1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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
2
$\begingroup$

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 $f[X]$).

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.