Understanding residual sets A set is called residual if it is the complement of a meager set (which is a countable union of nowhere dense subsets). I can't really picture how big (or dense) is a residual set.
As I understand it, a residual set should be a countable union of subsets that are "weakly dense"(the complement of nowhere dense subset), so is residual set itself dense? Also, is being residual stronger than dense?
e.g. do we have the following statements:

*

*Any subset that contains a residual set is residual (or dense?).


*The intersection of an open subset with a residual set is residual (or dense) in this open set.
 A: 
A set $A$ is residual in $X$ iff $A= \bigcap_n D_n$ for $D_n \subseteq X, n \in \Bbb N$ such that $\operatorname{int}(D_n)$ is dense in $X$.

(in that case the $X\setminus D_n$ are nowhere dense so $X\setminus A$ is a countable union of nowhere dense sets,etc.)
If $X$ is a Baire (so completely metrisable or locally compact Hausdorff) then all residual sets are dense (this is what Baire's theorem says). In general spaces this need not hold: e.g. in $\Bbb Q$ all subsets are residual, so the notion is pointless in such non-Baire spaces. Mostly it's considered in completely metrisable spaces like the $\Bbb R^n$.
By the above reformulation both  1 and 2 are clear. If $A \subseteq B$ we can just and $B \setminus A$ to all $D_n$ (still dense interior of course) to get $B$ as the intersection, and if $D_n$ has dense interior, $D_n \cap O$ also has dense interior in $O$, when $O$ is any non-empty open subset of $X$, so $A$ residual, $O$ open, implies $O \cap A$ residual in $O$.
A: It depends on the space. If $X$ is a countable $T_1$ space with no isolated points (e.g. $X=\Bbb Q$) then every subset of $X$ is both meager and residual. If X is a non-empty complete metric space with no isolated points (e.g. $X=\Bbb R$) then a residual $A\subset X$ is dense in $X$ and the cardinal of $A$ is at least $2^{\aleph_0}. $
Addendum: Theorem 1: If $Y$ is a non-empty completely metrizable space with no isolated points then $Y$ has a subspace homeomorphic to the Cantor Set $C,$ so $|Y|\ge |C|=2^{\aleph_0}.$ Theorem 2: Let $X$ be a completely metrizable space and let $Y$ be a subspace of $X.$ Then $Y$ is completely metrizable iff $Y$ is a $G_{\delta}$ subset of $X.$
So if $X$ is a non-empty complete metric space with no isolated points and if $A$ is residual in $X,$ then $A\supset Y=\cap_{n\in \Bbb N}Y_n$ where each $Y_n$ is dense in $X$ and open in $X$. By Theorem 2, $Y$ is completely metrizable. By the Baire Category Theorem, $Y$ is dense in $X$. So by Theorem 1, $|A|\ge |Y|\ge 2^{\aleph_0}.$
