# What kind of a property implies (sequentially compact $\iff$ compact)?

It is well known that the topology of metric spaces and topological notions in metric spaces are pretty much entirely determined by sequences. For example one has:

1. The closure of a set is the same as the limit points of sequences in the set.
2. A map whose domain is our space is continuous iff it is sequentially continuous.
3. A subset is compact iff it is sequentially compact.

In first countable spaces or, more generally, sequential spaces the first two statements are still valid. The third is not necessarily true, for example the long ray is first countable and sequentially compact but certainly not compact.

In what kind of spaces is property 3 valid?

This question is a bit vague. For example property 1 is valid in sequential spaces, but the definition of sequential is exactly that property 1 holds. A prettier answer, if the question would have been about property 1, would be first countable spaces. The definition of first countable is not artificial in this context and covers a very large class of spaces.

If you have any further examples, it would be nice to know:

Are there other sequential notions that are equivalent to a topological notion in metric spaces but not necessarily equivalent in first countable spaces?

• One class of spaces in which a subset is compact if and only if it is sequentially compact are Banach spaces in their weak topology. – Daniel Fischer Jun 12 '17 at 14:58
• Thats pretty cool, especially since the weak topology on a Banach space is never sequential (except in finite dimensional cases), so point 3. is completely different from the other two, and not a stronger condition! – s.harp Jun 12 '17 at 15:02
• @DanielFischer even for locally convex topological vector spaces that have a weaker metric vector space topology. IIRC $C_p(X)$ spaces for $X$ compact Hausdorff are among those as well. – Henno Brandsma Jun 12 '17 at 20:43
• I think in first countable Lindelöf spaces 3. holds. But as @DanielFischer already implies this is not necessary. – Henno Brandsma Jun 12 '17 at 20:45
• @HennoBrandsma Yes, in first countable Lindelöf spaces compactness is equivalent to sequential compactness. Sequential compactness implies countable compactness, and since a sequentially compact set is closed in a first countable space, we can first extract a countable subcover by the Lindelöf property, and then from that a finite subcover by countable compactness. Err, to deduce closedness we need the space to be Hausdorff. Not sure if the argument can be extended to work in non-Hausdorff spaces. – Daniel Fischer Jun 12 '17 at 20:54

Second countability is enough. The following proof is adapted from the proof of Proposition 1.6.23 from the book Convergence Structures and Applications to Functional Analysis by R.Battie and H.-P Butzmann, a fantastic book on convergence spaces.

Theorem:If $$X$$ is a second countable topological space, then $$X$$ is compact if and only if $$X$$ is sequentially compact.

Proof: Compact $$\Longrightarrow$$ Sequentially Compact: Assume that $$X$$ is compact. Take a sequence $$\xi:\mathbb N\to X$$ in $$X$$ and an ultrafilter $$\mathcal{G} \supseteq<\xi>.$$ Here $$<\xi>$$ is the associated filter of the sequence $$\xi.$$ Then $$\mathcal{G}$$ converges to a point $$x \in X$$ and thus $$N_x\subset\mathcal G,$$ where $$N_x$$ is the neighborhood filter at $$x.$$ Since $$X$$ is second countable, $$X$$ is also first countable, thus $$N_x$$ has a countable basis, say $$\mathcal F:=\{F_1,\cdots,F_n,\cdots\}.$$ Then $$F_{n} \cap\left\{\xi_{i} : i \geq k\right\}$$ is not empty for all $$n, k$$ and so one can choose a subsequence $$\xi^{\prime}$$ of $$\xi$$ such that $$<\xi^{\prime}>\supseteq \mathcal{F} .$$ This means that $$\xi^{\prime}$$ converges to $$x .$$ Therefore $$X$$ is sequentially compact.

Sequentially Compact $$\Longrightarrow$$ Compact: Let $$\mathcal B$$ and $$\mathcal C$$ be a countable basis and open cover for $$X,$$ respectively. Define a collection by $$\mathcal B_0:=\{U\in \mathcal B|\ U\ \text{is contained in a member of \mathcal C}\},$$ then

we claim that $$\mathcal B_0\cap F\neq\emptyset$$ for each convergent filter $$F.$$ In fact, since $$\mathcal C$$ is an open cover, for each $$x\in X$$ there is some $$V\in\mathcal C$$ with $$x\in V.$$ Since $$\mathcal B$$ is a basis for $$X,$$ there is some $$U\in\mathcal B$$ with $$x\in U\subset V,$$ thus $$U\in\mathcal B_0\cap N_x.$$ Hence $$\mathcal B_0\cap N_x\neq\emptyset,$$ therefore $$\mathcal B_0\cap F\neq\emptyset$$ for each filter $$F$$ convergent to $$x,$$ which proves our claim.

Let $$\mathcal B_0=\{B_1,B_2,B_3,\cdots\},$$ then we claim that there must be a $$n$$ such that $$\bigcup_{i=1}^n B_i=X.$$ We use proof by contradiction. Assume that $$\bigcup_{i=1}^n B_i\subsetneqq X$$ for all $$n\in\mathbb N,$$ then for each $$n$$ choose $$\xi_n\in X-\bigcup_{i=1}^n B_i,$$ then $$\xi_n$$ is a sequence in $$X,$$ thus has a convergent subsequence $$\eta_k=\xi_{n_k}.$$

Since $$\langle \eta_k\rangle$$ is a convergent filter, it follows that $$\mathcal B_0\cap\langle \eta_k\rangle\neq\emptyset,$$ thus $$\eta_k$$ eventually falls into some $$B_n.$$ Thus there is some $$N\in\mathbb N$$ such that $$\eta_k\in B_n$$ for all $$k\geq N.$$ However, let $$N':=\max(N,n),$$ then we have $$\eta_k=\xi_{n_k}\in X-\bigcup_{i=1}^{n_k}B_i\subset X-\bigcup_{i=1}^{k}B_i$$ for all $$k\geq N',$$ which is a contradiction. Thus there must be some $$n_0$$ with $$\bigcup_{i=1}^{n_0}B_i=X.$$

Note that for each $$i$$ there is a $$C_i\in\mathcal C$$ with $$B_i\subset C_i,$$ it follows that $$C_1, C_2,\cdots,C_{n_0}$$ is a finite subcover for $$\mathcal C,$$ thsu $$X$$ is compact, which completes the proof.

• Very nice. I find the result satisfying, it suggests an explicit continuation of the first two types of conditions being naturally fulfilled in first countable spaces and the third condition requiring a stronger condition. One is led to the question "if every sequentially compact subset in a first countable space is compact, is the space second countable?" In that sense this answer, in addition to being interesting in its own right, also works to make the comment by Daniel Fischer about the weak topology more interesting than it already it is. – s.harp Apr 20 at 7:05