Is there a countably compact sequential non-$T_2$ space that is not sequentially compact?

Question

Let $X$ be a topological space.

Definitions:

• $X$ is countably compact if every countable open cover of $X$ has a finite subcover or equivalently, every sequence in $X$ has a cluster point.
• $X$ is sequentially compact if every sequence in $X$ has a convergent subsequence
• $X$ is sequential if every sequentially closed set is closed.

It is known that if $X$ is countably compact + sequential + $T_2$ then $X$ is sequentially compact (see e.g. Engelking).

The proof goes like this: Let $x_n$ be a sequence in $X$. Since $X$ is countably compact $x_n$ has a cluster point $x \in X$. If $\{ n \mid x_n = x \}$ is infinite then we have a constant subsequence of $x_n$, thus convergent. So assume that $\{ n \mid x_n = x \}$ is finite such that there is some $n_0$ and $x_n \neq x$ for all $n \geq n_0$. Consider the set $A := \{ x_n \mid n \geq n_0 \} \setminus \{ x \}$. Then $A$ is not closed and since $X$ is sequential, $A$ is not sequentially closed. Thus, there is a sequence $y_k \in A$ and $y \in X \setminus A$ such that $y_k \to y$. Since $X$ is $T_2$ it follows that $y_k$ is not eventually constant since otherwise $y_k \to y_N \in A$ for some $N \in \mathbb{N}$ and $y_k \to y \in X \setminus A$ implies $y_N = y$ which is a contradiction. Thus, we have infinitely many $y_k$ in $A$ which can be finally used to construct a convergent subsequence of $x_n$.

There are also other properties $\varphi$ such that countable compactness + $\varphi$ imply sequential compactness. As an example, $\varphi$ can be taken to be first-countable or even Fréchet-Urysohn (cluster points of injective sequences $x_n$ are accumulation points of the corresponding sets $x(\mathbb{N})$, thus lying in the closure and thus being able to be approximated by a sequence in $x(\mathbb{N})$ which can be used to generate a convergent subsequence of $x_n$). There is no need for an additional separation property.

In my eyes, the Fréchet-Urysohn property is not "too far" away from the sequential property and thus it is a little bit "strange" that sequentialness needs an additional separation property. By "too far" I mean that typical spaces that are sequential but not Fréchet-Urysohn are a little bit pathological (e.g. Arens-Fort space).

Questions:

1. Is there some deeper insight, why we need a separation property for sequentialness but not for Fréchet-Urysohn?
2. Is the separation property really needed, i.e. is there some sequential space which is countably compact but not sequentially compact?

Remark: In fact, for the uniqueness of the sequential limit we can reduce the $T_2$ separation property to the $US$ separation property (i.e. $X$ is sequentially Hausdorff) which lies strictly between $T_1$ and $T_2$. This gives a hint, that $T_1$ should be not enough.

Answer 1

The argument in Engelking requires only $T_1$ separation: that ensures that finite sets are closed and hence that $\{y_k:k\in\Bbb N\}$ cannot be finite if $\langle y_k:k\in\Bbb N\rangle$ converges to $y$. Thus, we can assume that $\langle y_k:k\in\Bbb N\rangle$ is injective, and the rest of the argument goes through.

I’ll have to think more about the non-$T_1$ case.

Answer 2

Theorem: If $X$ is countably compact and sequential (without any additional separation property and thus not necessarily $T_1$) then $X$ is sequentially compact.

I found a proof in T. P. Kremsater, "Sequential Space Methods" (Master of Arts thesis). In the $T_1$ case the singleton sets $\{ x \}$ are closed. In the non-$T_1$ case consider instead the closure $\overline{\{ x \}}$ of the singleton. So, if $x_n$ is a sequence then in the $T_1$-case it is enough to consider the underlying set $\{ x_n \mid n \in \mathbb{N} \} = \bigcup_{n \in \mathbb{N}} \{ x_n \}$ whereas in the general case one should rather consider $\bigcup_{n \in \mathbb{N}} \overline{\{ x_n \}}$.

Here are the details for the proof:

Lemma 1: Let $X$ be a topological space and $x, y \in X$. Then the following are equivalent:

1. $x \in \overline{\{ y \}}$ (i.e. $x$ is smaller than $y$ in the specialization preorder)
2. for all $C \subseteq X$ closed: $y \in C \Rightarrow x \in C$
3. for all $U \subseteq X$ open: $x \in U \Rightarrow y \in U$.

In particular, if $x_n, x \in X$ with $x \in \overline{\{ x_n \}}$ for all $n$ then $\{ x_n \mid n \in \mathbb{N}\} \subseteq U$ for every open neighborhood $U$ of $x$ and thus $x_n \to x$.

