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There is a question here about whether every compact subset $K$ of Banach space $E$ is such that $\forall \epsilon >0$, $\exists \, V \subset E$ finite dimensional subspace with $d(x,V) < \epsilon$, $\forall x \in K$. Is the converse also true? That is, being $K$ a closed, bounded set of a Banach space $E$ such that $\forall \epsilon >0$, $\exists \, V \subset E$ finite dimensional subspace with $d(x,V) < \epsilon$, $\forall x \in K$, is $K$ compact?

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    $\begingroup$ I think the answer is yes (I could be wrong). Try to show K is totally bounded. $\endgroup$ – Tim kinsella Sep 21 '16 at 8:52
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Yes, this is true. It suffices to show $K$ is totally bounded. Fix $\epsilon>0$ and choose a finite-dimensional subspace $V\subseteq E$ such that $d(x,V)<\epsilon$ for all $x\in K$. Since $K$ is bounded, there is in fact a closed bounded subset $L\subseteq V$ such that $d(x,L)<\epsilon$ for all $x\in K$. Then $L$ is compact, so there are finitely many points $y_1,\dots,y_n\in L$ such that every point of $L$ is within $\epsilon$ of some $y_i$. For each $i$, choose a point $x_i\in K$ such that $d(x_i,y_i)<2\epsilon$, if any such point exists. Now for any $x\in K$, there exists $y\in L$ such that $d(x,y)<\epsilon$, and then some $y_i$ such that $d(y,y_i)<\epsilon$, and then $d(x,y_i)<2\epsilon$. Thus $x_i$ is defined, and $d(x,x_i)\leq d(x,y_i)+d(x_i,y_i)<4\epsilon$.

That is, any point of $K$ is within $4\epsilon$ of one of the points $x_i$. Since $\epsilon>0$ was arbitrary, we can conclude that $K$ is totally bounded.

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