# Intuition: Why is weak/weak sequential compactness more strict

It is known that for a metrizable space $$(X,\vert \vert \cdot \vert \vert)$$ where $$\vert \vert \cdot \vert \vert$$ is a norm on $$X$$, that:

sequential compactness $$\iff$$ compactness

And further for the weak topology on $$X$$, if $$X$$ is Banach space then:

weak sequential compactness $$\iff$$ weak compactness

I am struggling to find an intuition as to why we need the additional condition of $$X$$ Banach.

My thinking: Since by definition $$\tau_{\operatorname{weak}} \subseteq \tau_{\operatorname{strong}}$$ in terms of topologies, surely the condition of weak sequential compactness being equivalent to weak compactness is a lot "easier" than showing the same for a finer topology $$\tau_{\operatorname{strong}}$$?

I mean surely if $$X$$ compact then $$X$$ has to be weakly compact, given the coarser topology of $$\tau_{\operatorname{weak}}$$

• Your very last sentence doesn't make sense to me. Compactness means that every open cover has a finite open subcover. If you have less open sets (i.e. a coarser topology), then you have "less chance" to find this finite open subcover. Anyways, the unit ball in a reflexive banach space need not be compact but is weakly compact. So there's that as well. – Tony Jul 26 at 14:01
• I understand the first part, but I do not see the relevance of stating the unit ball in a reflexive banach space is weakly compact but need not be compact. I am asking whether the relation: compact $\Rightarrow$ weakly compact holds. And you just showed that weakly compact does not necessarily imply compact? – SABOY Jul 26 at 14:49
• Indeed I kind of misunderstood what you were asking at first. A more direct answer to your question is the following. Although we know that metrizable spaces have this property of equivalence between compactness and sequential compactness, a space with its weak topology is not metrizable in general; this property doesn't do anything for us. We must instead pass to the Eberlain-Smulian theorem which is a statement for Banach spaces, hence the qualification about being a Banach space. – Tony Jul 26 at 14:58
• Yes, basically what @TonyS.F. said, most Banach Spaces are a-priori not metrizable, generally, we have a theorem telling us that if $X^*$ is (norm) separable, then the unit ball of $X$ is weakly metrizable. For this reason, we need the Eberlein Smulian theorem. Specifically, the proof I have seen of E-S requires use of the uniform boundedness principle, a theorem that only holds in Banach Spaces (as well as some locally convex spaces, although this is not related to the discussion at hand). – rubikscube09 Jul 26 at 15:27

Consider any set $$X$$ with two topologies $$\tau, \tau'$$. If $$\tau$$ is coarser than $$\tau'$$, then obviously the following is true:
If $$C$$ is a compact subset of $$(X, \tau')$$, then it also a compact subset of $$(X, \tau)$$.
Consider any open cover $$\mathcal U$$ of $$C$$ in $$(X, \tau)$$. Then $$\mathcal U$$ is also an open cover in $$(X, \tau')$$ and thus has a finite subcover.
Be aware that in general there will be compact subsets of $$(X, \tau)$$ which are not compact in $$(X, \tau')$$.
Similarly it is obviuos that each convergent sequence in $$(X, \tau')$$ is also convergent in $$(X, \tau)$$ (with the same limit).