On dense sets and isolated points

Question

I need to prove that if $D$ is dense in $X$ and $X$ has no isolated points, then $D$ has no isolated points. I managed to prove that if $A\subset X$ is open, then $A$ has no limit points, but can't seem to make use of that. My attempt was to prove that $X = D'$, since $\overline{D} = X$, so that would imply $D \subset D'$, but it's not working out... Any tips?

Answer 1

For a topological space $X$, the following are equivalent:

1. $\{x\}$ is nowhere dense for every $x \in X$.

2. Every finite subset of $X$ is nowhere dense.

3. $D \cap U$ is infinite for every dense $D \subset X$ and every nonempty open $U \subset X$.

Proof:

$[1 \implies 2]$: A finite union of nowhere dense sets is still nowhere dense.

$[2 \implies 3]$: $D \cap U$ is dense in $U$. Therefore $D \cap U$ is infinite.

$[3 \implies 1]$: $\{x\} \cup (X\setminus \overline{\{x\}})$ is dense in $X$ and its intersection with $\operatorname{int} \overline{\{x\}}$ is a subset of $\{x\}$. Therefore $\operatorname{int} \overline{\{x\}} = \emptyset$. $\square$

In a T1 space, a singleton is nowhere dense unless it is open, so a T1 space without isolated points satifies 1-3.

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Suppose that $D$ is dense and $U$ is open, choose some $x,y\in U$ and find some $V,V'\subseteq U$ which are disjoint, with $x\in V$ and $y\in V'$.

Then by density $V\cap D$ and $V'\cap D$ are both non-empty. And therefore $U\cap D$ contains at least two points. By induction this shows that the intersection of $D$ and $U$ contains infinitely many points.