1
$\begingroup$

In point-set topology class we proved that if a topological space $X$ is first countable, then for any function $f:X\rightarrow Y$, where $Y$ is some topological space, the condition $x_n\rightarrow x \Rightarrow f(x_n)\rightarrow f(x)$ implies $f$ is continuous.

It is known that $\mathbb{R}^{\omega}$ with box topology is not first countable. An exercise is to find an example $f:\mathbb{R}^{\omega}\rightarrow Y$, such that $x_n\rightarrow x \Rightarrow f(x_n)\rightarrow f(x)$ but $f$ is not continuous. I have been struggling to find an example. Any ideas?

$\endgroup$
2
$\begingroup$

I will sketch the construction and verification of one example. There may be others that are simpler, but this is the first that occurs to me, and filling in the many missing details is a good exercise.

Let $z$ be the zero sequence in $\Bbb R^\omega$, and let $\langle x_n:n\in\Bbb N\rangle$ be a sequence in $\Bbb R^\omega$. Suppose that for each $n\in\Bbb N$ there are $m(n),k_n\ge n$ such that $x_{m(n)}(k_n)\ne 0$. Then it is not hard to construct an open nbhd $U$ of $z$ such that $x_{m(n)}\notin U$ for each $n\in\Bbb N$, thereby showing that $\langle x_n:n\in\Bbb N\rangle$ does not converge to $z$. Thus, if $\langle x_n:n\in\Bbb N\rangle\to z$, there must be an $n_0\in\Bbb N$ such that $x_m(k)=0$ for all $m,k\ge n_0$. This extends easily to show that if $\langle x_n:n\in\Bbb N\rangle\to x$, then there is an $n_0\in\Bbb N$ such that $x_m(k)=x(k)$ for all $m,k\ge n_0$. This isn’t quite as straightforward as saying that the only convergent sequences are the ones that are eventually constant, but it’s similar, and for a suitable choice of $Y$ we can use it in much the same way.

Define an equivalence relation $\sim$ on $\Bbb R^\omega$ as follows:

$$x\sim y\text{ iff }\{n\in\Bbb N:x(n)\ne y(n)\}\text{ is finite.}$$ (Equivalently, $x\sim y$ iff there is an $n_0\in\Bbb N$ such that $x(k)=y(k)$ for all $k\ge n_0$.) For each $x\in\Bbb R^\omega$ let $[x]$ be the $\sim$-equivalence class of $x$. Let $Y=\Bbb R^\omega/\!\sim$ with the discrete topology, so that the points of $Y$ are the $\sim$-classes $[x]$ for $x\in\Bbb R^\omega$. Let

$$f:\Bbb R^\omega\to Y:x\mapsto[x]\,;$$

then $f$ is not continuous, but $\langle f(x_n):n\in\Bbb N\rangle\to f(x)$ in $Y$ whenever $\langle x_n:n\in\Bbb N\rangle\to x$ in $\Bbb R^\omega$.

$\endgroup$
4
  • $\begingroup$ The map $f$ is precisely the one that sends $x$ to its connected component. $\endgroup$ Mar 8 '21 at 7:00
  • $\begingroup$ @Henno: Yep; if anyone wants to see a proof, there’s one in this answer. $\endgroup$ Mar 8 '21 at 7:26
  • $\begingroup$ This is great. I can convince myself the argument holds. Thanks! $\endgroup$
    – Jun Xu
    Mar 8 '21 at 10:22
  • $\begingroup$ @JunXu: You’re welcome! $\endgroup$ Mar 8 '21 at 20:08

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.