# Application of Baire category theorem

1. Oct 30, 2013

### mahler1

The problem statement, all variables and given/known data.

Let $(X,d)$ be a complete metric space without isolated points and let $D$ be an enumerable dense subset of $X$. Prove that $D$ is not a $G_δ$ (countable intersection of open sets).

The attempt at a solution.

Suppose that $D$ is a $G_δ$. So $D=\bigcap_{n \in \mathbb N} A_n$ , $A_i$ open for every $i \in \mathbb N$. Then $D^c=\bigcup_{n \in \mathbb N} X \setminus A_n$ is a countable union of closed sets. Here comes my question: How could I prove that this sets are nowhere dense? I know that $\overline {D}=X$ by hypothesis; then $\overline {D}^c=\emptyset$, but I want to prove that $\overline {D^c}^\circ=\emptyset$.
If I could prove that the complement is nowhere dense, then $X=A \cup B$ where $A=\bigcup_{n \in \mathbb N, d_n \in D} \{d_n\}$ and $B=D^c$. Since X has no isolated points, each $\{d_n\}^\circ=\emptyset$. Then $X$ is the union of nowhere-dense closed sets, which is absurd by the Baire category theorem.

Last edited: Oct 30, 2013
2. Oct 31, 2013

### Dick

$X \setminus A_n$ is already closed. So to prove it's nowhere dense you just have to show it has empty interior. I.e. it doesn't contain any open subsets.

3. Oct 31, 2013

### mahler1

You're right. $\overline D^c=D^c$. So, I have to prove ${D^c}^\circ=\emptyset$. I have to show that for every $x \in D^c$ and for every $δ>0$, the ball $B(x,δ) \not\subset D^c$. So, take $x \in D^c$ and $δ>0$ arbitrarily. Suppose $B(x,δ) \subset D^c$. $D$ is dense in $X$, so $B(x,δ) \cap D≠\emptyset$. But then, there exists $y$ such that $y \in D$ and $y \in D^c$, which is absurd. It follows that ${D^c}^\circ=∅$.

4. Oct 31, 2013

### Dick

That's a little round about. I would just say that an open subset of $X \setminus A_n$ wouldn't contain any elements of D because they are all in $A_n$. But that's impossible because D is dense. But I think it's the same idea.

5. Oct 31, 2013

### mahler1

I have a doubt: with your idea, I would have proved that $(X \setminus A_n)^\circ=\emptyset$. Then $\bigcup_{n \in \mathbb N} (X \setminus A_n)^\circ=\emptyset$. But I am not sure if $\bigcup_{n \in \mathbb N} (X \setminus A_n)^\circ=(\bigcup_{n \in \mathbb N} X \setminus A_n)^\circ={D^c}^\circ$. It's easy to see that given two sets $A$ and $B$, it's always true that $A^\circ \cup B^\circ \subset (A \cup B)^\circ$, but I don't know if the other inclusion always holds.

6. Oct 31, 2013

### Dick

The other inclusion doesn't hold in general. But you don't need that. You are going off on some kind of tangent here. If you prove that $(X \setminus A_n)$ has no interior, doesn't that show $(X \setminus A_n)$ is nowhere dense? Doesn't that in turn mean X is the countable union of nowhere dense sets? It's the union of all of those sets plus the union of the countable singletons in D.

7. Oct 31, 2013

### mahler1

Yes, sorry. I forgot the original problem.

8. Nov 1, 2013

### Dick

No problem. Actually, I finally see what you are trying to do. I was skipping over some parts since they were so obviously wrong, I though that they were just bad notation. If you were actually trying to prove that $X \setminus D$ is nowhere dense think of the example where D is the rationals (Q) and X is the reals (R). The $R \setminus Q$ is not nowhere dense, in fact, it's dense. A countable union of nowhere dense sets doesn't have to be nowhere dense. It often helps to pick a concrete example like Q in R to hold in your mind when you are doing proofs like this.

9. Nov 1, 2013

### mahler1

Thanks for the remark.