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

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.

• I don't have Engelking at hand, but doesn't he define countably compact as Hausdorff plus ... like he does with compactness and paracompactness? – Henno Brandsma Feb 16 '16 at 21:43
• Also, in the absence of any separation axioms, do you use limit point compactness, or the cover definition? – Henno Brandsma Feb 16 '16 at 21:50
• @HennoBrandsma Engelking defines countable compactness for Hausdorff spaces by the cover property, i.e. every countable open cover has a finite subcover. A general topological space $X$ (not necessarily $T_2$) is countably compact $\Leftrightarrow$ every sequence in $X$ has a cluster point $\Leftrightarrow$ every infinite subset of $X$ has an $\omega$-accumulation point. $X$ is limit point compact $\Leftrightarrow$ every infinite subset has an accumulation point. Thus, $X$ countably compact $\Rightarrow$ limit point compact and the converse holds in $T_1$ spaces. – yada Feb 17 '16 at 9:30
• @yadaddy But why do you call this property (every sequence has a cluster point) "countable compactness"? The cover definition is much more natural and is also used for other cardinals. It also always follows from compactness. The cluster point property that you use clearly implies the countable compactness but for the reverse $T_1$ seems to be needed (but maybe not). So, e.g., compactness + sequential => sequential compactness still seems to need $T_1$. Or not? – ald.li Mar 13 '18 at 17:04
• @ald.li In [Steen, Seebach, "Counterexamples in Topology", Section 3] you can find four properties (including the above mentioned countable covering property and the cluster point property) that are all equivalent (at least in ZFC) for a general topological space - no separation axioms are needed. Also, the implication "countable compactness + sequential $\Rightarrow$ sequential compactness" does not need any separation properties as well (see the proof in my answer below). Hence, the implication "compact + sequential $\Rightarrow$ sequentially compact" holds for all topological spaces. – yada Mar 14 '18 at 7:26

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.

• I have added a proof in an answer for the non-$T_1$ case. – yada Apr 8 '16 at 7:07
• @yadaddy: Thanks for letting me know; I’ll take a look tomorrow when I’m awake enough to appreciate it. – Brian M. Scott Apr 8 '16 at 7:31

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.

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.

• Your space is sequential. If $A \subseteq X$ is sequentially closed, then given any $(n,i) \in A$ as the constant sequence $k \mapsto (n,i)$ converges to $(n,1-i)$ it follows that $(n,1-i) \in A$. Thus $A = B \times \{0,1\}$ for some $B \subseteq \mathbb N$, so $A$ is closed. – fəˈnɛtɪk Feb 17 '16 at 7:29
• I refer to $X$ as countably compact if "every sequence has a cluster point" which in non-$T_1$-spaces implies that "every infinite set has a limit point" (i.e. limit point compactness) but is not necessarily equivalent. – yada Feb 17 '16 at 9:34