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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​. –  Dylan Moreland Jul 23 '11 at 5:03
    
@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.) –  Martin Sleziak Jul 23 '11 at 7:21
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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. –  Martin Sleziak Jul 23 '11 at 7:30
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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. –  Nate Eldredge Jul 23 '11 at 13:46
    
@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.) –  Martin Sleziak Jul 23 '11 at 13:49

3 Answers 3

Construct a space which has a point which is not the limit of any sequence of points different from it, but which can be reached by a net (in other words, which is not isolated)

For example, let $\Omega$ be the smallest uncountable ordinal, let $X=\Omega+1$, and put the order topology on $X$. The biggest element $p$ in $X$ is such a point. Let $f:X\to X$ coincide with the identity map on $\Omega$ but with $f(p)=0$.

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The characterization of continuous functions in terms of preservation of limits of sequences works well in a large class of topological spaces, but not all spaces. Better than restricting the class of spaces under consideration is to replace sequences with something better equipped to deal with the general case, namely nets or filters.

For this topic I recommend my notes on convergence in topological spaces. (Often I recommend my own notes just because I know what's in them and that they are available on the internet. This is an instance where I wouldn't really know where else to point to, as the standard texts all seem to tell different portions of the full story.)

Some specifics:

Proposition 2.3 shows that if $f: X \rightarrow Y$ is a map between topological spaces with $X$ first countable, then $f$ is continuous iff it preserves all limits of sequences.

In $\S 2.2$ I mention that the previous fact holds more generally when $X$ is sequential, which is the class of spaces studied in that section.

[This leaves open the question of giving a clean necessary and sufficient condition on a space $X$ such that for all spaces $Y$, a map $f: X \rightarrow Y$ is continuous iff all limits of sequences are preserved. I haven't really thought about this, but it would be interesting if there were a nice answer.]

In $\S 3$ I discuss nets. Proposition 3.2 shows that a map $f: X \rightarrow Y$ is continuous iff it preserves all limits of nets in $X$.

In $\S 5$ I discuss filters. Proposition 5.14 gives the analogous characterization of continuity in terms of preservation of convergent filters (actually "prefilters", which are more often called filter bases, but it is easy to see that this result remains true with all instances of "pre" omitted).

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I think that being sequential is also necessary condition. For a space $(X,\tau)$ that is not sequential, it suffices to take $id:(X,\tau)\to(X,\tau')$, where $\tau'$ denotes the sequential coreflection (=closed subsets are precisely the sequentially closed subsets of the original topology). Since $X$ has sequentially closed subset that is not closed, this map is not continuous.\\I hope I did not miss something. –  Martin Sleziak Jul 23 '11 at 6:18
    
I should have also written in my previous comment that $id: (X,\tau)\to(X,\tau')$ preserves convergent sequences. Hopefully, it was clear from the context. (Since I wanted to show that there is a map with domain $(X,\tau)$ that is sequentially continuous but not continuous.) –  Martin Sleziak Jul 23 '11 at 9:33

EDIT: My post is about the spaces fulfilling the condition: If $x_n$ converges to $x$, then $f(x_n)$ converges to $f(x)$.

It seems, the you were asking about weaker condition: If $x_n$ is convergent, then $f(x_n)$ is convergent (as pointed out by Nate Eldredge).

I'll leave my answer, since it might be interesting for you anyway. (Any counterexample to the stronger condition is a counterexample to the weaker condition as well.)


This is true if $X$ is a sequential space. This paper gives a good introduction into the topic. Example 3.6 in this paper is an example of space that is not sequential. You can also find some useful information in this blog and its continuation. This question is also related to sequential spaces.

Your claim is true for arbitrary topological spaces if you replace sequences with nets.

There are many examples of spaces that are not sequential. Any non-discrete topological space with no non-trivial convergent sequences will do, such as cocountable topology on an uncountable set (see wikipedia article) or Stone-Čech compactification of countable discrete space (see this question).

For a space $X$ where every convergent sequence is eventually constant, you can take a discrete topological space $Y$ having at least 2 points. Then every function $f:X\to Y$ preserves convergence of sequences. But all such functions are continuous only if $X$ is discrete.

I will give the following counterexample (again, in this space all convergent sequences are eventually constant):

Counterexample

In this picture every arrow represents a convergent sequence (i.e. a topological space on a countable set, that has unique accumulation point and the neighborhoods of this point are precisely the cofinite sets; e.g. $\{0\}\cup\{1/n; n\in\mathbb N\}$ as the subspace of real line has this topology). If we make a quotient of a topological sum of such spaces (in the way given in this picture), we get a sequential space. The picture shows a subspace of a sequential space that is not sequential. (There is only one point that is not isolated. No sequence of isolated points converges to this point. Showing this is basically an exercise in working with quotient topology.)

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