Let $A=\{\text{isolated points of } X\}$. $X$ is a countable complete metric space. Show that $A$ is dense in $X$.
My attempt: Basically we want to show that $\bar A = X$. First, we show that $\bar A \subset X$. We know that $A\subset X$. For all those points that are in $\bar A$, but not in $A$, we can construct a sequence in $A$ (thus in $X$) such that the sequence converges to that point. Since $X$ is complete, then that point has to be in $X$ as well.
Then I tried to show that $X\subset \bar A$. Take $x\in X$. If $x$ is already in $A$ then we are done.
Suppose $x \notin A,$ then it is not an isolated point. Then by definition of isolated point, $\forall \epsilon>0$, $B(x;\epsilon)\cap X\ne\{x\}$. But we know that $x$ is there, so that means there is some other point in that intersection. Say that point is $y$.
Then I got stucked. I guess I really want to say that $\forall \epsilon>0$, the intersection will always have an isolated points, so that I can construct such sequence that is in $A$. But I don't know how to conclude that? I believe that must be true, because otherwise there will be an $\epsilon$ ball such that every point in that ball is a dense point, which seems like a contradiction…? (since I didn't actually use the fact that $X$ is a countable metric space)
Best Answer
I just answered this question for a homework assignment, so I've copied my solution below.
For each $x \in X\setminus A$, define $U_x := X\setminus\{x\}$. Then $\{x\}$ closed $\implies U_x$ open. Moreover, $\overline{U}_x = X$ as $\{x\}$ is not an open since $x \notin A$. Since $X$ is countable, we have \begin{equation*} \bigcap\limits_{x \in X \setminus A} U_x = A \end{equation*} i.e. we can write $A$ as a countable intersection of dense open sets. By the Baire-Category Theorem, this implies that $A$ is dense, i.e. $\bar{A} = X$.