# Baire Category Theorem without Completeness

Let $(X,d)$ be a metric space.

We say that $Y \subseteq X$ is dense in $X$ if for any non-empty open set $U\subseteq X,$ we have $U \cap Y \neq \emptyset.$

Baire Category Theorem states that

If $(X,d)$ is a complete metric space with $(U_n)_{n \in \mathbb{N}}$ being a sequence of open dense sets in $X,$ then their intersection $\bigcap_{n \in \mathbb{N}}U_n$ is dense in $X.$

If we remove openness in the condition, then the theorem will not hold anymore, simply let $X = \mathbb{R},$ $U_1 = \mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}.$ Clearly $X$ is complete and $U_1$ and $U_2$ are dense in $X,$ but $U_1 \cap U_2 = \emptyset$ is not dense in $X.$

Question: Give an example such that $X$ is not complete with $(U_n)_{n \in \mathbb{N}}$ a sequence of open dense sets but their intersection $\bigcap_{n \in \mathbb{N}}U_n$ is not dense in $X.$

I have been trying to come out with an example that satisfies the question above. Since finite dimensional space is always complete, I have to let $X$ be infinite dimensional. One example that comes to my mind is $C[0,1],$ the set of continuous functions on $[0,1].$

However, I do not know which set is dense in $C[0,1].$

Any hint would be appreciated.

• It would be easier if you took something like $X = \mathbb{Q}$. – Daniel Fischer Jul 21 '17 at 14:46
• And if you take an infinite-dimensional vector space, you still need to choose something that isn't complete. In its usual topology, $C[0,1]$ is complete. You could take $C[0,1]$ and endow it e.g. with an $L^p$-norm for $1 \leqslant p < +\infty$. – Daniel Fischer Jul 21 '17 at 14:49
• @DanielFischer: Thanks. But I still could not obtain a counterexample to the Baire theorem if $X = \mathbb{Q}.$ Any hint? – Idonknow Jul 21 '17 at 17:11
• @Idonknow enumerate the rationals, and then, in that ordering, construct a suitably small interval around each rational. You'll get an open covering of the rationals. Choose the length of each interval in such a way to assure the intersection will be empty. – Ittay Weiss Jul 21 '17 at 17:23
• Or simply $U_n = \mathbb{Q}\setminus \{r_n\}$, where $\{r_n : n \in \mathbb{N}\}$ is an enumeration of the rationals. Then $\bigcap U_n = \varnothing$. – Daniel Fischer Jul 21 '17 at 20:25

## 1 Answer

Let $X$ be a countable $T_1$ space without isolated points.

Then $D_x = X\setminus \{x\}$ is open (as $\{x\}$ is closed by $T_1$-ness), and dense (it's not closed as $\{x\}$ is not open, so its closure is $X$).

Then $\emptyset = \cap\{D_x: x \in X\}$ is a countable intersection of dense open sets of $X$, and it's far from dense.

The only metrisable example of such a space is $\mathbb{Q}$ (in its metric topology, inherited from $\mathbb{R}$).

• This is a good example. It would be better if we could have a simpler example, say, in $\mathbb{R}$. – Idonknow Oct 11 '17 at 23:34
• @Idonknow $\mathbb{Q}$ is a subset of $\mathbb{R}$, right. The reals itself are complete so cannot be an example. – Henno Brandsma Oct 12 '17 at 5:12