# From injective map to continuous map

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.

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## 1 Answer

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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thank you. Brain M.Scott – yaoxiao May 29 '12 at 12:28
@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. – user12014 May 30 '12 at 6:54
Please why $K_n$ is compact , (it is closed ok) but why compact ? – Vrouvrou Feb 23 '15 at 17:32
@yaoxiao ? can you answer me ?? – Vrouvrou Feb 23 '15 at 17:38
If a sequence is not eventually constant it doesn't necessarily consist of distinct points. – egreg Feb 23 '15 at 18:32