Let $X$ be a metric with $D\subset X$ I have to prove that the following senteces are equal
Could you please help me on this one with a hint or something i completely lost. I know that since D is dense every $x\in X $ and for every $ε>0$ there is $y\in D$ so that $d(x,y)<ε$
I will appeal to the following useful lemma in topology:
Lemma: A subset $D\subset X$ is dense if and only if for any non-empty open subset $U\subset X$, one has $D \cap U \neq \emptyset$.
Proof: $(\Rightarrow)$ Let $D$ be dense and $U\subset X$ be any arbitrary (non-empty) open subset. Pick any $x\in U$ and since $U$ is open, there exists $\delta>0$ such that $B(x,\delta) \subset U$. But by your definition of $D$ being dense, there exists $y\in D$ such that $d(x,y) <\delta$, i.e. $y\in B(x,\delta)\subset U$. So $D\cap U \neq \emptyset$.
$(\Leftarrow)$ I will leave this as an exercise to you. Q.E.D.
With this, we can now proof the entire statement. Assume $D$ is dense. If $(X-D)^O\neq \emptyset$, then the it is an nonempty open subset of $X$. In particular $D \cap (X-D)^O \neq \emptyset$.
But $D \cap (X-D) = \emptyset$ already and so certainly $D \cap (X-D)^O $ cannot be non empty. So we conclude $(X-D)^O = \emptyset$.
On the other hand, if $(X-D)^O$ is empty, we have $\emptyset =(X-D)^O = X - \overline{D}$. So $\overline{D}=X$ and we conclude $D$ is dense.