The proof is clear.

Lemma 2: Let $X$ be a topological space and $x_n \in X$. If $x_n$ has no convergent subsequence then $\bigcup_{n \in \mathbb{N}} \overline{\{ x_n \}}$ is sequentially closed.

Proof: Assume that $A$ is not sequentially closed. Then there is a sequence $y_k \in A$ and $y \in X \setminus A$ such that $y_k \to y$. From $y \in A$ it follows that there is a sequence $n_k$ such that $y_k \in \overline{\{ x_{n_k} \}}$. We show that the sequence $n_k$ is bounded. Otherwise, there is an increasing sequence $k_l$ such that $n_{k_l}$ is increasing. Then $y_{k_l}$ is a subsequence of $y_k$ and $x_{n_{k_l}}$ a subsequence of $x_n$. Thus $y_{k_l} \to y$ and from $y_{k_l} \in \overline{\{ x_{n_{k_l}}\}}$ for each $l$ it follows by Lemma 1 that $x_{n_{k_l}} \to y$. (Indeed, for every open neighborhood $U$ of $y$ there exists $l_0$ such that $y_{k_l} \in U$ for all $l \geq l_0$ and thus $x_{n_{k_l}} \in U$ for all $l \geq l_0$.) Thus, $x_n$ has a convergent sequence which is a contradiction to the premise in the Lemma. Therefore, $n_k$ is bounded. It follows that there is $m$ such that $n_k = m$ infinitely often and thus $y_k$ is frequently in $\overline{\{ x_m \}}$. Thus, there exists a subsequence $y_{k_l}$ such that $y_{k_l} \in \overline{\{ x_m \}}$ for all $l$ and since $y_{k_l} \to y$ and and $\overline{\{ x_m \}}$ is closed it follows that $y \in \overline{\{ x_m \}} \subseteq A$. But $y \in X \setminus A$, contradiction. Thus, $A$ is sequentially closed.

Proof of Theorem: Let $x_n$ be a sequence in $X$ and assume that $x_n$ has no convergent subsequence. By Lemma 2 it follows that $A := \bigcup_{n \in \mathbb{N}} \overline{\{ x_n \}}$ is sequentially closed and since $X$ is sequential, $A$ is closed. Since $X$ is countably compact the closed subset $A$ is also countably compact. Therefore, $x_n$ has a cluster point $x \in A$. We show that $B := \{ n \mid x \in \overline{\{ x_n \}} \}$ is finite. Otherwise, if $B$ is infinite, there is an increasing sequence $n_k$ such that $x \in \overline{\{ x_{n_k}\}}$ for all $k$. Thus, $x_{n_k}$ is a subsequence of $x_n$ and by Lemma 1 it follows that $x_{n_k} \to x$. But this contradicts the assumption that $x_n$ has no convergent subsequence. It follows that there is $N$ such that $x \not\in \bigcup_{n \geq N} \overline{\{ x_n \}}$. Since the sequence $(x_n)_{n \in \mathbb{N}}$ has no convergent subsequence it follows that the sequence $(x_n)_{n \geq N}$ has also no convergent subsequence. By Lemma 2 it follows that $\bigcup_{n \geq N} \overline{\{ x_n \}}$ is sequentially closed and since $X$ is sequential, $A$ is closed. But $x$ is a cluster point of $(x_n)_{n \geq N}$ which implies that $x \in \overline{\{ x_n \mid n \geq N\}}$. Since $\bigcup_{n \geq N} \overline{\{ x_n \}}$ is closed it follows that $x \in \overline{\{ x_n \mid n \geq N \}} = \overline{ \bigcup_{n \geq N} \{ x_n \}} \subseteq \bigcup_{n \geq N} \overline{ \{ x_n \} }$, contradiction. Thus, the assumption that $x_n$ has no convergent subsequence does not hold which finally implies that $X$ is sequentially compact.

Answer 3

If you (as you seem to do) use the definition of countably compact as "every infinite set has a limit point", then $X= \mathbb{N} \times \{0,1\}$, where the first space is discrete and the second space has the indiscrete topology (so we double the points of the integers), is limit point compact trivially (if $(n,i)$ is in the set, then $(n,i')$ where $i' \neq i$, is a limit point of that set) but not sequentially compact (as $((n,0))_n$ has no convergent subsequence). The space $X$ is not countably compact in the definition using countable open covers, of course. So at least you need to specify what version you are using.

I also think this $X$ is a sequential space (sequentially closed sets are closed, I think), but correct me if I'm wrong. So this might be a candidate example as under 2, if you are using the limit point definition.