Does the preservation of sequential limits imply continuity in non-metrizable spaces? I have a question concerning the following theorem in Munkres's Topology:

Theorem 21.3. Let $f : X \rightarrow Y$. If the function $f$ is continuous, then for every convergent sequence $x_n \rightarrow x$ in $X$, the sequence $f(x_n)$ converges to $f(x)$. The converse holds if $X$ is metrizable.

My question is: Does the converse holds if $X$ not metrizable?
I have no idea about where to start, but I know some spaces which are not metrizable. E.g. $\mathbb{R}^\omega$ with the box topology, the topological vector space of all functions $f : \mathbb{R} \rightarrow \mathbb{R}$, the uncountable product of $\mathbb{R}$.
 A: The class of spaces for which "sequential continuity" always coincides with continuity is the sequential spaces.
Definition. A subset $A$ of a topological space $X$ is called sequentially closed if whenever $\langle x_n \rangle_{n \in \mathbb{N}}$ is a sequence in $A$ and $x_n \rightarrow_n x_\infty$, then $x_\infty \in A$.
Fact. For any topological space $X$, all closed subsets of $X$ are sequentially closed.
Definition. A topological space $X$ is called sequential if every sequentially closed set is closed.
Fact. All first countable spaces (and hence all metric spaces) are sequential.

Proposition. Suppose that $X$ is a sequential space, $Y$ is any space, and $f : X \to Y$ has the property that given any sequence $\langle x_n \rangle_{n \in \mathbb{N}}$ in $X$ with $x_n \rightarrow_n x_\infty$, then $f(x_n) \rightarrow_n f(x_\infty)$. Then $f$ is continuous.
Proof. If $f$ is not continuous, then there is a closed $E \subseteq Y$ such that $f^{-1} [ E ]$ is not closed in $X$. Since $X$ is sequential $f^{-1} [ E ]$ is not sequentially closed, and so pick a sequence $\langle x_n \rangle_{n \in \mathbb{N}}$ in $f^{-1} [ E ]$ such that $x_n \rightarrow_n x_\infty$ for some $x_\infty \in X \setminus f^{-1} [ E ]$. Note that $\langle f(x_n) \rangle_{n \in \mathbb{N}}$ is a sequence in $E$, and $f(x_\infty) \notin E$, and since $E$ is sequentially closed it cannot be that $f(x_n) \to_n f(x_\infty)$.

Proposition. If $X$ is not sequential, then there is a function $f : X \to S$ where $S = \{ 0,1 \}$ with the Sierpiński topology $\{ \emptyset , \{ 1 \} , \{ 0,1 \} \}$ such that


*

*$f$ is not continuous, and

*whenever $\langle x_n \rangle_{n \in \mathbb{N}}$ is a sequence in $X$ and $x_n \to_n x_\infty$, then $f(x_n) \to_n f(x_\infty)$.


Proof. If $X$ is not sequential, then there is a subset $A \subseteq X$ which is not closed, but is sequentially closed. Define a mapping $f : X \to S$ by: $$f(x) = \begin{cases}
0, &\text{if }x \in A \\
1, &\text{if }x \notin A.
\end{cases}$$ Note that $f$ is not continuous, as $f^{-1} [ \{ 0 \} ] = A$ is not closed, but $\{ 0 \} \subseteq S$ is closed.
Suppose that $\langle x_n \rangle_{n \in \mathbb{N}}$ is a sequence in $X$ and $x_n \to_n x_\infty$.


*

*If $\langle x_n \rangle_{n \in \mathbb{N}}$ is eventually not in $A$ (i.e., there is an $N$ such that $x_n \notin A$ for all $n \geq N$), it follows that $f(x_n) = 1$ for all $n \geq N$. Regardless of whether $f(x_\infty) = 0$ or $f(x_\infty) = 1$, we can show that $f(x_n) \to_n f(x_\infty)$.

*Otherwise there are infinitely many $n$ such that $x_n \in A$. By passing to the subsequence $\langle x_{n_i} \rangle_{i \in \mathbb{N}}$ consisting of these $n$ it follows that $x_\infty \in A$ (since the subsequence must also converge to $x_\infty$). Thus $f(x_\infty) = 0$, and since all sequences in $S$ converge to $0$ it follows that $f(x_n) \rightarrow_n f(x_\infty)$.



Here are some examples of non-sequential spaces.


*

*The co-countable topology on an uncountable set. (The only convergent sequences are the eventually constant sequences which converge only to their eventually constant value, so every subset is sequentially closed.)

*The Arens-Fort space. (The only convergent sequences are the eventually constant sequences, which converge only to their eventually constant value.)

*The Stone-Čech compactification of $\mathbb{N}$. (Again, the only convergent sequences are the eventually constant sequences, and only converge to their eventually constant value.)

*The ordinal space $\omega_1 + 1$. ($[0,\omega_1)$ is sequentially closed, but not closed.)

*Uncountable products of $\mathbb{R}$.

A: Let $Y=\Bbb{R}$ with the usual topology and $X=\Bbb{R}$ with the lower limit topology (which is not metrizable). Then take $f$ to be the identity map on $\Bbb{R}$. (This is continuous since the topology on $Y$ is finer than the topology on $X$.) Now if $x_n=1-\frac{1}{n}$, then $f(x_n)\rightarrow 1$ (in $Y$) but $\{x_n\}$ does not converge (in $X$).
