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In Munkres' Topology, Theorem 21.3 says

Let $f:X \to Y$. If the function, $f$ is continuous, then for every convergent series $x_n \to x$ in $X$, the sequence $f(x_n)$ converges to $f(x)$. The converse holds if $X$ is metrizable.

I don't understand what they mean by "the converse" here. It certainly can't mean "if $f(x_n)$ converges to $f(x)$ then $x_n \to x$" since this is patently false (e.g. if $f$ is constant or periodic). I think what it means is something like:

Let $y_n \to y$ where the $y_n$ and $y$ are in the image of $f$. Then, there exists points $x_n$ and $x$ in $X$ such that $x_n \to x$, $f(x_n) = y_n$, and $f(y) = y$.

If this is right, I don't understand the proof though. The book proves that $f(\overline{A}) \subset \overline{f(A)}$ for $A \subset X$ but I don't see how that implies my statement above.

Keen-ameteur
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roundsquare
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2 Answers2

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The converse is:

Let $f\colon X\longrightarrow Y$, where $X$ is a metrizable space. If the function $f$ is such that, for any sequence $(x_n)_{n\in\Bbb N}$ of elements of $X$ converging to some $x\in X$, you always have $\lim_{n\to\infty}f(x_n)=f(x)$, then $f$ continuous.

And this is true because if $A\subset X$ and $x\in\overline A$, then there is a sequence $(x_n)_{n\in\Bbb N}$ of elements of $A$ such that $\lim_{n\to\infty}x_n=x$. But then $\lim_{n\to\infty}f(x_n)=f(x)$ and therefore, since each $f(x_n)$ belongs to $f(A)$, $f(x)\in\overline{f(A)}$. And this proves that $f\left(\overline A\right)\subset\overline{f(A)}$. Since this takes place for each $A\subset X$, $f$ is continuous.

José Carlos Santos
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I think the converse means here that you can verify continuity by sequential continuity. Namely, if $f(x_n)\to f(x)$ for any sequence $x_n\to x$, then $x$ is continuous. Something like verifying continuity of a function using sequences.

If you look at this thread, you will see that $f$ is continuous if and only if $f(\overline{A})\subseteq \overline{f(A)}$ for any $A\subseteq X$.

Keen-ameteur
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