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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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    @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

3 Answers 3

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