A set $A$ in a metric space $(M, d)$ is said to be totally bounded if, given any $\epsilon>0$, there exist finitely many points $x_1,\ldots,x_n\in M$ such that $A\subset\bigcup_{i=1}^nB_\epsilon(x_i)$. That is, each $x\in A$ is within $\epsilon$ of some $x_i$.

The author then goes on to say:

In the definition of a totally bounded set $A$, we could easily insist that each $\epsilon$-ball be centered at a point of $A$.

Indeed, given $\epsilon>0$, choose $x_1,\ldots,x_n\in M$ so that $A\subset\bigcup_{i=1}^nB_{\epsilon/2}(x_i)$.

We may certainly assume that $A\cap B_{\epsilon/2}(x_i)\ne\varnothing$ for each $i$, -------- HOW??

and so we may choose a point $y_i\in A\cap B_{\epsilon/2}(x_i)$ for each $i$.

By the triangle inequality, we then have $A\subset\bigcup_{i=1}^nB_\epsilon(y_i)$. That is, $A$ can be covered by finitely many $\epsilon$-balls, each centered at a point in $A$.

What is the justification for the line marked "HOW??" above?

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    $\begingroup$ Because you care only about the balls that cover your set. If the intersection is empty then you can throw it off and still have an $\epsilon$ finite cover. If I understood correctly. $\endgroup$ – Leo Oct 10 '18 at 1:32

What happens when $A\cap B(x_j)=\emptyset$ for some $j$? Then

$$A=A\backslash B(x_j)\subseteq\bigg(\bigcup B(x_i)\bigg)\backslash B(x_j)\subseteq\bigcup_{i\neq j} B(x_i)$$

In particular we can refine our covering $\{B(x_i)\}$ by removing $B(x_j)$ and still preserving all required properties.


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