# Finding Supremum of Real Numbers

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? –  joriki Oct 1 '12 at 10:11
Huh? What is your condition (2) meant to say, and what is your definition of 'sup A'? –  Billy Oct 1 '12 at 10:12
@robjohn: I'm pinging you for Brian's comment to Michael (since you were the one approving it). –  Asaf Karagila Oct 1 '12 at 10:43
@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. –  Michael Albanese Oct 1 '12 at 10:47
@Michael: You’re right. Very weird. Anyhow, I’ve restored your edit. Sorry about all the confusion. –  Brian M. Scott Oct 1 '12 at 11:02

Whatever it is, compare it to the definition given for the $\sup$, try to deduce one from the other.

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Step 1: write down the definition of $\sup$ of a set.

Step 2: Prove the $\implies$ direction. To do this, assume $s = \sup A$. Then show that $s$ satisfies 1. and 2.

Step 3: Prove the $\Longleftarrow$ direction. To do this, assume that $s$ satisfies 1. and 2. and show that $s = \sup A$.

Once you have done that you have solved the question, that's all you need to do.

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Hint: If $t<s$, then prove that there exists $h>0$(however small), such that $(t+h)<s$, and this $(t+h)$ will be your required $a$.You can refer Analysis by Terence Tao for detailed explanation.

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