# Compact subspace of a Banach space .

The following statement doesn't make sense to me, can someone justify it to me ?

If $K$ is a compact subset of a Banach space $Y$ then there exists for $\epsilon > 0$ a finite dimensional subspace $Y'$ of $Y$ such that $d(x, Y') < \epsilon$ for every $x \in K$ , where as usual , $d(x, Y')=\inf \{\|x-y\|, y \in Y'\}$ .

This is pretty surprising ! (if it's true)

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If $K$ is compact, it is totally bounded. Hence for any $\epsilon>0$ there exists a finite $\epsilon$-net. Let $Y'$ be the span of this $\epsilon$-net (which must be finite dimensional). – copper.hat Jan 9 '13 at 5:59

## 1 Answer

Cover $K$ by balls of radius $\varepsilon$ and pick a finite subcover of these balls. Let $x_1,\dots,x_n$ be the centers of these balls and let $V$ be the linear subspace generated by $x_1,\dots,x_n$. Then for any $x \in K$ we have some $i$ such that $x \in B(x_i,\varepsilon)$ so that $d(x,x_i)<\varepsilon$ implying that $d(x,V)<\varepsilon$.

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