I am studying Baire Theorem on Functional Analysis and I am dealing with the following definitions (translated from portuguese):
A subset of a topological space $ X $ is said to be rare in $ X $ if the interior of its closure is empty and is said thin in $ X $ if it is contained in a countable union of rare sets. A subset is said generic if it is the countable intersection of open sets and is dense on $X$.
Considering these definitions, I have to prove that:
If $H$ is thin in the Baire space $X$, then $X\setminus H$ contains a generic subset $G$ in $X$.
Particularly, this implies that $\mathbb{R}\setminus \mathbb{Q}$ is generic.
I cannot see it as a countable intersection of open sets.
Many thanks for a help.
The standard terminology in English is
rare is called "nowhere dense",
thin is called "meagre" ("meager" in the US), or first category.
generic is the same in both.
So $X$ is Baire (which means that a countable intersection of open and dense sets is dense) and $H$ is meagre, then $X\setminus H$ contains a generic subset.
This is quite clear if you realise that $N$ nowhere dense is equivalent to $X\setminus N$ contains a dense and open subset (namely $X\setminus \overline{N}$), so if $H$ is meagre, so $H \subseteq \bigcup_n N_n$ where all $N_n$ are nowhere dense, then by de Morgan we have:
$$\bigcap_n (X\setminus N_n) \subseteq X\setminus H$$
and all $X\setminus N_n$ contain the dense and open set $G_n = X\setminus \overline{N}$ and so $\bigcap_n (X\setminus N_n)$ is generic (because it's dense as $X$ is Baire). So $X\setminus H$ indeed contains a generic set.
Finally,
$$\mathbb{R}\setminus{\Bbb Q} = \bigcap_{q \in \Bbb Q} \mathbb{R}\setminus \{q\}$$
showing the irrationals are quite clearly a generic set, as the complement of a point is open and dense in the reals, and there are only countably many rationals, and the irrationals are dense (which we could already conclude from the reals being complete and hence a Baire space).