# Definition of totally bounded set

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?

• 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. – Leo Lerena 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.