0
$\begingroup$

Let $A\subseteq \mathbb{R}$. We say that a real number $M\in\mathbb{R}$ is a maximum of $A$ if $M$ is an upper bound for $A$ and $M\in A$.

If $A$ has a maximum, prove that it only has one; and prove that if $\max A$ exists, then so does $\sup A$, and $\sup A=\max A$.

To prove that $A$ has only one maximum, I'm pretty sure I have to assume that there are two maximums but then show that they are the same. I'm just confused as to how to do this.

Any help is greatly appreciated. Thank you!

$\endgroup$
5
$\begingroup$

Let $M$ and $M'$ two maximum of a set $A$ then $M,M'\in A$ and

$$\forall a\in A,\; a\le M$$ so in particular and since $M'\in A$ then we have $M'\le M$. Similarly we get $M\le M'$ hence $M=M'$.

$\endgroup$
1
  • 1
    $\begingroup$ Similarly, since $\sup A$ is the least upper bound for $A$ and $M$ is an upper bound for $A$, we have $\sup A \leq M$. But $\sup A$ is an upper bound for $A$ and $M \in A$, so $M \leq \sup A$. $\endgroup$ – Alex Wertheim Jan 20 '15 at 21:48
0
$\begingroup$

Let $M_1$ and $M_2$ both be maxima of $A$, and suppose $M_1 > M_2$. Then since $M_2$ is an upper bound of $A$, $M_1 \not \in A$, contradicting the definition of a maximum which requires $M \in A$.

Then suppose $M$ is a maximum of $A$. Then $\forall a \in A: a \leq M$, so that part of the definition of $\sup$ is met by $M$. Now suppose there were some other value $M' < M$ also meeting that definition. Since $M\in A$, $M$ cannot be less than or equal to than $M'$. Amd that contradicts the statement about $M'$ also meeting that part of the definition of $\sup A$. So $M$ is the lowest number such that all members of $A$ are less than or equal to $M$, completing the requirements for $M = \sup A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.