is it true that $a < \sup A \implies \exists b \in A \; \text{such that} \; a $A \subset \mathbb{R}$
'$a < \sup A \implies \exists b \in A \; \text{such that} \; a < b$' ?
let denote $\sup A$ by $s$
then $\forall x \in A, \; x \leq s$
if the sup is attained then clearly $b = \max A$
but I'm stuck in the case where $\sup A \notin A \implies \forall x \in A, \; x < s$
can such a $b$ exist ?
 A: Assuming that $\sup A$ exist then if $a < \sup A$ then $a$ is not an upper bound of $A$ because $\sup A$ is the least upper bound.  
Since $a$ is not an upper bound then, by definition, there is a $b \in A$ so that $a < b$.
That's it.  
That is what the words "upper bound" and "least" mean.  It's simply logically not possible for anything less than the $\sup $ to be an upper bound and it is not logically possible for anything that is not an upper bound to be equal or larger than all elements.
A: Suppose for the sake of insanity that 
$$
\forall b\in A,\;a\geq b
$$
then $a$ is an upper bound for the set $A$. But it's smaller than the $\sup$, which is the least upper bound! This is silly.
edit: this can be reformulated to be just the contrapositive. 
A: $\sup A$ is the least upper bound of $A$. If $a<\sup A$, then $a$ is not an upper bound of $A$. Namely, the following assertion is true: $$\neg(\forall b\in A, b\le a)$$
Id est $$\exists b\in A,\ \neg( b\le a)$$
By the trichotomy property of total orderings, $\neg ( b\le a)\iff a<b$. Thus $$\exists b\in A,\ a<b$$
