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Suppose that $X$ is a Banach space (or more generally, Frechet space) such that $X$ is the closure of the span of a compact (in the original topology) subset $K$. Do we know anything "nice" about $X$, from this information alone?

If $X$ is the span of $K$, without needing take a closure, then a Baire category argument shows that $X$ is locally compact, hence finite dimensional. Can something like this be made to happen in the dense case?

More disconcertingly, could all separable Banach spaces be of this form?

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As pointed out by Yemon Choi, your condition is equivalent to separability. However, if you weaken it in the natural way by replacing the condition of compactness to weak compactness, you get an interesting class of spaces (called WCG spaces for obvious reasons) which have remarkable properties and have been studied intensively. The initial paper was a seminal one by Amir and Lindenstrauss in the Annals (available on line) and it is an easy task to locate more recent developments. – jbc Feb 21 '13 at 9:03
The phrase "compactly generated" has another meaning in general topology, where continuity of maps out of $X$ can be probed by testing continuity of their restrictions to compact subsets. This would seem to be a condition much different from the one of the OP! – Todd Trimble Oct 15 '14 at 2:32
up vote 11 down vote accepted

I think not. Suppose X has a countable subset S whose span is dense in X. Enumerate S as a sequence, then by rescaling you can assume the sequence converges to zero in norm. This gives a countable relatively compact set whose linear span is dense in X.

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And thus we do know something "nice" about $X$: it is separable! – Nik Weaver Feb 21 '13 at 5:56
@Nik: Isn't the Argyros-Haydon space, being an isomorphic predual of $\ell^1$, separable? ;) – Yemon Choi Feb 21 '13 at 6:16
Yemon's answer shows that every separable Frechet space is compactly generated. The converse is also true: The compact generator is contained in the closed absolutely convex hull of a sequence converging to $0$, and the countable set of all rational (finite) linear combinations of that sequence is dense. – Jochen Wengenroth Feb 21 '13 at 10:18
Thanks for the quick explanation! – Iian Smythe Feb 21 '13 at 15:22
@Yemon: I believe that space is not bad, it is misunderstood. – Nik Weaver Feb 21 '13 at 18:32

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