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?
Best Answer
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.