The book "Metric Spaces" by Babu Ram says this about the proof of Cantor's Intersection Theorem:
Create nested intervals $F_{n+1}\subset F_n$ such that $\lim_{n\to\infty}\text{diameter}(F_n)=0$. Choose points $x_i\in F_i$. $\{x_i\}$ is a cauchy sequence. Hence, by the completeness property of the metric space, it has a limit point in the space. Let it be $a$. Our claim is that $a\in\bigcap_{i=1}^{\infty} F_i$.
Let us assume to the contrary. Then there is an $F_k$ such that $a\notin F_k$, and hence $a\notin F_j$ for all $j\geq k$. Let $r=d(a,F_k)$. Then $r>0$. Hence, $B(a,\frac{r}{2})$ does not contain any of the points $x_e$, $e\geq k$. Therefore, it can't be the limit point of $\{x_i\}$.
My question
Why can't $r=0$? For example, if $F_k=(1,2)$, and $a=1$, then $$d(a,F_k)=\inf\{ d(a,y): y\in F_k\}=0$$ If $r=0$, then doesn't this whole proof fall flat? Clearly checking whether $B(a,0)$ contains any of the points of $\{x_i\}$, and hence checking whether it is a convergent point of $\{x_i\}$ makes no sense.
Best Answer
The intervals $F_n$ are closed intervals, and hence, are a positive distance from any point outside of them. More generally, in any metric space, closed sets are a positive distance from any point outside them, since the closed set's complement is an open neighborhood of the point.