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Let $A$ be a subset of a metric space $(X,d)$, I want to show equivalence of

(i) For every $\varepsilon > 0$ there is a finite set $E = \{ x_1, x_2, \ldots, x_n \} \subset X$ with $$ A \subset \bigcup_{i = 1}^n B(x_i, \varepsilon) $$ and

(ii) For every $\varepsilon > 0$ there is a finite set $E = \{ x_1, x_2, \ldots, x_m \} \subset A$ with $$ A \subset \bigcup_{i = 1}^m B(x_i, \varepsilon) $$

The proof of $(ii) \Rightarrow (i)$ is trivial I think, for the other direction let $\varepsilon > 0$ and $E = \{x_1, \ldots, x_n \} \subset X$ with $$ A \subset \bigcup_{i = 1}^n B(x_i, \varepsilon) $$ Without loss of generality I can assume that every $B(x_i, \varepsilon)$ contains points of $A$, so all I need to do now is select a finite collection of points $a_1,\ldots,a_m \in A$ for every $B(x_i, \varepsilon)$ such that $B(x_i, \varepsilon) \cap A \subset \bigcup_{i=1}^m B(a_i, \varepsilon)$ and then take the union for every $B(x_i, \varepsilon)$. I have an intuitive feel that this is possible and I draw some pictures which convince me that this is always possible, but i have no idea how to make this formal?

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up vote 1 down vote accepted

Almost solution for the implication $(i)\implies(ii)$

  1. consider $\varepsilon/2$-net $\{x'_1,\ldots,x'_m\}\subset X$.

  2. For each $x'_i$ find $a_i\in B(x'_i,\varepsilon/2)\cap A$.

  3. For each $a\in A$ we can find $k$ such that $d(a,x_k')<\varepsilon/2$, so $$ d(a,a_k)\leq d(a,x_k')+d(x_k',x_k)\leq \ldots $$

  4. The rest is clear

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that was my idea, but it don't need to be the case that $B(x_i',\varepsilon/2)\cap A \subset B(x_i, \varepsilon)$ so i might need to select other $x_i$'s, but my problem is to show that a finite collection of points $x_i$ for every $x_i'$ is enough. – Stefan Oct 17 '12 at 23:59
You do not need this inclusion. Just show that every point of $A$ situated from some $x_i$ not far than $\varepsilon$ – Norbert Oct 18 '12 at 0:03

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