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Let $X$ and $Y$ metric spaces, $f$ is an injective from $X$ to $Y$, and $f$ sets every compact set in $X$ to compact set in $Y$. How to prove $f$ is continuous map?

Any comments and advice will be appreciated.

1 Answers 1

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Since $X$ and $Y$ are metric spaces, it suffices to show that if $\langle x_n:n\in\Bbb N\rangle$ is a convergent sequence in $X$ with limit $x$, then $\langle f(x_n):n\in\Bbb N\rangle$ is a convergent sequence in $Y$ with limit $f(x)$; in words, f preserves convergent sequences.

Suppose that $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ in $X$. If there is an $n_0\in\Bbb N$ such that $x_n=x$ for all $n\ge n_0$, it’s trivially true that $\langle f(x_n):n\in\Bbb N\rangle\to f(x)$, so assume (by passing to a subsequence if necessary) that $\langle x_n:n\in\Bbb N\rangle$ is a sequence of distinct points. (Since $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ and is not eventually constant at $x$, it cannot have a constant infinite subsequence: for each $n\in\Bbb N$ there must be an $m>n$ such that $x_k\ne x_n$ whenever $k\ge m$.)

For each $n\in\Bbb N$ set $K_n=\{x\}\cup\{x_k:k\ge n\}$; each $K_n$ is compact and infinite. (Why?) By hypothesis, therefore, each $f[K_n]$ is compact.

For convenience let $y=f(x)$, and let $y_n=f(x_n)$ and $H_n=f[K_n]$ for $n\in\Bbb N$. By hypothesis each $H_n$ is compact and infinite, so each contains a limit point. Fix $n\in\Bbb N$. For each $k\ge n$, $Y\setminus H_{k+1}$ is an open nbhd of $y_k$ that contains only finitely many points of $H_n$ (why?), so $y_k$ can’t be a limit point of $H_n$. Thus, for each $n\in\Bbb N$ the only possible limit point of $H_n$ is $y$ itself. From here you should be able to prove without too much trouble that $\langle y_n:n\in\Bbb N\rangle\to y$ and hence that $f$ is continuous.

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    @yaoxiao If the answer was useful to you, you should accept it by clicking the green checkmark on the left. This lets the rest of the community know that the question has an answer deemed as good by the person who asked and also awards the person who answered the deserved reputation points.2012-05-30
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    Please why $K_n$ is compact , (it is closed ok) but why compact ?2015-02-23
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    @yaoxiao ? can you answer me ??2015-02-23
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    If a sequence is not eventually constant it doesn't necessarily consist of distinct points.2015-02-23
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    @egreg: No, but since it’s convergent, it has a subsequence that does. Back then I thought this fairly obvious, but I suppose that I should make it explicit.2015-02-23
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    @Vrouvrou: To see that $K_n$ is compact, suppose that $\mathscr{U}$ is an open cover of $K_n$. There is some $U_0\in\mathscr{U}$ that contains $x$. $U_0$ is an open nbhd of $x$, so there is an $m\in\Bbb N$ such that $x_k\in U_0$ for all $k\ge m$. Thus, $U_0$ contains every point of $K_n$ except possibly the points $x_k$ such that $n\le k. There are only finitely many of those points, so we need only finitely many members of $\mathscr{U}$ to cover them. Those finitely many, together with $U_0$, are a finite subcover.2015-02-23
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    @BrianM.Scott I don't think it's really necessary to assume $(x_n)$ consisting of distinct points.2015-02-23
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    @egreg: It’s not; it’s a convenience in picturing what’s going on. Specifically, it lets one see $K_n$ as a simple sequence, a copy of $\omega+1$. I’m quite prepared to believe that for some it just gets in the way; I know that for others it doesn’t.2015-02-23