Not Homework just Personal Study
Reference text: Kolmogorov and Fomin - Introductory Real Analysis
I am working on understanding why a complete metric space with no isolated points is uncountable. The hint given is that in such a metric space, every singleton set is nowhere dense in the space. I am having trouble with this idea.
Kolmogorov and Fomin define a set $S$ to be nowhere dense in a metric space $(X,d)$ if $S$ is dense in no open ball in $X$. In mathematical language this should be equivalent to for any open ball $B(x,\epsilon)$ there must exist $y \in B(x,\epsilon)$ and $r>0$ such that
$$B(y,r)\cap S= \emptyset$$
So if $\\{x\\}$ is nowhere dense it must be true that for any open sphere $B(x,\epsilon)$ there must exist $r>0$ and $y \in B(x,\epsilon)$ such that $$B(y,r)\cap \{x\} = \emptyset.$$
For $x$ to be an isolated point of a metric space $(S,d)$ there must exist an $r>0$ such that $B(x,r)\cap S = \emptyset.$ So a space $X$ with no isolated points means that for all $x\in X$ and all $\epsilon>0$ the open ball
$$B(x,\epsilon)\cap(X \setminus \{x\})\neq \emptyset.$$
I see how proving that a space $R$ with no isolated points satisfies $\{x\}$ is nowhere dense in $R$ paired with Baire Category Theorem, which I have proven, is enough. I just do not see how to prove, using the above definitions, that $\\{x\\}$ must be nowhere dense in a space with no isolated points. I am not permitting myself to use that nowhere dense is the same as the interior of the closure is empty, as the authors did not introduce this, and is thus not what they had intended for an argument. Thank you in advanced for any help.