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Sequential continuity is equivalent to continuity in a first countable space $X$. Look at the quotient projection $g:X\to Y$ to the space of equivalence classes of an equivalence relation with the quotient topology and a map $f:Y\to Z$. I want to test if $f$ is continuous.

Can I do this by showing $$\lim_n\;f(g(x_n))=f(g(\lim_n\;x_n))\;?$$ Can I do this at least if $X$ is a metric space? Can I do this by showing $$\lim_n\;f(g(x_n))=f(\lim_n\;g(x_n))\;?$$

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Note that in non-Hausdorff spaces a sequence can have more than one limit. Therefore sometimes you can see something notations like $a \in \lim x_n$ instead of $a=\lim x_n$ and $f(\lim x_n) \subseteq \lim f(x_n)$ instead of $\lim f(x_n) = f(\lim x_n)$ in cases when you do not know that the space you work with is Hausdorff. – Martin Sleziak Nov 7 '11 at 22:53
up vote 2 down vote accepted

You certainly can’t do it in general if $X$ isn’t a sequential space, i.e., one whose structure is completely determined by its convergent sequences. $X$ is sequential iff it’s the quotient of a metric space, and the composition of two quotient maps is a quotient map, so if $X$ is sequential, $Y$ is also sequential, therefore $f$ is continuous iff $\lim\limits_n\;f(y_n) = f(\lim\limits_n\;y_n)$ for every convergent sequence $\langle y_n:n\in\omega\rangle$ in $Y$.

However, you can’t always pull this back to $X$.

Edit (24 April 2016): Take $X=[0,1]$, and let $Y$ be the quotient obtained by identifying $0$ and $1$ to a point $p$, $g$ being the quotient map. Suppose that $f:Y\to Z$, and you want to test the continuity of $f$. For $n\in\mathbb{Z}^+$ let

$$x_n=\begin{cases} \frac2{n+4},&\text{if }n\text{ is even}\\ \frac{n+1}{n+3},&\text{if }n\text{ is odd}\;, \end{cases}$$

so that

$$\langle x_n:n\in\Bbb Z^+\rangle=\left\langle\frac12,\frac13,\frac23,\frac14,\frac34,\frac15,\frac45,\ldots\right\rangle\;.$$

Then $\langle g(x_n):n\in\mathbb{Z}^+\rangle$ converges to $p$ in $Y$, so you need to test whether $\langle f(g(x_n)):n\in\mathbb{Z}^+\rangle$ converges to $f(p)$ in $Z$, but you can’t do this by asking whether $$\lim_n\;f(g(x_n))=f(g(\lim_n\;x_n))\;,$$ because $\langle x_n:n\in\mathbb{Z}^+\rangle$ isn’t convergent in $X$. Thus, the answer to your first question is no even if $X$ is metric.

(This replaces a flawed example with one that actually works, borrowed from this answer by Eric Wofsey.)

End edit.

The answer to your second question, however, is yes. Let $\langle y_n:n\in\omega\rangle$ be a convergent sequence in $Y$ with limit $y$. Then there are $x_n\in X$ such that $y_n = g(x_n)$ for $n\in\omega$, so checking that $$\lim_n\;f(y_n)=f(y)$$ is checking that $$\lim_n\;f(g(x_n))=f(\lim_n\;g(x_n))\;.$$ Note, though, that you have to check all sequences in $X$, not just convergent ones.

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I believe that if I understand this equality $\lim_n\;f(g(x_n))=f(g(\lim_n\;x_n))$ as equality of sets (i.e., every limit of $f(g(x_n))$ is of the form $f(g(y))$ for some limit $y$ of $x_n$ and vice-versa), then the claim works whenever $X$ is sequential. (I pointed out that $\lim x_n$ can mean a set, particularly for non-Hausdorff spaces, in my comment above.)\\I believe this is basically reformulation of what you wrote, but I wanted to mention this possible point of view. – Martin Sleziak Nov 8 '11 at 8:05
So I was wrong in my above comment - it seems that we have only inclusion, not equality. (See the proposition from Engliking I've edited into my answer.) – Martin Sleziak Nov 11 '11 at 11:13
@Martin: I’m sorry; I meant to mention those lemmas in response to your other comment and completely forgot to do so. – Brian M. Scott Nov 11 '11 at 11:26
@BrianM.Scott I don't see, why $x_n = n + 2^{-n}$ converges to $p$ in $Y$. Consider the set $V := \bigcup_{k \in \mathbb{N}} (k - 2^{-(k+1)}, k + 2^{k+1})$. Then $p = \mathbb{N} \subseteq V$ and $V$ is open in $\mathbb{R}$ and saturated w.r.t. $g$ (i.e. $g^{-1}(g(V)) = V$) which implies that $g(V)$ is open neighborhood of $p$ in $Y$. But $x_n$ is not contained in $g(V)$ for any $n$. – yadaddy Apr 24 at 7:23
@yadaddy: You are of course correct; I must have been asleep when I wrote that. I’ll think about it and either simply delete that part or find an actual example. – Brian M. Scott Apr 24 at 21:36

Sequential spaces are the spaces in which closed sets are precisely the sets which are closed under limits of sequences. (I.e., $V$ is closed if and only if for every convergent sequence $v_n$ of elements of $v$ every limit of this sequence belongs to $V$, too.)

  • Quotient of a sequential space is again sequential.

  • A map defined on a sequential space is continuous if and only if it is sequentially continuous.

  • Every first-countable (hence also every metric space) is sequential.

For a good introduction in sequential spaces, you can have a look at Other references from the wikipedia article might be useful too. There are also several interesting entries at this blog.

I'll include two results relevant to this question as given in Engelking's General Topology. Note that by $\lim x_{\sigma}$ he denotes the set of all limits of net (or sequence) $(x_\sigma)$. (This distinction is important only in non-Hausdorff case.)

Proposition 1.6.15: A mapping $f$ of a sequential space $X$ to a \tsp $Y$ is continuous if and only if $$f[{\lim x_i}] \subset \lim f(x_i)$$ for every sequence $(x_i)$ in the space $X$.

This is basically sequential variation of the "net version" for arbitrary topological spaces.

Proposition 1.6.6: A mapping $f$ of a topological space $X$ to a topological space $Y$ is continuous if and only if $$f[{\lim\limits_{\sigma\in\Sigma} x_\sigma}] \subset \lim\limits_{\sigma\in\Sigma} f(x_\sigma)$$ for every net $\{x_\sigma, \sigma\in\Sigma\}$ in the space $X$.

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