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.