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It's been a few years since I studied point-set topology, and I'm a bit rusty on the basics. Would appreciate help with the following question.

Suppose $f:X\rightarrow Y$ is a map between two topological spaces, and I know that for any sequence $x_n\rightarrow x$ in $X$, there is a subsequence $x_{n_k}$ such that $f(x_{n_k})\rightarrow f(x)$. Does it follow that $x_n\rightarrow x\Rightarrow f(x_n)\rightarrow f(x)$? In the case when $X$ is first countable, the first condition is enough to imply that $f$ is continuous (I believe), and so the second condition must hold. I suspect this isn't true for general $X$, but I can't come up with a counterexample.

Thanks.

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What do you mean $x\to x_n$? Did you mean $x_n\to x$? –  Thomas Andrews Jul 23 '12 at 16:42
    
Yes, I did. Fixed. –  user17945 Jul 23 '12 at 17:29
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up vote 7 down vote accepted

If $x_n\to x$ and $f(x_n)$ doesn't converge to $f(x)$, take $V$ a neighborhood of $f(x)$ such that for infinitely many $k$, $f(x_k)\notin V$. We can write this infinitely many as a subsequence $x_{n_k}$. We have $x_{n_k}\to x$, but we can't extract a subsequence of $\{f(x_{n_k})\}$ which is convergent to $f(x)$.

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Wow, I really am rusty :) This is so straightforward. Thanks a lot for clearing up my confusion. –  user17945 Jul 23 '12 at 16:28
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