How can a finite set not have a maximum or minimum? Consider a set of real numbers such that $0\le x<1$. My book (Tom Apostol) says that this set do not have any maximum. How is it possible? Isn't the number just below $1$ the maximum of the set? Isn't that the least upper bound (supremum)? Why is $1$ the supremum?
 A: *

*The set of real numbers in $[0,1)$ isn't finite, it is bounded.

*There is no real number "just below" 1. Give any real number $\alpha < 1$
and $\alpha + \frac{(1-\alpha)}{2}$ is also strictly less than 1, and strictly greater than $\alpha$.
A: First of all, every finite set does have a maximum and a minimum element. This is not a finite set, though; it's infinite. I think you've confused "finite" with "bounded."

Secondly, there is no maximum element. If $q$ were the maximum of the set, then $(1 + q)/2$ would be another element of the set, strictly greater than $q$. This is a contradiction, so no maximum exists. 
A: A maximum element $k$ in the set $S=\{x:0\le x<1\}$ is such that no element in $S$ is greater than $k$. Since $k\in S$, we have that $k<1$, so $k\neq 1$. By the Density Property, there is always a number between $k$ and $1$, so $k\neq\max(S)$. This is a contradiction, so there is no maximum.
Can you now follow the method given here to show that $1$ is a supremum?
A: Your argument is based on a serious misunderstanding of the real numbers. 
There is no number "just before $1$", just as there is no point on the number line "right next to" a given point. You can always go halfway from any point to any other. Zeno's dichotomy paradox relies on this idea.
It has taken mathematicians centuries  to figure out how to do calculus and real analysis rigorously in the face of that difficulty. That's what the epsilons and deltas and least upper bounds are for.
The good news is that you now have some motivation to learn to reason with those notions. 
A: For any real number a which is in S.
There exists another number b which is larger than a: b=(a+1)/2.
So there is no maximum.
