Suppose a set $A$ is non-empty and bounded above. Given $\epsilon>0$, prove that there is an $a ∈ A$ such that $\sup{A}-\epsilonThe second inequality is not difficult to prove. All I use is the definition of an upper bound. So $\sup{A}\geq a$ for all $a∈A$. But I'm finding the first inequality difficult to prove.
 A: I suppose that this is your definition of supremum: 

A real number $s$ is the least upper bound for a set $A \subset \mathbb{R}$ if the following hold:
  
  
*
  
*$s$ is an upper bound for $A$;
  
*if $b$ is any upper bound for $A$, then $s \le b$.
  

Now, we need to prove that given $\epsilon>0$, there is an $a ∈ A$ such that $\sup{A}-\epsilon<a\leq\sup{A}$.
First, let's see what it means for a number $r \in \mathbb{R}$ to not be an upperbound for $A$. A number $r \in \mathbb{R}$ is not an upperbound for $A$ if there exists an $a\in A$ such that $r<a$. This is just the negation of the definition of upperbound.
Now, we can begin the proof. Let $s=\sup A$ and $\epsilon > 0$ be given. Now, we note that $s-\epsilon$ is no longer an upperbound for $a$ (This is due to the contrapositive of the second part of the definition). Since, $s-\epsilon$ is not an upperbound for $A$, there must be an $a\in A$ such that $s-\epsilon < a$. Now, that $a\in A$ and $s$ is an upperbound for $A$, then $a\le s$. Hence, we have $s-\epsilon < a \le s$.
A: There are two aspects to being $\sup A$.
1.$\sup A$ is an upper bound
This means i) for all $a \in A$ then $a \le \sup A$ and ii) for all $b > \sup A$, $b\not \in A$.


*$\sup A$ is the least upper bound.


This means i) if $b$ is an upper bound of $A$ then $b \ge A$; ii) if $b < \sup A$ then $b$ is not an upper bound.
Now $\sup A - \epsilon < \sup A$.
So $\sup A - \epsilon$ is not an upper bound of $A$.
So it is not true that all $a \in A$ are less than or equal to $\sup A - \epsilon$.
So there must be at least one $a \in A$ so that is more than $\sup A - \epsilon$.
Let $a$ be that element.
$\sup A - \epsilon < a$.
And as $\sup A$ is an upper bound.
$\sup A - \epsilon < a \le \sup A$.
....
I think part of your problem was that you were assuming the $a$ was chosen from the beginning.  Obviously not all $a$ have to be greater than $\sup A - \epsilon$.  We just had to find one that was greater.  If you pick an arbitrary $a$ you can't prove it is $\sup A - \epsilon < a$ because that might not be true.
Instead becuase $\sup A - \epsilon$ is not an upper bound, we know there must be one.  We don't actually no anything about this $a$.  We just know it must exist.
