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 $\tau_{\operatorname{weak}}$

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    $\begingroup$ 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. $\endgroup$ – Tony Jul 26 at 14:01
  • $\begingroup$ 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? $\endgroup$ – SABOY Jul 26 at 14:49
  • $\begingroup$ 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. $\endgroup$ – Tony Jul 26 at 14:58
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    $\begingroup$ 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). $\endgroup$ – 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).

However, in your case the non-trivial theorem is the equivalence of weak sequential compactness and weak compactness. You cannot expect to get that for free.


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