# If maximum of an ordered, set exists, this is equal to the supremum.

Suppose we have an ordered set $$A \subset M$$, which has a maximum, denote it by $$\max(A)$$, then this maximum is equal to the supremum (the smallest of upper bounds) of $$A$$.

I will focus on explaining my reasoning very meticulously to see if I really understand the concepts well, I am asking for any improvements or logical errors.

The maximum has the property that for all $$a \in A$$, we know $$a \leq \max(A) \in A$$. Since $$A$$ is bounded from above (as it has a maximum), there are upper bounds, the smallest of which is the supremum, we know that $$\max(A) \leq \sup (A)$$ as the supremum is an upper bound so it is greater than or equal to any element in $$A$$, in particular the elements $$\max(A)$$.

Now notice that $$\max(A)$$ is greater than or equal to any element in $$A$$, so it is in fact an upper bound. There is no smaller upper bound than the supremum, so we must have that $$\sup (A) \leq \max (A)$$. By anti-symmetry of the order relation $$\leq$$,

we have that $$\sup(A)= \max(A)$$.

• This is just fine. Moreover, if the maximum exists you do not have to assume separately that the set is bounded above. – Ethan Bolker Nov 8 '18 at 17:47
• It looks well to me. – RScrlli Nov 8 '18 at 17:51
• That's a good addition, thank you. – Wesley Strik Nov 8 '18 at 22:07