I'm trying to understand the proof of this theorem: $X$ is first countable and $f:X\rightarrow Y$ is a function. Then $f$ is continuous at $x\in X$ iff for every sequence $(x_n)$ which converges to $x$, $f(x_n)$ converges to $f(x)$.

Fix $x\in X$. To prove the "only if" statement, it suffices to show that $f(\overline{A})\subseteq\overline{f(A)}$ for each $A\subset X$ such that $x\in\overline{A}$.

Since $x\in\overline{A}$, by a previously proven lemma, there is a sequence $(x_n)$ in $A$ which converges to $x$. By assumption, the sequence $f(x_n)$ in $Y$ converges to $f(x)$. Since $f(x_n)\in f(A)$ for each $n\in\mathbb{N}$, we have $f(x)\in\overline{f(A)}$.

The last line of the proof seems to imply that if $X$ is first countable, then so is $f(X)$. But why is this true? It think it is only true if $f$ is continuous at $x$, but obviously we can't assume that.

  • $\begingroup$ Why do you say it implies that $f(X)$ is first countable? $\endgroup$ Sep 8 '17 at 4:22
  • $\begingroup$ I understood why. $f(X)$ does not have to be first countable. The result follows from an equivalent definition of $\overline{f(A)}$. $\endgroup$ Sep 8 '17 at 4:29
  • $\begingroup$ what do you mean by "first countable"? $\endgroup$
    – Mark Joshi
    Sep 8 '17 at 4:43
  • $\begingroup$ X is First countable if every x in X has a countable neighborhood basis. $\endgroup$ Sep 8 '17 at 5:34
  • $\begingroup$ The point is (see my answer) that sequence limits from a set are always in the closure of the set, in any space. It's the reverse that uses (but not implies!) first countability. $\endgroup$ Sep 8 '17 at 9:00

Firstly: $f[X]$ is not necessarily first countable, e.g. take $X = \mathbb{R}$ in the usual topology and let $\sim$ be the equivalence relation with classes $\{x\} , x \notin \mathbb{Z}$ and $\mathbb{Z}$ (We identify the integers to a point). Then the quotient space $Y = X/\sim$ in the quotient topology, induced by the standard map $q$ mapping $x$ to its class, is a continuous image (of $q$) which is not first countable at the class/point $[\mathbb{Z}]$.

Following your proof, which is quite correct: we assume $f$ preserves sequence limits. Then we want so show that for any $A \subseteq X$, $f[\overline{A}] \subseteq \overline{f[A]}$, which is indeed is one of the characterisations of continuity.

So let $f(x) \in f[\overline{A}]$, so $x \in \overline{A}$, which implies that there is a sequence $x_n$ from $A$ that converges to $x$ (this holds in first countable spaces! this is where we use it). Then $f(x_n) \to f(x)$ by the assumption on $f$, and as all $f(x_n) \in f[A]$, $f(x) \in \overline{f[A]}$ as required.

The last fact is true in all spaces: if $x_n \in B$ and $x_n \to x$ in the space, then $x \in \overline{B}$: let $O$ be any open neighbourhood of $x$. Then $O$ contains all $x_n$ for $n \ge N$ for some $N$. But in particular: $x_N \in O \cap B \neq \emptyset$, so $x \in \overline{B}$. But for the reverse (that being in the closure means that we have a sequence from the set that converges to it), uses (part of ) first countability.

Also, the direction $f$ continuous implies $f$ preserves all sequence limits, holds for all spaces $X,Y$ and $f$ between them. Only the sufficiency needs some assumption on $X$, but no assumptions on $Y$ (or $f[X]$).


Here is a continuous counter example to the question of the title.
Let Y be any space that is not first countable.
Let X be the points of Y with the discrete topology.
Let f be the the identity map from X to Y.

To show f is continuous iff f preserves sequence limits use
the fact that f is continuous iff for all A,
$f(cl A) \subseteq cl f(A)$ and the theorem K is closed iff for all x, (x in K iff some sequence within K converges to x).

cl = closure.

  • $\begingroup$ But $X$ is assumed to be first countable, so this is not a counterexample. $f$ is not continuous of course. $\endgroup$ Sep 8 '17 at 8:06
  • $\begingroup$ @HennoBrandsma. Why is the discrete space X not first countable? Is not every function from a discrete space continuous? It is Y that is not first countable. $\endgroup$ Sep 8 '17 at 9:03
  • $\begingroup$ Then the direction is wrong: $f$ goes from $X$ to $Y$. You choose $Y$ discrete, not $X$ , in your answer. $\endgroup$ Sep 8 '17 at 9:06
  • $\begingroup$ And if you choose $X$ discrete the statement holds as $f$ is continuous and $f$ also preserves convergent sequences (which are all eventually constant). $\endgroup$ Sep 8 '17 at 9:08

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