elementary-set-theory
I have 4 different sets:
a) $\{ [x]: x \in \mathbb{R}\}$
b) $\mathbb Q \cup (2,+\infty)$
c) $\{n^2: n \in \mathbb N\}$
d) $\mathbb Q \cap (2,+\infty)$
A set is numerable if exists a bijective function from $\mathbb N$ to $A$
the c) is numerable because exists $f:\mathbb N \rightarrow A$ so that $f(n)=n^2$
but in the other cases?
We consider the equation
$\bigcup_{{i=0}}^\infty\left(\bigcap_{{j=0}}^\infty A_{{ij}}\right)=\bigcap_{{h:\mathbb N\to\mathbb N}}\left(\bigcup_{{i=0}}^\infty A_{{ih(i)}}\right)$
Suppose $x$ is in the LHS, then for some $i$, $x\in A_{ij}$ for all $j$. In particular, $x\in A_{ih(i)}$ for all $h:\mathbb N\to\mathbb N$. Hence $x$ is in the RHS.
Now suppose $x$ is not in the LHS.
Then for all $i$ there is $j$ such that $x\not\in A_{ij}$. Define $h(i)$ as the least such $j$. Now $x\not\in\bigcup_{i=0}^\infty A_{ih(i)}$.
Hence $x$ is not in the RHS.
Since this holds for all sets $x$, we have the equality.
Halmos' Naive Set Theory gives a hint which is the following Lemma. It took me awhile to understand the hint's usefulness for proving that the union of two finite sets is finite.
The following is the proof that you may be interested in.
[sorry for posting pictures, but I prefer typing in a LaTeX editor for its macros + \newcommands + autocompletion]
Best Answer
$a)$ is numerable because the set is identical with $\mathbb Z$
$b)$ is not numerable : Already the interval $(2,3)$ is uncountable.
$d)$ is numerable because it is an infinite subset of $\mathbb Q$