I have a question in the proof of that there is no countable dense set is a $G_\delta$ (that is a set of all countable intersection of open set) in a complete metric space $X$ which has no isolated points.
The proof is here:
Let $x_n$ be the points of a countable dense set $E$ in $X$. Assume that $E$ is a $G_\delta$. Then $E=\bigcap V_n$, where each $V_n$ is dense and open since if each $V_n$ is not dense, $E\subset X\subset\bar{E}$ does not hold. Let
$$W_n=V_n-\bigcup_{k=1}^{n} \{x_k\}.$$
Then each $W_n$ is a dense open set, but the intersection of $W_n$ is empty, it contradicts Baire's category theorem.
Question: Why $W_n$ is open and dense?
To check density, we need to check $W_n\subset X\subset \bar{W_n}$, but I couldn't work well. Should I use other definition of density: $E$ is dense in $X$ if every point of $X$ is a limit point of $E$ or a point of $E$?
Best Answer
Note that each $V_n$ is dense simply because $E \subseteq V_n$ and $E$ is dense so $X=\overline{E}\subseteq \overline{V_n} \subseteq X$, so $\overline{V_n}=X$ too.
Every finite set is closed (in a metric space for sure, but this holds for a large class of spaces, the so-called $T_1$ spaces), so each set of the form $X\setminus \{x\}$ is open.
A finite intersection of dense open sets is (open and) dense and any set of the form $X\setminus \{x\}$ is dense (or else $x \notin \overline{X\setminus \{x\}}$ which means $\{x\}$ is open, contradiction)
Now $W_n = V_n \cap (X\setminus \{x_1\}) \cap \ldots \cap (X\setminus \{x_n\})$ is open and dense as a finite intersection of open and dense sets.