The relation of type $A \subset V \subset \overline{V} \subset U$ in Banach space

Question

Let $X$ be a Banach space, $A$ is a compact subset of $X$, $U$ is an open set containing $A$.

Question: Does there exist a relatively compact open set $V$, such that $A \subset V \subset \overline{V} \subset U$?

If, moreover, $X$ is locally compact, then the answer is definitely yes. But the local compactness is too strong for a Banach space, since all locally compact Banach spaces must be finite dimensional...

Answer 1

No compact set in an infinite dimensional Banach space can have non-empty interior.

So no non-empty open set can be relatively compact.

These are standard results: local compactness implies finite-dimensionality and even one compact set with non-empty interior implies (by homogeneity) that all points have compact neighbourhoods and thus the space is locally compact.

Answer 2

There never is (in infinite dimension), because then any ball inside $V$ would be relatively compact. More generally, an infinite-dimensional Banach space has no relatively compact open set.

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Funny thing is that, if you read it bottom-to-top, the post can be rephrased as

I know that an infinite-dimensional Banach space is not locally compact, but are all Banach spaces locally compact?