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.

  • 5
    $\begingroup$ By the way, if $\max A$ exists, then $\max A=\sup A\in A$, so you would be done. I think you are assuming too much at the start, though. Prove it exists! (Hint: $|A|\in\omega$, so try induction...) $\endgroup$ Dec 16, 2012 at 10:59
  • $\begingroup$ Why not just let r = (max A + sup A)/2? (Though I don't really understand what your definition of 'max A' is here. Perhaps there's an even simpler proof if you tell me what you mean by that.) $\endgroup$
    – Billy
    Dec 16, 2012 at 11:01
  • $\begingroup$ It is supposed that I can not conclude that $\max A = \sup A$ immediately, so I need to use supremum properties... Should I order elements of $A$ insted of using $\max A$? $\endgroup$
    – user50554
    Dec 16, 2012 at 11:03

2 Answers 2


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 $A$.


This is not a complete proof since you haven't shown that $\max A$ exists in the first place. The reason why this is important is that not all subsets of $\mathbb{R}$ have a maximum element (take the open interval $(0,1)$ for instance). So, maybe, this set $A$ (subset of $\mathbb{R}$), finite though it is, doesn't have a maximum element? The crux of this proof should be to show that the finiteness guarantees the existence of a maximum element. Also, don't get me wrong—an infinite subset of $\mathbb{R}$ can also have its maximum element (consider the closed interval $[0,1]$, for instance), but not every infinite subset of $\mathbb{R}$ need contain its least upper bound, i.e., it doesn't need to have a maximum element.


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