Every finite set contains its supremum: proof improvement.

Every finite subset of $\mathbb R$ contains its supremum (and its infimum)

Proof Let $A=\{a_1,...,a_n\}$ be a finite subset of $\mathbb{R}$. Since it is non-empty and it is bounded ($\max A$ is an upper bound), it has supremum, that is $\exists \sup A$ and by definition $\forall a \in A \;\, a \leq \sup A$. Let's suppose that $\sup A \not\in A$ then, since $\max A \in A$ we have that $\max A < \sup A$. But considering that $\mathbb Q$ is dense in $\mathbb R$ we can conclude that $\exists r \in \mathbb Q$ s.t. $\max A < r < \sup A$, but this is absurd since $r$ is an upper bound of $A$ and it is lower than the supremum. Necessarily, $\sup A\in A$.

Is there anything wrong? Is there any way to prove this without using density of $\mathbb Q$ or another property? Thanks in advance.

Solution 1:

Your proof starts right away with using $\max A$. But if you know the existence of $\max$ then it is automatically also the $\sup$. So this is probably not the way you are expected to proceed.

If $A=\{a\}$ is a singleton set, then clearly $\sup A=\max A = a$.

If $A$ has cardinality $n>1$ and we know as induction hypothesis that all sets of cardinality $<n$ have a maximal element, let $a\in A$ be an arbitrary element and let $A'=A\setminus\{a\}$. Since $A'$ has less than $n$ elements, let $a'= \max A'$. If $a'\ge a$, then $a'$ is a maximal element of $A$. If $a'< a$, then $a$ is a maximal element of $A$.