Here we are given different sets, and we can know the range of elements in the set by the least upper bound (LUB) and the greatest lower bound (GLB).
An upper bound of a set $\mathbf{S}$ is an element of k which is greater than or equal to every element of $\mathbf{S}$.
For example: 7 is a upper bound of the set {5,6,7}. So are 7, 8, and 9.
A set which has an upper bound is said to be bounded above .
The Least Upper Bound (LUB) is the smallest element in upper bounds.
For example: 7 is the LUB of the set {5,6,7}.
The LUB also called supermun (SUP), whihc is the greatest element in the set.
LUB needs not be in the set.
Any element that is greater than LUB, does not belong to the set.
A set may have infinite upper bound, but have at most one LUB.
Sets with no upper bound have no LUB.
Not every subset of $\mathbf{Q}$ has a LUB in $\mathbf{Q}$.
Consider a set $\mathbf{x}: \{x | x \in \mathbf{Q} \text{ and } x^2 < 2 \}, $ prove that $\frac{3}{2}$ is a upper bound of $\mathbf{x}$.
Prove: if $x = \frac{3}{2}$, $x^2 = \frac{9}{4} > 2. $ Thus, $\frac{3}{2}$ is a upper bound.
Consider a set $\mathbf{x}: \{x | x \in \mathbf{Q} \text{ and } x^2 < 2 \}, $ prove that $\mathbf{x}$ has an upper bound, but does not have a LUB in \mathbf{Q}.
Prove: let $y = \frac{p}{q}$ be a value, so that all $x < y, x \in \mathbf{X}$.
1. Let y is the first rational number so that $y^2 > 2.$
Consider $\hat{y} = \frac{4+3y}{3+2y}.$
$\hat{y}^2 - 2 = \frac{y^2 - 2}{(3+2y)^2} > 0$
$\hat{y}^2 > 2$
Also $\hat{y} < y$, so we cannot have a y so that $y^2 > 2$.
2. What about $y^2 < 2?$
Again $\hat{y} = \frac{4+3y}{3+2y}.$
$\hat{y}^2 - 2 = \frac{y^2 - 2}{(3+2y)^2} < 0$
By similiar way, we cannot find such a y.
Thus, $\mathbf{x}$ has an upper bound, but does not have a LUB in \mathbf{Q}.
Since we have upper bound, lower bound works the opposite way.