Kreyszig Functional Analysis book presents the following definition.
I'm trying to get some examples.
(a) The cantor set $K$ is rare in $\mathbb{R}$ because it's closed and has empty interior so that $$\operatorname{int}\left(\overline{K}\right)=\operatorname{int}(K)=\varnothing.$$
(b) The set $\mathbb{Q}$ of rationals numbers is meager in $\mathbb{R}$ because it's enumerable so that we can write $$\mathbb{Q}=\{r_1,r_2,…,r_n,…\}=\bigcup_{i=1}^\infty{\{r_i}\},$$ where $\{r_i\}$ is rare for all $i\in\mathbb{N}$.
(c) I think that $(0,1)$ (like any other non-degenerate interval) is a nonmeager set in $\mathbb{R}$. My question is: how to prove it?.
Thanks.
Best Answer
Your first two examples are spot on.
Actually, so is the third.
The proof of this relies on the Baire Category Theorem, mentioned in the comments above.
To see this, recall that any non-degenerate open interval $(a,b)$ is homeomorphic to the real line $\mathbb{R}$. As $\mathbb{R}$ is a complete metric space, it follows that $(a,b)$ is completely metrizable (of course, the usual metric is not complete on $(a,b)$, but some other metric — one derived from a homeomorphism from $\mathbb{R}$ — makes it complete). Therefore $(a,b)$ is nonmeagre-in-itself by the Baire Category Theorem.
If $(a,b) = \bigcup_n Z_n$ where each $Z_n$ is nowhere dense in $\mathbb{R}$, it would follow that each $Z_n$ is also nowhere dense in the subspace $(a,b)$, which would then contradict the fact that $(a,b)$ is nonmeagre-in-itself.
A quick comment about your second example: It is good that you note that the singleton sets in $\mathbb{R}$ are nowhere dense. For instance, $\mathbb{N}$ is countable, but not meagre-in-itself, because $\mathbb{N}$ is discrete (all subsets are open) and so the singleton sets are not nowhere dense in $\mathbb{N}$. (Of course, $\mathbb{N}$ is meagre in $\mathbb{R}$.)