# 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 $\mathbb{R}$.

• No, the converse does not hold in general. You can find in-depth discussions of this point under the topic of "uniform spaces". The concepts of "nets" and "uniform spaces" were invented (in part) in order to generalize the concepts of "sequences" and "metric spaces" in a manner which allows the converse to hold in a broader setting. Feb 17 '17 at 14:44
• You can have a look at this question and other posts linked there. Mar 28 '17 at 10:49

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

1. $f$ is not continuous, and
2. 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 $\mathbb{R}$.

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 $X$).