1. ## supremum

2. Originally Posted by suliman
Suppose:

$
\sup_{m\in \mathbb{N}}(b_m) > \sup_{m\in \mathbb{N}}(a_m)
$

This implies that there exists a $k\in \mathbb{N}$ such that:

$
b_k > \sup_{m\in \mathbb{N}}(a_m)
$

which in turn implies that there exists a $j \in \mathbb{N}$ such that:

$
X_{k,j} > \sup_{m\in \mathbb{N}}(a_m)
$

But:

$
\sup_{m\in \mathbb{N}}(a_m) \ge X_{u,v}
$

for all $u,v \in \mathbb{N}$ .

So we have a contradiction; so the premiss that:

$
\sup_{m\in \mathbb{N}}(b_m) > \sup_{m\in \mathbb{N}}(a_m)
$

is false.

In a similar manner we can show that:

$
\sup_{m\in \mathbb{N}}(b_m) < \sup_{m\in \mathbb{N}}(a_m)
$

is false, so we are forcesd to conclude that:

$
\sup_{m\in \mathbb{N}}(b_m) = \sup_{m\in \mathbb{N}}(a_m)
$

RonL