# Relative compactness of metric space

I know that in a metric space $X$ compactness, countable compactness and sequential compactness of a subspace $X'$ are equivalent using the definition of countable compactness as every infinite subset of $X'$ has an accumulation point in $X'$ and of sequential compactness as every sequence in $X'$ has a subsequence converging to a point in $X'$.

If we define relative countable compactness as every infinite subset of $X'$ has an accumulation point in $X$ and of relative sequential compactness as every sequence in $X'$ has a subsequence converging to a point in $X$, do I correctly understand if I desume, in a straightforward way by taking into account the properties of closed sets in a metric space, that in a metric space $X$ relative compactness, countable relative compactness and sequential relative compactness of a subspace $X'$ are equivalent?

$\infty$ thanks!

I already showed in this answer that $X'$ relatively compact implies $X'$ relatively countably compact. Also, if $X'$ is relatively compact, the $\overline{X'}$ is compact, and thus sequentially compact (this holds in particular in metric spaces, but also more broadly). So any sequence from $X'$ has a convergent subsequence with limit in $\overline{X'}$, so $X'$ is then relatively sequentially compact as well.

In any space, $X'$ relatively sequentially compact implies $X'$ relatively countably compact: any infinite subset $A$ of $X'$ contains some sequence with all different elements, which has a convergent subsequence to some $x \in X$, and this $x$ is an accumulation point of $A$.

If $X'$ is relatively countably compact, and $X$ is metric (first countable and $T_1$ will already do), let $(x_n)$ be a sequence from $X'$. If $A = \{x_n: n \in \mathbb{N}\}$ is finite, some value occurs an infinite number of times, and yields a convergent subsequence. So assume $A$ is infinite, so it has an accumulation point $p \in X$. Because $X$ is $T_1$, this means that every neighbourhood of $p$ intersects $A$ in infinitely many points. Pick $x_{n_1}$ in $B(p, 1)$, $x_{n_2}$ with $n_2 > n_1$ in $B(p, \frac{1}{2})$, and so on, by recursion. This defines a convergent subsequence of $(x_n)$ that converges to $p$. So $X'$ is relatively countably compact.

If $X'$ is relatively countably compact, and $X$ is metric, then we do get that $X'$ is relatively compact (this is due to Hausdorff, IIRC). I cannot reconstruct a proof right away, but there is one here, e.g. (but this involves Cauchy filters etc.)

It turns out that these notions are also equivalent in some topological vector spaces (spaces of the form $C_p(X)$ where $X$ is compact (Grothendieck), and weak topologicals on normed vector spaces (Eberlein-Smulian)), but these are a bit more involved, I think.

• I heartily thank you!!! So doesn't countable compactness imply sequential compactness in general? For weak* topology? I wonder because my book seemed to me to mean that in this issue: math.stackexchange.com/questions/929630/… – Self-teaching worker Sep 13 '14 at 23:03
• No, there are even compact (and thus countably compact) spaces that are not sequentially compact, like $[0,1]^\mathbb{R}$ in the product topology, or the Čech-Stone compactification of the integers ($\beta\mathbb{N}$). So there is something to prove! – Henno Brandsma Sep 14 '14 at 11:35
• ...and the text calls it an "immediate consequence"...! $\aleph_1$ thanks! – Self-teaching worker Sep 14 '14 at 12:19
• It could be equivalent for weak*, but it's not equivalent in general. My functional analysis is somewhat rusty.. – Henno Brandsma Sep 14 '14 at 13:05
• In math.stackexchange.com/a/27439/4280 example (i): the unit ball in $\ell_\infty^\ast$ is weak*-compact but not weak*-sequentially compact. In the other answer there it is shown that $X$ separable will make it true, as then this unit ball is metrisable, and sequential compactness and compactness are equivalent again. – Henno Brandsma Sep 14 '14 at 13:11