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I want to show that if $X$ is first-countable and $x_n \to x$ implies $f(x_n) \to f(x)$ then $f: X \to Y$ is continuous.

That is for a neighborhood $V$ of $f(x)$ there is a neighborhood $U$ of $x$ such that $f(U) \subset V$.

So let $V$ be a neighborhood of $f(x)$. How to proceed from here?

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In your last sentence you are already assuming $f$ to be continuous... You cannot start like this. –  Olivier Bégassat Apr 24 '12 at 23:25
    
You need to prove the last sentence. In other words, you need to prove that if $V\subset Y$ is open, then $f^{-1}(V) \subset X$ is open. –  Nicholas Stull Apr 24 '12 at 23:26
    
@OlivierBégassat Right, thank you. I'll edit the question. –  Matt N. Apr 24 '12 at 23:29
    
It may help to first prove that $U\subset X$ is open if and only if whenever $x_n\rightarrow x\in U$, then $(x_n)$ is eventually in $U$. –  David Mitra Apr 24 '12 at 23:34

1 Answer 1

up vote 2 down vote accepted

Lemma: Let $X$ be first countable. A set $U\subset X$ is open if and only if whenever $x_n\rightarrow x\in U$, then $(x_n)$ is eventually in $U$ (that is, there is a positive integer $n$ so that $x_m\in U$ for each $m\ge n$).

Proof: As the forward implication is trivial, we only provide the proof for the reverse implication. Towards this end, suppose $U$ is not open. Let $x$ be a point in $U$ such that no nhood of $x$ is contained in $U$. Let $\{ U_i\mid i\in\Bbb N\}$ be a countable nhood base at $x$ such that $U_1\supset U_2\supset U_3\supset\cdots$ (replace $U_n$ with $\cap_{k=1}^n U_n$ if necessary). For each $i$, choose $x_i\in U_i$ such that $x_i\notin U$. Then $(x_i)$ converges to $x\in U$ but is not eventually in $U$.


Now to show that your statement is true:

Let $V$ be open in $Y$. Set $U=f^{-1}(V)$ and suppose $x\in U$.

Let $(x_n)$ be a sequence in $X$ converging to $x$. Then $(f(x_n))$ converges to $f(x)\in V$. Since $V$ is open the the sequence $(f(x_n))$ is eventually in $V$ by the Lemma. From this, it follows that the sequence $(x_n)$ is eventually in $U$. Thus, by the Lemma again, $U$ is open in $X$.

As $V$ was an arbitrary open set in $Y$, it follows that $f$ is continuous.

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Thank you. When you wrote "is eventually in $V$ by the lemma" you meant "is eventually in $V$ by definition of convergence", right? –  Matt N. Apr 25 '12 at 8:22

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