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Let $X$ be a Banach space and $A\subset X$ be bounded. Suppose that, for any $\varepsilon>0$, $\exists\; F_\varepsilon$ a subspace of $X$ of finite dimension such that $$\text{dist}(x,F_\varepsilon)\leq\varepsilon,\quad\forall x\in A.$$

Then prove that $A$ is relatively compact in $X$, i.e. its closure is compact in $X$.

Can anybody give me an hint?

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2 Answers

up vote 2 down vote accepted

I would reason as follows. Fix $\varepsilon>0$ and accordingly the finite dimensional subspace $F_\varepsilon$ granted by the hypotheses of the problem. Then $F_\varepsilon$ is closed. In particular $\forall x\in A$ exists $v_x\in F_\varepsilon$ such that $\|x-v_x\|=\text{dist}(x,F_\varepsilon)\leq \varepsilon$. Since $A$ is bounded, the set of all such $v_x$ is bounded, because $\|v_x\|\leq \|x-v_x\|+\|x\|\leq \varepsilon+M$. Then it istotally bounded because bounded in a finite dimensional space. Suppose $v_p$ is an $\varepsilon$ net in this set. Then we may choose, for any $p$, $x_p$ as (one of) the element(s) of $A$ such that $\|x_p-v_p\|=\text{dist}(x_p,F_\varepsilon).$ Notice that $x_p$ exists by construction of the set of the $v_x's$. Then the $\{x_p\}$ are a $3\varepsilon$ net in $A$ because if we choose $x\in A$, let $v_p$ be such that $\|v_x-v_p\|\leq \varepsilon$ by definition of $\varepsilon$ net. Then $\|x-x_p\|\leq \|x-v_x\|+\|v_x-v_p\|+\|v_p-x_p\|\leq 3\varepsilon.$

Hope this is correct.

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Fix $\varepsilon>0$. Let $B$ be a ball containing $A$. The set $G_\varepsilon=F_\varepsilon\cap B$ is non-empty bounded subset of finite dimensional subspace, so you can find here finite $\varepsilon$-net $M_\varepsilon$. It is easy to prove that $M_\varepsilon$ is a $2\varepsilon$-net for $A$. The rest is clear.

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what if $F_\varepsilon\cap A$ is empty? –  guido giuliani Aug 24 '12 at 15:25
    
Well, I think we can salvage this proof –  Norbert Aug 24 '12 at 16:26
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