and $(Y, d_Y)$ is said to be proper if for every compact set $K \subset Y$, the set $f^{-1}(K)$ is compact. Suppose a continuous $f: (0,1) \to (0,1)$ is proper and $\{x_n\}$ is a sequence in $(0,1)$ that converges to $0$. Show that $\{f(x_n)\}$ has no subsequence that converges in $(0,1)$.

I know that if $f: X \to Y$ is continuous and $K \subset X$ is compact, then $f(K)$ is compact. I think that $x_n$ converging to $0$ implies that this sequence is defined in an open interval, and this leads to $\{f(x_n)\}$ having no subsequence in some way (?), but I am not sure. I appreciate if you give some help.

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2 Answers 2

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If there is a convergent subsequence $(f(x_{n_i}))$ with limit $l$ in $(0,1)$ the $K=\{l, f(x_{n_1}),f(x_{n_2})...\}$ is a compact set in $(0,1)$. By hypothesis the sequence $(x_{n_k})$ in the compact set $f^{-1}(K)$ must have a subsequence which converges in $(0,1)$. This is false because $x_n \to 0$.

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Suppose some $p \in (0,1)$ exists such that $(f(x_{n_k}))_k \to p$, so striving for a contradiction. Then $K=\{f(x_{n_k}): k \in \Bbb N\} \cup \{p\}$ is compact (standard argument) and so $(x_{n_k})_k$, which lies in the compact (!) set $f^{-1}[K]$, must have a convergent subsequence too. But that cannot happen as $0 \notin f^{-1}[K]$ and it's the only candidate for its limit by unicity of limits in $\Bbb R$.

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