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From Peter Lax Functional analysis page 104:

Show that a weakly sequentially compact set is bounded.

Definition. A subset $C$ of a Banach space $X$ is called weakly sequentially compact if any sequence of points in $C$ has a subsequence weakly convergent to a point of $C$.

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I did not down vote but perhaps you might want to show some of your thoughts. – Rudy the Reindeer Nov 13 '12 at 16:07
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What do you mean by weakly sequential compact? Is it in $\Bbb R^n$ or in any metric space? – Berci Nov 13 '12 at 16:48
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@Berci: Sequentially compact in the weak topology on $X$. – Brian M. Scott Nov 13 '12 at 19:39
up vote 3 down vote accepted

The key is the following result:

If $X$ is a normed space, and $\{x_n\}\subset X$ a sequence which converges weakly to $X$, then $\sup_{n\in\Bbb N}\lVert x_n\rVert<\infty$.

It's a consequence of Baire's categories theorem applied to $X^*$, the topological dual of $X$.

We show that an unbounded set cannot be weakly compact. If $S$ is not bounded, let $\{x_n\}\subset S$ such that $\lVert x_n\rVert\geqslant n$. If $\{x_{n_k}\}$ is a subsequence of $\{x_n\}$, then for each $k$, $\lVert x_{n_k}\rVert\geqslant n_k$. In particular, $\{x_{n_k}\}$ is not bounded hence cannot be weakly convergent.

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