I have a question first, what is this intersection symbol ?
Given a collection \(\displaystyle X\) of sets, the intersection
\(\displaystyle \displaystyle
\bigcap_{A\in X}A
\)
simply means the set whose elements are precisely those elements belonging to every set in the collection \(\displaystyle X\). For a finite set this would just be the intersection \(\displaystyle A_1\cap A_2\cap\dots\cap A_n\) of the sets in \(\displaystyle X\). Likewise, if \(\displaystyle I\) is some set of indices, then
\(\displaystyle \displaystyle
\bigcap_{i\in I} A_i
\)
is the intersection of each \(\displaystyle A_i\) with \(\displaystyle i\in I\), which consists of those elements belonging to \(\displaystyle A_i\) for every \(\displaystyle i\in I\).
I did not understand anything of this inequality
I'm not too used to seeing the backwards-bracket notation for an open interval, but
\(\displaystyle \displaystyle
[a,b]=\bigcap_{n\in\mathbb{N}^*} \left]a-\frac{1}{n},b+\frac{1}{n}\right[\)
is just saying that the interval \(\displaystyle [a,b]\) can be written as the intersection of open intervals \(\displaystyle ]a-1/n, b+1/n[\) for all positive integers \(\displaystyle n\).
As for the proof, show that the set on the left has the same elements as the set on the right. This can be done in two steps: to prove a set \(\displaystyle A\) is equal to a set \(\displaystyle B\), first show \(\displaystyle A\subseteq B\) and then show \(\displaystyle B\subseteq A\).
