On the proof of
If $X$ is a complete metric space, then any dense $G_\delta$ in $X$ is residual in $X$.
I'm confused on how to prove this statement. $G_\delta = \cap_{n=1}^\infty U_n$ where every $U_n$ is an open set. If every $U_n$ where dense, then it would be easy to see that the complement of $G_\delta$ is of first category. However, if $X=\overline{G}_{\delta}=\overline{\cap_{n=1}^\infty U_n}$, it doesn't seem correct to conclude that every $U_n$ by itself is dense.
On the proof of
Let $X$ be a (nonvoid) complete metric space and $E$ a residual set in $X$. Then $E$ is of second category in $X$. IN particular, $X$ is of second category in itself.
Proof: Assume that $E$ is of first category in $X$. Then $X = E\cup E^c$, being a countable union of nowhere dense sets (proof continues)
Why $E^c$ has to be of first category in this case? For example in $\mathbb{R}$, the rationals are of first category but the irrationals are of second category and they are the complement of $\mathbb{Q}$.
Perhaps what is meant is $E\cup \lbrace {\rm limit\; points\; of\; } E\rbrace$? In this case either why these limit points could be a nowhere dense set?
Best Answer
If $D\subseteq E\subseteq X$, and $D$ is dense in $X$, then $E$ is dense in $X$: if $x\in X$, and $U$ is an open nbhd of $x$, then $U\cap E\supseteq U\cap D\ne\varnothing$, so $x\in\operatorname{cl}E$, and hence $\operatorname{cl}E=X$. (This is even easier to see if you realize that a set $D$ is dense in $X$ iff $U\cap D\ne\varnothing$ for each non-empty open $U\subseteq X$.) Thus, all of the open sets $U_n$ in your setting are dense in $X$.
For your second question, the residual sets are by definition the complements of first category sets, and $E$ is residual, so $X\setminus E$ is first category. You’re quite right that in this setting $X\setminus E$ cannot in fact be of first category, but that is the point of the proof: we assume that $E$ is of first category and derive a contradiction.