Relation between continuous maps and convergence of sequences

Question

I am studying metric spaces and I know that in a normed space $E$ a map $T:E \to E$ is contínuous if and only if $T(x_n) \to T(x)$ for every convergent sequence $x_n \to x$ in $E$. In my notes there is a comment stating that this result is not valid in a general topological space.

Does anyone know which are the minimal hypotheses on a topological space so that the previous statement is true?

Answer 1

The minimal condition on a space $X$ such that every function from $X$ is continuous if and only if it is sequentially continuous is that $X$ be sequential.

First countable spaces are sequential. Thus a function from a first-countable space is continuous if and only if it is sequentially continuous.

A space is called sequential if every sequentially open set is open. And a subset of a topological space is called sequentially open if every converging sequence with its limit in the subset is eventually in the subset.

Here's a proof of the equivalence:

• Let $X$ be sequential.

Every continuous function is trivially sequentially continuous. Now let $f:X\rightarrow Y$ be sequentially continuous. Let $U\subseteq Y$ be an open set. We need to show that $f^{-1}(U)$ is open. By the assumption that $X$ is sequential, it suffices to show that $f^{-1}(U)$ is sequentially open. Let $\{x_n\}$ be a converging sequence with limit $x$ such that $x\in f^{-1}(U)$. We need to show that $\{x_n\}$ is eventually in $f^{-1}(U)$. Since $f$ is sequentially continuous, we have that $\{f(x_n)\}$ converges to $f(x)$. Thus for every neighborhood $V$ of $f(x)$ (including $U$) there is an $N$ such that for every $n\geq N$ we have $\{f(x_n)\}_{n\geq N}\subseteq V$. In particular, there is an $N$ such that for all $n\geq N$, we have $f(x_n)\subseteq U$. Thus $$\{x_n\}_{n\geq N}\subseteq\{f^{-1}(f(x_n))\}_{n\geq N}\subseteq f^{-1}(U)$$ Thus $f^{-1}(U)$ is sequentially open and open. Hence $f$ is continuous.

• Now suppose $X$ is not sequential.

Then $X$ contains a sequentially open set $U$ that is not open. Let $Y=\{0,1\}$ and let $\mathcal T=\{\varnothing, \{0\}, Y\}$. Define the function $f:X\rightarrow Y$ by $$f(x)=\begin{cases}0 & \text{ if } x\in U\\1 &\text{ if } x\in X\setminus U\end{cases}$$ Then $f$ is sequentially continuous, since $f^{-1}(\{0\})$ is sequentially open, but $f$ is not continuous since $f^{-1}(\{1\})=X\setminus U$ is not closed in $X$. Thus there is a sequentially continuous function out of $X$ that is not continuous.

Answer 2

It's good enough to have $E$ be first countable, i.e. at each $x\in E$ we need a countable basis $\{U_i\}_{i\in\mathbb{N}}$ of open neighborhoods of $x$. The easiest proof uses that a function $f$ is continuous just if $f\left( \overline{A}\right)\subset \overline{f\left(A\right)}$ for every $A\subset E$. In a first countable space $\bar A$ is exactly the limit points of sequences in $A$.

Answer 3

You need countable basis of neighboroods for any points.

Indeed, suppose $E$ has countable basis $\{U_n^x\}$ of neighborhoods for any point $x$.

1) $T$ is sequence continuous implies $T$ is continuous.

Proof. Let $A$ be an open set. If $B=T^{-1}(A)$ is not open there is $x\in B$ which is not interior. Thus, for any $U_{k}^x$ containing $x$ there is $x_k\in U_k^x$, $x_k\notin B$. Clerarly $x_k\to x$ but $T(x_k)\notin A$ while $T(x)\in A$. So we cannot have $T(x_k)\to T(x)$.

2) $T$ is continuous implies $T$ sequence continuous. (for that no additional hypothesis is needed)

Proof. Let $x_n\to x$. For any open set $U$ containing $x$, $T^{-1}(U)$ is open, thus $x_n\in T^{-1}(U)$ for $n$ sufficiently large. Since this holds for any $U$, this implies that $T(x_n)\to x$.

3) on the other hand, if $E$ has no countable basis, sequences indexed by $\mathbb N$ say nothing about continuity

4) you can use spaces with "basis of type order $\alpha$" together with $\alpha$-sequences for continuity and the proof works the same.