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Is it true that:

For a map $f:X\rightarrow Y$, between two topological spaces. If the image of every convergent sequence in $X$ is also convergent in $Y$. Then $f$ is continuous.

If it is true, how to prove it? Or if it is false, what is the counter-example? I guess it is false, because it is usually insufficient to characterize topological space with sequences. But I can't construct a counter-example. So I ask for help here.


Thanks for all the answers. Using nets or filters to characterize convergence seems to be a big topic such that I will spend some more time to digest. Before that, I seem to find an easy counter-example by myself.

Let $X=\{\{a\},\{a,b\},\emptyset\}$, every sequence in $X$ converges. The function $f$ from $X$ to $Y=\{\{f(a),f(b)\},\{f(b)\},\emptyset\}$. Then the image of every convergent sequence in $X$ is convergent in $Y$ but $f^{-1}(\{f(b)\})=\{b\}$ is not open in $X$, so $f$ is not continuous.

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    In certain kinds of spaces these things are the same; in general, you want to look at nets. See [this discussion](http://en.wikipedia.org/wiki/Continuous_function#Sequences_and_nets).2011-07-23
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    @pluskid: Note that your counterexample does not work: The constant sequence $x_n=a$ converges to *b* (since every neighborhood of *b* contains *a*) but the constant sequence $f(x_n)=f(a)$ does not converge to $f(b)$ in *Y* (You are working with non-Hausdorff spaces - a sequence can have more than one limit.)2011-07-23
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    As a side note: If you are trying to find a counterexample, it has to be infinite. Finite => first countable => Frechet-Urysohn => sequential.2011-07-23
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    @Martin: The hypothesis is only that for every convergent sequence $x_n$ in $X$, $f(x_n)$ is a convergent sequence in $Y$. pluskid's counterexample *does* have this property, trivially, because every sequence in $Y$ converges. We are not assuming anything about what $f(x_n)$ converges *to*, and so this is not enough to force continuity.2011-07-23
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    @Nate: You are right. I thought that we want $x_n\to x$ $\Rightarrow$ $f(x_n)\to f(x)$, since this is what is usually required. (This is what works for nets.)2011-07-23
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    @pluskid: Sorry if I confused you by claiming that your counterexample is not correct. See Nate's comment above. It seems that I misunderstood your question. (But I guess some of the answerers understood the question in the same way I did, judging by the answers.)2011-07-24
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    Similar question for nets: http://math.stackexchange.com/questions/360419/f-brings-convergent-nets-to-convergent-nets-is-it-continuous2014-06-17

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