# Counter example for Baire's Theorem

Theorem: Let $$(X,d)$$ be a complete metric space, and let $$D_n, n\in \mathbb N$$ be open, dense subsets of $$X$$. Then also $$\bigcap_{n\in\mathbb N} D_n$$ is dense in $$X$$.

This statement is false if $$X$$ is not complete. Take $$X=\mathbb Q=\{q_1,q_2,\dots\}$$ and $$D_n=X\setminus \{q_n\}$$.

$$\bullet$$ $$\mathbb Q$$ is not complete

$$\bullet$$ $$D_n$$ is open since $$X\setminus D_n=\{q_n\}$$ and singeltons are closed sets.

$$\bullet$$ $$D_n$$ is dense in X. This is what I do not understand. The closure of $$D_n$$ is equal $$\mathbb R$$, but not $$\mathbb Q$$. But $$\mathbb R$$ is no subset of $$X$$. So is actually here the closure of $$D_n=\mathbb Q$$? And if, why?

$$\bullet$$ $$\bigcap_{n=1}^\infty D_n=\emptyset$$ which is not dense in $$\mathbb Q$$.

Could someone elaborate on the third bullet?

• $\mathbb{R}$ contains $X$. Density w.r.t. $\mathbb{R}$ means for any point $x$ in $\mathbb{R}$ we can find a converging sequence of points in $D_n$ which have limit $x$. In particular, we can choose any point of $X$. – Tony S.F. Jun 16 at 19:25
• Whenever you talk about "closure" of a subset, it is with respect to some ambient metric space. While the closure of $D_n$ in $\mathbb{R}$ is $\mathbb{R}$, the closure of $D_n$ in $\mathbb{Q}$ is $\mathbb{Q}$, since every element of $\mathbb{Q}$ is a limit of points in $D_n$. – mathworker21 Jun 16 at 19:25
• @TonyS.F. your comment seems irrelevant – mathworker21 Jun 16 at 19:26
• @mathworker21 it's not irrelevant; i edited it anyway to be more precise but the point is that if you are dense w.r.t. some set you are also dense w.r.t. any subset of it. – Tony S.F. Jun 16 at 19:27

The metric space $$X$$ under consideration here is the set of the rational numbers with the metric inherited from the real numbers. That is, given two rational numbers $$p,q\in X$$, the distance between them is $$|p-q|$$. Now if you take any rational number, $$p$$, then $$\mathbb{Q}\backslash\{p\}$$ is dense in $$X$$, because given any rational number $$r$$, there exists a sequence in $$\mathbb{Q}\backslash\{p\}$$ converging to $$r$$. This is of course obvious if $$r\neq p$$, and if $$r=p$$, then the sequence $$\{p-\frac{1}{n}\}_{n=1}^{\infty}$$ belongs to $$\mathbb{Q}\backslash\{p\}$$ and converges to $$r$$, so that $$\mathbb{Q}\backslash\{p\}$$ is dense.
In that example, $$X$$ is $$\mathbb Q$$. So, it makes no sense to assert that $$\overline{D_n}=\mathbb R$$. Your universe here is $$\mathbb Q$$ and therefore $$\overline{D_n}$$ must be a subset of $$\mathbb Q$$. And, in fact, it is equal to $$\mathbb Q$$.