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.

Let $\emptyset \neq A \subseteq B \subseteq \mathbb R,$ and $B$ is bounded above. Then there is a real number $u$ such that $b \leq u, \forall b \in B.$ In particular, $a \leq u, \forall a \in A.$ This shows that $A$ is bounded above.
In this particular case, $\cap A_n \subseteq A_n$ and each $A_n$ is bounded above (since they  have supremum). So $\cap A_n$ is bounded above. Now the completeness property of real numbers says that every non-empty subset of $\mathbb R,$ which is bounded above has a supremum.
Note: Since OP didn't mentioned anything about $A_n,$ I assume that the base space is $\mathbb R.$ If it's not $\mathbb R,$ then the statement is no longer true (see other answer for an counter example).
