Application of Baire category theorem

[mahler1]

Homework Statement .

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.

 

[Dick]

Homework Statement .

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.

$X \setminus A_n$ 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.

 

[mahler1]

$X \setminus A_n$ 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.

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 ${D^c}^\circ=∅$.

 

[Dick]

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 ${D^c}^\circ=∅$.

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.

 

[mahler1]

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.

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.

 

[Dick]

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.

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.

 

[mahler1]

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.

Yes, sorry. I forgot the original problem.

 

[Dick]

Yes, sorry. I forgot the original problem.

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.

 

[mahler1]

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.

Thanks for the remark.