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.
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 $\mathbb{R}$).