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I was asked to prove this in my 3rd weak in college.

"Let $A$ be a nonempty set of real numbers, bounded above. Suppose exists $K>0$ so that for every two different numbers $x,y\in A: |x-y|>K$. Prove: $A$ has a maximum element."

Thank you in advance.

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    Try this, let $\alpha = \sup A$. Then either $\alpha \in A$ or not. If it is you are finished. If not, use the property above to find a contradiction. Draw a picture. And take your vitamins.2012-11-05

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HINT: Since $A\ne\varnothing$, let $a\in A$. Let $u$ be an upper bound for $A$, so that $x\le u$ for every $x\in A$. The distance from $a$ to $u$ is $u-a$. Divide the interval $[a,u]$ into pieces of length $K$, perhaps with one shorter piece left over. Could there be more than one element of $A$ in each piece?

Alternatively, look at the sequence $\langle a,a+K,a+2K,\dots\rangle$; each interval $\big[a+nK,a+(n+1)K\big)$ can contain at most one element of $A$, and intervals beyond $u$ are irrelevant.

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    @David: $u$ can in fact be a member of $A$, though it need not be. If u-a (no absolute values needed), then $[a,u]$ is the ‘one shorter piece left over’. Of course if $u-a\le K$, then $(a,u]\cap A=\varnothing$, and $a=\max A$.2012-11-05
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Not sure this is particularly a calculus problem, but...

Let $\beta \in \mathbb{R}$ be the least upper bound for $A$.

Suppose $a \in A$, with $|a - \beta| \leq K$. We'll prove by contradiction that $a$ is the maximum element of $A$.

(Assume the law of the excluded middle, and) suppose $A \ni b > a$. Then $b > K + a \geq \beta$. But $b \in A \Rightarrow b < \beta$. This is a contradiction, so there cannot exist a $b > a$ in $A$.

(Also, should this perhaps be tagged "homework"?)

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    If $a$ is the maximum element of $A$ it must be equal to sup $A$ which you define $\beta$ to be. Do you mean to say that $\beta$ is *an* upper bound? If $\beta$ is the least upper bound of $A$ do you not need to establish that $a = \beta$? Or are you leaving that to the OP?2012-11-05