Definition of Least Upper Bound Definition $1$. From my lecture note, 


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*Let $A$ be a nonempty subset of $\mathbb{R}$ with a upper bound and $s$ is a real number. We say that $s$ is a least upper bound of $A$ if $s$ is an upper bound of $A$ and if $b$ is any upper bound of $A$ then $s\leq b$.


Is there a mistake? I think that last sentence is a mistake.
Definition $2$. My definition of SUP:


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*Let $A$ be a nonempty subset of $\mathbb{R}$ with a upper bound and $s$ is a real number. We say that $s$ is a least upper bound of $A$ if $s$ is an upper bound of $A$ and if $b$ is not a upper bound of A and $b\leq s$.

 A: The second definition is wrong. In particular the part that says " if $b$ is not a upper bound of A and $b\le s$." If you use this piece as your definition then you are saying that every upper bound is the least upper bound.
EX: Let $(0,1)\subset \mathbb{R}$. Then we'd like to say that $1$ is the least upper bound. But as per the second definition you have, we find that $2$ is an upper bound and since $2$ is not not an upper bound the second condition you have is vacuous. Hence $2$ is a least upper bound. Which is not at all true.
A: Consider the closed interval [0,1].  The definition of upper bound of this set in R is any real number, p, such that if x $\in$ [0,1], then x $\le$ p. Therefore, 1, 5, 34, anything greater than or equal to 1 is an upper bound. 
The least upper bound will be the smallest upper bound of all of these.  If s is the least upper bound of this set, then it must be less than or equal to any other upper bound, b, of [0,1].
It is easy to show that s = 1. 
Proof:
If x $\in$ [0,1], then 0 $\le$ x $\le$ 1. So 1 is an upper bound of [0,1].  Suppose b is an upper bound of [0,1], such that 0 < b < 1.  Then b < 2b < b + 1 < 2, which implies b < (b + 1)/2 < 1.  Thus b cannot be an upper bound of [0,1], since there exists an element of [0,1], (b + 1)/2, greater than b. Hence there is no upper bound less than 1, and s =1. 
