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Let $A$ be any non-empty subset of $\mathbb{R}$. Then $s = \sup A$ iff $s$ has the following properties:

  1. $s \geq a$ for every $a \in A$,

  2. if $t < s$, then there exists an $a \in A$ such that $a > t$.

Prove it? having problem in proving 2.

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    What does "If tt" mean?2012-10-01
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    Huh? What is your condition (2) meant to say, and what is your definition of 'sup A'?2012-10-01
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    @robjohn: I'm pinging you for Brian's comment to Michael (since you were the one approving it).2012-10-01
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    @BrianM.Scott: I didn't make any conscious changes to the mathematical content, I only Latexed the symbols and put (1) and (2) into a list. There was certainly a lot more than 'If tt' written for (2) when I edited the question.2012-10-01
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    @Michael: That’s odd: the edit record shows what’s up there now as the original version.2012-10-01
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    @BrianM.Scott: I know what the problem is. The original version is what you see above, but when I went to edit it, there were several words and symbols the OP had typed which did not show up. By Latexing the symbols, both the words and symbols show up. If you click edit and select the original version, you should be able to see text like "If $t, then there exists an $a\in A$ such that $a>t$" (without the Latexed symbols) which doesn't appear in the actual post.2012-10-01
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    @Michael: You’re right. Very weird. Anyhow, I’ve restored your edit. Sorry about all the confusion.2012-10-01
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    @BrianM.Scott: No problem.2012-10-01
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    @user43165: What do you mean by "having problem in proving 2"? This is an if and only if proof which means you need to show that the supremum of $A$ satisfies 1 and 2, and also that if $s$ satisfies 1 and 2, then s is the supremum of $A$. **Some comments on the conditions:** The first condition implies that $s$ is an upper bound for $A$; the second shows that if $t$ is less than $s$, then $t$ is not an upper bound for $A$ (therefore $s$ is the least upper bound).2012-10-01

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