Proof or counterexample : Supremum and infimum If $($An$)_{n \in N}$ are sets such that each $A_n$ has a supremum and $∩_{n \in N}$$A_n$ $\neq$ $\emptyset$ , then $∩_{n \in N}$$A_n$ has a supremum.
How to Prove This.

The statement is false.
I bring you a counter example.
consider $\mathbb{Q}$ as the universal set with the following set definitions:
$$A_1=\{x\in \mathbb{Q}|x^2<2\}\cup{\{5\}}$$
$$A_2=\{x\in \mathbb{Q}|x^2<2\}\cup{\{6\}}$$
$$A_n=\{x\in \mathbb{Q}|x^2<2\}\cup{\{4+n\}}$$
In this example, sumpermum of $A_1$ is 5 and for $A_2$ is 6.
However, $A_1 \cap A_2$ has no supremum:
$$A_1 \cap A_2=\{x\in \mathbb{Q}|x^2<2\}$$
And so on:
$$A_1 \cap A_2 \cap ... =\{x\in \mathbb{Q}|x^2<2\}$$
However, in $\mathbb{R}$ as a universal set, any bounded non-empty set has a supremum.
