Let $A$ be any non-empty subset of $\mathbb{R}$. Then $s = \sup A$ iff $s$ has the following properties:
$s \geq a$ for every $a \in A$,
if $t < s$, then there exists an $a \in A$ such that $a > t$.
Prove it? having problem in proving 2.
Let $A$ be any non-empty subset of $\mathbb{R}$. Then $s = \sup A$ iff $s$ has the following properties:
$s \geq a$ for every $a \in A$,
if $t < s$, then there exists an $a \in A$ such that $a > t$.
Prove it? having problem in proving 2.
Whatever it is, compare it to the definition given for the $\sup$, try to deduce one from the other.
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.
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.