I am learning functional analysis, and following definition for Category is introduced to prove Uniform Boundedness Theorem.
A subset $M$ of a metric space $X$ is said to be
(a) rare in $X$ if its closure has no interior points
(b) meager (of first category) in $X$ if $M$ is the union of countably many sets each of which is rare in $X$
(c) nonmeager (of the second category) in $X$ if $M$ is not meager in $X$
And here is my questions, some of which I don't know the answer:
Q1. Is a first category always rare?
A1. No. $\mathbb Q$ is of first category in $\mathbb R$, but it is not rare in $\mathbb R$.
Q2. Is it true that for every countable union of a first category subset in $X$, each term is rare?
Q3. Can a superset of a non-meager be meager in $X$?
Best Answer
Terminology point: "rare" is usually called "nowhere dense" in English ("nergens dicht" in Dutch etc.), "rare" is the word used in French. Stick to "nowhere dense" when writing English.
Meagre (British spelling) sets are closed under subsets: a subset of a meagre set is meagre, because the same holds for nowhere dense sets. (If $N$ is nowhere dense and $A \subseteq N$ we have $\operatorname{int}(\overline{A}) \subseteq \operatorname{int}(\overline{N}) = \emptyset$). So definitely NO to Q3.
Meagre sets are closed under countable unions, almost by definition. The combination of being closed under subsets and under countable unions is called being a $\sigma$-ideal, the Lebesgue measure $0$ subsets in Euclidean spaces are another example. It sort of formalises the idea of a family of "small subsets" (in some, e.g. topological or measure-theoretic sense). See Oxtoby's nice book "Measure and Category" for more on the parallels and differences between these two example $\sigma$-ideals.
Q2 is obviously no: a meagre set $M$ (which is not nowhere dense) we can write as a trivial countable union of copies of itself, and then we have a trivial counterexample. But it's also possible to write down a countable dense subset $M$ of the reals that is a disjoint union of dense (so not nowhere dense) sets, or use $\Bbb Q = \cup_{n \in \Bbb Z} \left(\Bbb Q \cap [n,n+1)\right)$ as the terms of the union.