Proof or counterexample : Supremum and infimum

Question

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.

Answer 1

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.

Answer 2

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).

Answer 3

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.