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$\left(X,\|\cdot\|\right)$ is a normed vector space. $\textbf{Riesz's Theorem of compactness}$ says that $$ \{x \in X \colon \|x\| \leq 1 \} \ \text{compact} \ \Longleftrightarrow \ \text{Each bounded sequence in $X$ has a convergent subsequence}.$$

I am looking for the proof of this theorem.

Some thoughts to $"\Rightarrow"$: If $\{x \in X \colon \|x\| \leq 1 \}$ is compact, it's clear that each sequence in $\{x \in X \colon \|x\| \leq 1 \}$ has a convergent subsequence. But how can I conclude that each bounded sequence in $X$ (!?) has a convergent subsequence?

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Where does the term "Riesz's Theorem of compactness" come from? – Michael Greinecker Apr 20 '14 at 11:35
In my lecture notes at university Würzburg this theorem is called "Kompaktheitssatz von Riesz" and I thought "Riesz's Theorem of compactness" would be the best translation. Is it misleading? – user138765 Apr 20 '14 at 12:08
I was just surprised. The compactness of the unit-ball is equivalent to the normed space being finite dimensional, so the statement seemed odd to me. Also, Riesz did thins more deserving of putting his name on. – Michael Greinecker Apr 20 '14 at 12:10
@MichaelGreinecker : I believe this is a reference to Riesz' Lemma, which is usually used to prove that this is equivalent to finite dimensionality. – Prahlad Vaidyanathan Apr 20 '14 at 13:25
up vote 1 down vote accepted

A bounded set is contained in $B_r(0) = \{x \in X : \|x\| \leq r\}$. The map that sends $x\to rx$ is a homeomorphism of $X$, so if $B_1(0)$ is compact, so is $B_r(0)$. Hence, your bounded sequence is contained in a compact set, and so has a convergent subsequence.

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I forgot that $B_r(0)$ is compact, too. So I think "$\Rightarrow$" is clear. "$\Leftarrow$" seems to be more difficult. – user138765 Apr 20 '14 at 11:29
$\Leftarrow$ is trivial isn't it? Assuming you are using sequential compactness. – Prahlad Vaidyanathan Apr 20 '14 at 13:27

HINT: If $x_n$ is a bounded sequence, and $M$ is a bound on $\|x_n\|$, consider the sequence $\frac1Mx_n$.

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