If $A\subset X$ is a $G_\delta$ set and dense in $X$, then what I want to show is that $X\setminus A$ is of the first Baire category.
Where the meaning of first category is that I can write
$$X\setminus A= \bigcup_n^{\infty}E_n$$
and the sets $E_n \subset X$ are nowhere dense ($X\setminus \overline{E_{n}}$ is dense or $\forall \epsilon>0 $ and $x \in X$ $B_{\epsilon }(x) \nsubseteq \overline{A}$)
My attempt
Then we can write
$$A=\bigcap_{n=1}^{\infty}C_{n} $$
where $C_{n}$ is open for all $n$, this is because $A$ is $G_{\delta}$ so what I was triying to show that each of the $C_n$ should be dense and then conclude that since I can write the complement using DeMorgan's laws as
$$X\setminus A= \bigcup X\setminus C_n$$
I can get that the complements $X\setminus (X\setminus C_n)$ are dense but So I think it won't help because this is in fact $C_n$.
So can someone help me to fix this idea (If I am in the right track of course) or provide a different proof of this please?
Edition
I want to conclude that I can
$$X\setminus A= \bigcup_n^{\infty}E_n$$
where the sets $E_n \subset X$ are nowhere dense. To do that we use the fact that since $A$ is $G_{\delta}$
$$A=\bigcap_{n=1}^{\infty}C_{n} $$
and given that $A$ is dense we have that eah $C_n$ is dense too. So now we may write (By DeMorgan's laws)
$$X\setminus A= \bigcup_n^{\infty} X \setminus C_n$$
So we only need to prove that $X \setminus (X \setminus C_n)$ is dense, but $X \setminus (X \setminus C_n)=C_n$ that is dense, therefore $X \setminus A$ is of first category.
Am I right?
Thanks a lot in advance for your help.
Let $A^c=X\setminus A$ be complement of $A$, $\overline{A}$ be closure of $A$ and $A^o$ be interior of $A$.
We define:
1). $A$ is dense iff $\overline{A}=X$.
2). $A$ is nowhere dense iff $\overline{A}^o=\varnothing$.
In this problem, we will prove that $C_n^c$ is nowhere dense.
First since $A\subset C_n$ and $\overline{A}=X$, there is $$ \overline{C_n}\supset \overline{A}=X\quad\text{Thus}\quad \overline{C_n}=X $$ So $C_n$ is dense. Clearly $\overline{C_n}^c=\varnothing$. Moreover $$ \overline{C_n}^c=(((C_n^{c})^o)^c)^c=(C_n^{c})^o=\varnothing $$ Since $C_n$ is open, $C_n^c$ is closed. So $\overline{C_n^c}=C_n^c$. Thus we have $$ \overline{C_n^c}^o=(C_n^c)^o=\varnothing $$ i.e. $C_n^c$ is nowhere dense.