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 $\sup(A)= \max(A)$.

  • 2
    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 at 17:47
  • 1
    It looks well to me. – Ramiro Scorolli Nov 8 at 17:51
  • That's a good addition, thank you. – WesleyGroupshaveFeelingsToo Nov 8 at 22:07

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