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The theorem is

Let $X$ be a locally convex topological vector space, and let $K ⊂ X$ be a non-empty, compact, and convex set. Then given any continuous mapping $f: K → K$ there exists $x ∈ K$ such that $f(x) = x$.

Suppose $X$ is a Banach space. When it says $K \subset X$ must be compact, is this the same as saying $K \subset X$ must be a compact embedding, i.e., for any bounded sequence in $K$, the sequence must have a subsequence which converges in $X$?

assa888
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  • No, it is the same as saying that $K$ is bounded and any (bounded) sequence in $K$ admits a subsequence which converges in $K$. EDIT: See here: http://math.stackexchange.com/questions/822236/compactness-and-sequential-compactness-in-metric-spaces – PhoemueX Sep 03 '14 at 12:53
  • Thanks. But isn't $K \subset X$ being compact the same as $i:K \to X$ being a compact map? – assa888 Sep 03 '14 at 14:06
  • If your definition of a map $f:X\to Y$ being compact is that every bounded sequence $(x_n)n$ in $X$ has a subsequence $(x{n_k})k$ such that $(f(x{n_k}))_k$ is convergent, then this is not the same. To see this, note that the identity map $\iota : \Bbb{R} \to \Bbb{R}, x\mapsto x$ is (with this definition) a compact map, but $\Bbb{R}$ is certainly not compact. – PhoemueX Sep 03 '14 at 15:01

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