Understanding the Least Upper Bound I just started studying Real Analysis, and I came across this website: http://mathonline.wikidot.com/proofs-regarding-the-supremum-or-infimum-of-a-bounded-set that talks about the supremum and infimum of a set.
It says...
For a bounded set S to have a least upper bound $Sup(A)=u$ for some $u \in \mathbb R $ you have to show two things:


*

*That $u$ is an upper bound to the set S. Meaning $ \forall a \in A$, $a \leq u$

*$u$ is the least upper bound. That is if $b \lt u$ then $\exists a \in A$ such that $b \lt a$
The first part is simple enough, but I'm not quite sure I understand the motivation behind this $b$. 
Why do we care that it's less than some $a$? Is it because it was stated earlier that $a \leq u$, and so if $b \lt a$ then it can't be greater than $u$? So $u$ must be a least upper bound?
 A: Part 2 says that nothing less than $u$ can be an upper bound--that is $u$ is the $\underline{\rm least}$ upper bound.
So if you take anything less than $u$, it won't be an upper bound (it will be exceeded by something in set $A$).
A: As a complimentary answer, it might be intuitive to see what $2$ is equivalent to.
Again given that $u$ is to be a least upper bound, $u$ is an upper bound.
$$2. \text{ If } b<u \text{ then } \exists a \in A,\, b<a $$
$$2'. \text{ If } x\in \Bbb{R}, \text{ and } \forall a\in A,\, a\leq x , \text{ then } u\leq x$$
$2'$ is saying exactly what you expect, it's less than (or equal to) any other upper bound.
Say that $u$ is a least upper bound using $2$ then supposing that $x$ is another upper bound. If it were that $x<u$, then there is an $a\in A$ such that $x<a\leq u$ by property $2$, so it must be that $u\leq x$.
Now instead let the definition of upper bound use $2'$. Suppose that we have a $b<u$: if there is no $a\in A$ with $b<a$, then $b$ is an upper bound to $A$ and by property $2'$ it would have to be that $u\leq b$. So it must be the case that there is an $a \in A$ with $b< a$. 
A: I will first summarize the conditions that need to be met in order for an upper bound to be a supremum of an arbitrary set A:
condition 1 says that u needs to be an upper bound (obviously this means ∀a∈A, a≤u)
condition 2 says that any number b < u cannot be an upper bound By saying that there must be at least one a in A that is bigger than b. This way b does not satisfy condition 1!
your reasoning is right: we want u to be the least upper bound, so any b < u should not be an upper bound of A. Therefore any b < u should not satisfy the requirement ∀a∈A, a≤b. So there must necessarily be some a > b! 
In a short summary (1) u must be an upper bound of set A and (2) any b < u cannot be an upper bound of set A. Therefore u is the least upper bound, or supremum, of set A. This is usually denoted sup(A)=u.  
Here is a picture

