How is completeness used in the Eberlein–Šmulian theorem? So for weakly closed subsets of Banach spaces compactness and sequential compactness coincide, but upon studying the proof I can't put my finger on what exactly would go wrong if we dropped the completeness assumption.
I figure the result doesn't hold true for general normed spaces as several sources explicitly require a Banach space. What part of the proof does necessarily require completeness? What would be a counterexample if it is dropped?
 A: Nothing would go wrong per se, but in an incomplete normed space, there are simply much fewer weakly compact sets.
Let $X$ be a normed space, and $\tilde{X}$ its completion. Then we have a canonical isometric isomorphism $\rho \colon \tilde{X}' \to X'$, namely the restriction of a functional to $X$. Thus we may identify the two spaces (that is habitually done tacitly), and note that the weak topology $\sigma(X,X')$ is the subspace topology induced by $\sigma(\tilde{X},X')$. Thus a subset $A\subset X$ is weakly compact if and only if it is weakly compact when regarded as a subset of $\tilde{X}$. The same holds for weak sequential compactness and weak limit point compactness, since these properties are all intrinsic properties of a topological space, they don't depend (in contrast to closedness or openness) on an ambient space.
Hence the Eberlein–Šmulian theorem holds for all normed spaces, for subsets of $X$, the properties of


*

*weak compactness,

*weak sequential compactness, and

*weak limit point compactness


all coincide.
But the case of Banach spaces is more interesting than the case of incomplete normed spaces, since in Banach spaces, one can have simple criteria for one of the forms of compactness. E.g. the closed unit ball in any reflexive Banach space is weakly compact by the Banach-Alaoglu theorem, and the Eberlein–Šmulian theorem then asserts its weak sequential compactness.
