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Let $A\subset\mathbb{R}$ a nonempty set of real numbers bounded above and $u$ be an upper bound of $A$. Prove that if $u\in A$, then $u=\sup A$.

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This is pretty straightforward; where are you stuck? –  Brian M. Scott May 1 '12 at 22:43
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What have you even tried? –  TMM May 1 '12 at 22:51
    
Assume $v$ is an upper bound and $v > u$. Arrive at a contradiction. Conclude it must be $v \leq u$. –  Pedro Tamaroff May 2 '12 at 2:34
    
Note that Brian Scott and TMM's comments are meant to help us formulate a response. We believe in helping people learn, but you should put in some effort to let us better help you. –  Willie Wong May 2 '12 at 8:07
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1 Answer 1

up vote 2 down vote accepted

Recall the definition of supremum: If the non-empty set of real numbers $S$ is bounded above then $u$ is the supremum of $S$ if both the following hold:

$\ \ \ $1) $u$ is an upper bound of $S$

and

$\ \ \ $2) if $v$ is any upper bound of $S$, then $u\le v$.

A hint for your problem: Suppose $v$ is another upper bound of $A$. Think about condition 2), keeping in mind that $u\in A$.

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