First countability and convergence

I am trying to prove the following statement:

Let $X$ and $Y$ be topological spaces. If $X$ is first countable and $f:X \rightarrow Y$ is a map such that $a_n \rightarrow a$ in $X$ implies $f(a_n)\rightarrow a$ in $Y$, then $f$ is continuous.

I am missing something critical, as my proof, which I include below, does not use the fact that $X$ is first countable. What is it that I am missing?

Flawed Proof:

Assume $f$ is such a map. Let $U$ be an open set in $Y$. To prove $f$ is continuous, we must show $f^{-1}(U)$ is an open set in $X$. Let $a_n$ be a sequence converging to $a \in f^{-1}(U)$. By hypothesis, this implies $f(a_n)$ converges to $f(a) \in U$. So, there exists some $N \in \mathbb{N}$ such that $f(a_i) \in U$ for all $i \geq N$. This implies $a_i \in f^{-1}(U)$ for all $i \geq N$. This implies $f^{-1}(U)$ is open since every sequence converging to a pont in $f^{-1}(U)$ is eventually contained in $f^{-1}(U)$.

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In the statement of the theorem you want $f(a_n)\to f(a)$, not $f(a_n)\to a$. – Brian M. Scott Feb 29 '12 at 14:09
What does your last argument mean? – Ilya Feb 29 '12 at 14:10
The characterization of openness in the last sentence (which should read "every sequence $(a_n)$  converging to an $a\in f^{-1}(U)$..." uses it. – David Mitra Feb 29 '12 at 14:11
In this particular case, I find that proof by contradiction is easier. Try assuming on the contrary and everything else should follow easily. – Stuck_pls_help Feb 29 '12 at 14:16

2 Answers

You are using the following fact:

Characterization of openness in first countable spaces:

In a first countable space, a set $U$ is open if and only if every sequence that converges to a point in $U$ is eventually contained in $U$.

You appeal to this in the last sentence of your proof; so this is where you are using first countability.

As for why the characterization of openness above relies on first countability, let's look at it's proof.

One way to prove the reverse implication of the Characterization of openness theorem (which is what you use in your proof) is to use the Characterization of closedness in first countable spaces:

Characterization of closedness in first countable spaces:

If $X$ is first countable and $A\subset X$, then $x\in\overline A$ if and only if there is a sequence $(x_n)$ contained in $A$ which converges to $x$.

The proof of the forward implication of Characterization of closedness theorem uses first countability in an essential way: If $x\in \overline A$, pick a countable decreasing nhood base $\{U_n\}$ at $x$. Then choosing $x_n\in U_n\cap A$ provides the required sequence (note that since $x\in \overline A$ the sets $U_n\cap A$ are nonempty).

Now, back to the prove of the reverse implication of the Characterization of openness theorem. We proceed with the proof of the contrapositive:

Let $A$ be a subset of $X$ that is not open. Then there is a point $x$ in $A$ that is also in the closure of $A^C$. From the Characterization of closedness theorem, there is a sequence in $A^C$ that converges to $x$. Such a sequence cannot be eventually contained in $A$. (One could also argue directly here: every nhood of $x$ contains points of $A^C$. So, select a decreasing countable nhood base at $x$ from which to find a sequence $(x_n)\subset A^C$ that converges to $x$.)

Note that Brian M. Scott's answer shows that first countability is necessary for the reverse implication in the Characterization of openness theorem.

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The problem is with your last assertion: the fact that every sequence converging to a point of $f^{-1}[U]$ is eventually inside $f^{-1}[U]$ does not in general imply that $f^{-1}[U]$ is open. In the ordinal space $[0,\omega_1]$, for instance, with the order topology, the set $A=\{\omega_1\}$ has the property that every sequence converging to $\omega_1$ is eventually inside $A$ (because every such sequence is eventually constant), but $A$ isn’t open. First countability is used to draw that final inference.

However, that’s doing it the hard way. Try showing instead that if $f$ isn’t continuous, there must be a point $a\in X$ and a sequence $\langle a_n:n\in\mathbb{N}\rangle$ in $X$ converging to $a$ such that $\langle f(a_n):n\in\mathbb{N}\rangle$ does not converge to $f(a)$ in $Y$. It should be much easier to see where to use first countability of $X$ in this approach.

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Upon looking at the proof a second time, I saw first countability was the condition needed to guarantee the final assertion. I used that condition because I had previously proven it. So you think proving the statement by contradiction is more illustrative in terms of showing the need for first countability? I will give it a go. Thank you. – Holdsworth88 Feb 29 '12 at 14:30
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