Complement of co-dense set.

Question

Asaf's argument : (ZF) If $\mathbb{R}^k$ is a countable union of closed sets, then at least one has a nonempty interior

Let $X$ be a separable complete metric space. Let $D$ be a countable debse subset. Let $F$ be a closed subset with empty interior.

Here, how do i show that $D\setminus F$ is dense?

Ive been trying to figure it out for a day, and still stuck here..

Answer 1

Let $U$ be a non-empty open subset of $X$. Since $F$ is closed and has an empty interior we have that $U\setminus F$ is non-empty and open.

By density of $D$ we know that $D\cap(U\setminus F)=(D\setminus F)\cap U = (D\cap U)\setminus F$ is non-empty, therefore $D\setminus F$ is dense.

Answer 2

Choose an arbitrary nonempty open set $U$. Then $U$ is also Polish, and $D\cap U$ is countable dense and $F\cap U$ is a closed subset with empty interior. So it is enough to show that $D\setminus F$ is nonempty. But if a closed set contains a dense set, it is the entire space, which is a contradiction.

Answer 3

Let $D$ be a countable dense subset. Let $F$ be a closed subset of $X$ such that $\text{Int}(F) = \emptyset$. Let $U$ be any nonempty open set. $U \cap (X - F)$ is a nonempty open set because if $U \cap (X - F) = \emptyset$, then $U \subset F$, which contradicts $\text{Int}(F) = \emptyset$. Since $U \cap (X - F)$ is nonempty open and $D$ is dense, there exists $d \in D$ such that $d \in U \cap (X - F)$. Hence there exists a $d \in D - F$ such that $d \in U$. $U$ is arbitrary so $D - F$ is dense.

Answer 4

Regardless of the topological space, $$\overline{ D \setminus F } = \overline{ D \cap ( X \setminus F ) } = \overline{ \overline{D} \cap ( X \setminus F ) } = \overline{ X \cap ( X \setminus F ) } = \overline{ X \setminus F } = X \setminus \mathrm{Int} ( F ) = X \setminus \emptyset = X.$$ (I use the result here.)