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.
 A: 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.
A: 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).
A: Let $x \in \cap_{n \in \mathbb N} A_n$. Since $x$ belongs to each $A_n$, $x \le sup\ A_n$ for any $n \in \mathbb N$. Then we see that $\{x : x \in \cap_{n \in \mathbb N} A_n\}$ is bounded above (by $sup\ A_1$ for example) and so by the Completeness Property of $\mathbb R$, $\cap_{n \in \mathbb N} A_n$ has a supremum.
