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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:

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

  2. The intersection of an open subset with a residual set is residual (or dense) in this open set.

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    Perhaps the nicest way to think of residual sets is this: a set $G$ in a space $X$ is residual if and only if there are sets $D_n\subseteq X$ for $n\in\Bbb N$ such that $G=\bigcap_{n\in\Bbb N}D_n$, and $\operatorname{int}A_n$ is dense in $X$ for each $n\in\Bbb N$. – Brian M. Scott Apr 20 '21 at 03:22
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    It's more appropriate to call the complement of a nowhere dense set "strongly dense", not "weakly dense". It applies dense after all, and not all dense sets have a nowhere dense complement etc. – Henno Brandsma Apr 20 '21 at 05:50
  • When avoiding generalities at the topological space level (and beyond), such as restricting ourselves to complete metric spaces with no isolated points, then every residual [= co-meager] set (in a fixed space) intersects each nonempty open subset in that space (equivalently, each nonempty open ball in that space) in a set of cardinality continuum (thus more than just "dense"), and there are sets with this property that are not residual (i.e. residual is in some sense an even stronger notion of being dense than "cardinality continuum in every open ball"; (continued) – Dave L. Renfro Apr 20 '21 at 06:21
  • see Proof B in this answer for a very strong example in ${\mathbb R}).$ For an overview of "residual" as meaning "large", see this answer. Also of possible interest: Generic Elements of a Set AND Is saying a set is nowhere dense the same as saying a set has no interior? AND Porous Sets: Why are they interesting? – Dave L. Renfro Apr 20 '21 at 06:23
  • In my first comment, "in a set of cardinality continuum" should be "in a set of cardinality at least that of the continuum". – Dave L. Renfro Apr 20 '21 at 06:27
  • I added to my A, as some results I stated are, I think, far from obvious. – DanielWainfleet Apr 20 '21 at 07:18

2 Answers2

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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 $O$.

Henno Brandsma
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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, $|A|\ge |Y|\ge 2^{\aleph_0}.$

DanielWainfleet
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