The Least Upper Bound and The Greatest Lower Bound I am taking math class, and I am not sure about LUB and GLB. I need someone to give this dummy a short explanation about them....
On interval (0,10), 0 is a lower bound and 10 is upper bound, but since they are not included in interval I thought 1 and 9 are the least and greatest. Is it right? 
If [0,10], 0 and 10 are in the interval, so they are the least and greatest. 
What is the best way to prove the least and greatest bound? I practiced few it is still not clear.
Thank you so much,
 A: The interval $(0,10)$ consists of all real numbers $x$ such that $0\lt x\lt 10$. In particular, $9.93$ is in $(0,10)$, so $9$ is not an upper bound for our set. Similarly, $1$ is not a lower bound. 
There is no general way to prove that $b$ is the least upper bound of a set $S$. But the problem naturally breaks up into two parts, which in general should be handled separately. 
We need to show that (i) $s\le b$ for every $s$ in $S$ and (ii) there is no "cheaper" upper bound than $b$. That means that for any $b'\lt b$, there is an $s$ in $S$ such that $s\gt b'$.
Let's apply that strategy to $S=(0,10)$.
(i) It is obvious from the definition of $S$ that any $s\in S$ is $\le 10$. In fact, $s\lt 10$.
(ii) Suppose that $b'\lt 10$. We show that $b'$ is not an upper bound of $S$. 
If $b'\le 0$, then clearly $b'$ is not an upper bound for $S$. If $b'\ge 0$,
let $s=\frac{b'+10}{2}$, the average of $b'$ and $10$. Then $s$ is strictly between $b'$ and $10$. This shows that $b'$ is not an upper bound of $S$.
Greatest lower bounds are handled in basically the same way. 
Remark: The little bit about the case $b'\lt 0$ was just pedantry. If $b'=-20$, and $s=\frac{b'+10}{2}$, then $s$ is not in $S$, and we needed $s$ to be in $S$.
A: The interval $(0,10)$ includes all real numbers $x$ satisfying the inequality $0<x<10$, not just the integers in that interval. For instance, $0<\frac12<10)$, so $\frac12\in(0,1)$; and $\frac12<1$, so $1$ is not a lower bound for the interval. Similarly, $9$ is not an upper bound: $9+\frac12$ is larger than $9$ and is still in the interval $(0,10)$.
In fact, if $x$ is any element of the interval $(0,10)$, then $0<\frac{x}2<x<10$, so $\frac{x}2$ is also in the interval and is smaller than $x$. This shows that $(0,10)$ has no smallest element. It also shows that no positive real number is a lower bound for the interval. On the other hand, $0<x$ for every $x$ in the interval, so $0$ is a lower bound for the interval; and since no positive number is a lower bound, $0$ must be the greatest lower bound of the interval.
The reasoning showing that $10$ is the least (smallest) upper bound of $(0,10)$ is similar. Every real number $u\ge 10$ is clearly an upper bound. On the other hand, if $x<10$, then either $x<9$, in which case $9$ is a member of $(0,10)$ bigger than $x$, or $$x<\frac12(x+10)<10\;,$$ and $\frac12(x+10)$  is a member of the interval bigger than $x$. ($\frac12(x+10)$ is just the average of $x$ and $10$, which is halfway between $x$ and $10$.) In each case $x$ cannot be an upper bound for $(0,10)$, because something in $(0,10)$ is bigger than $x$. Thus, no number smaller than $10$ is an upper bound for $(0,10)$, and $10$ is an upper bound, so it must be the least upper bound.
If a set has a smallest element, that element is always the greatest lower bound. Similarly, if it has a largest element, that will be the least upper bound. But with a set like $(0,10)$, which has no smallest or largest element, you have to look a bit more closely to see what’s happening.
A: i think  with ma understanding, having a  interval(0,10) it is a bounded sequence so to determine your upper bound and lower bound for a bounded sequence, draw a number line and indicate the numbers in the intervals , move to the right hand side, the highest no is your upper bound, and to the left the lowest number is your lower bound
