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:
- Is there some deeper insight, why we need a separation property for sequentialness but not for Fréchet-Urysohn?
- 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.
compactness + sequential => sequential compactness
still seems to need $T_1$. Or not? $\endgroup$