# Explain definition of least upper bound .

The text that I have says something like this,
A real number s is the least upper bound for a set $A \subseteq R$ if it meets the following two criteria:
(1) s is an upper bound of A
(2) if b is any upper bound for A, then $s \le b$
Its the second statement that creates a confusion. I read answers to same question and the people have explained it with an example,
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.

But consider the statement it says, "if b is any upper bound for A" and if A is a set $[0,1]$ then 2, 3, any number above 1 is not a part of set A.
Can anyone explain how the second statement is true and what exactly it wants to convey regarding least upper bound?

• An upper bound doesn't have to be an element of the set. In fact: it usually isn't. If an upper bound is an element of the set, it's necessarily the least upper bound (and also the maximum). – StackTD Jan 19 '17 at 17:00
• Thanks, I was thinking in the wrong direction. So, there are infinitely many upper bounds but the least upper bound is unique and is unique to the set in consideration, and turns out to be the maximum or supremum (precisely) of the set A. – user405401 Jan 19 '17 at 17:09
• Yes, a least upper bound is exactly what it says it is. It is a bound. Which means no points of the set go "beyond" that. It is an upper bound. Which means to points of the set are more than it. And it is the least upper bound which means of all the upper bounds it is the least. A more confusing but somewhat more useful way of looking at it (who cares if 5432 – fleablood Jan 19 '17 at 17:54
• A more confusing way but somewhat more useful way of looking at it. (who cares if $5,403$ is an upper bound of $[0,1]$) is that if $b$ is the least upper bound $S$, then any $a < b$ is not an upper bound of S (as $b$ is least) so there is an $s \in S$ so that $a < s \le b$ (because $a$ is not a bound there is an $s$ beyond it). So $b$ is the "first" element that is equal or larger than all the S, or that doesn't have a s that is larger than it. – fleablood Jan 19 '17 at 18:16
• @fleablood, thanks for example. It was very helpful. – user405401 Jan 19 '17 at 18:53

Condition (1) says that $s$ is an upper bound. Condition (2) says that $s$ is less than or equal to any other upper bound. Hence, least upper bound.
The second condition or guarantees that the least upper bound, if it exists, is unique. Suppose if ${s_1}$ and ${s_2}$ are both least upper bounds for a set $A$, then part (2) of the definition says that both ${s_1} \le {s_2}$ and ${s_2} \le {s_1}$. Thus $s_1 = s_2$. So there can be infinite number of upper bounds to a set, but it can only have one least upper bound.
In your example, $2$ is an upper bound of $[0,1]$. Condition (1) is satisfied by $s=2$. But since $1.5$ is also an upper bound of $[0,1]$, and $1.5 < 2$, condition (2) is not satisfied. So $2$ is not the least upper bound of $[0,1]$.
Now $1$ is an upper bound of $[0,1]$. And in contrast to $2$, any number less than $1$ is not an upper bound of $[0,1]$ (being less than $1$ precludes it from also being $\geq 1$). So $1$ is the least upper bound of $[0,1]$.
• That an that there is a specific least one. If you "universe" is only the rational numbers and your set is the rationals whose squares are less than 2 (i.e. the interval $(-\sqrt{2}, \sqrt{2})$ restricted to only rational numbers) there are plenty of upper bounds but no one of them is smaller than all the others. But if you allow your universe to include irrationals. $\sqrt{2}$ is an upper bound and it is the least of all possible upper bounds. Note: the l.u.b. need not be in the set either. – fleablood Jan 19 '17 at 18:05