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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?

TheSimpliFire
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Soumyaneel Manna
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    What's the name of the "the number just below 1"? Or a representation of it; can you write it down? –  Mar 22 '18 at 18:02
  • I don't know, but there has to be a number under 1, that is in the set and the maximum of the set, isn't it? – Soumyaneel Manna Mar 22 '18 at 18:03
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    No, there doesn't; if you name a number $r$ below $1$, then $(1 + r)/2$ is a bigger number below $1$. –  Mar 22 '18 at 18:04
  • Suppose we call that number $q$. Then it's certainly less than $1$. But then...so is $\frac{1+q}{2}$, which is therefore also in the set...but is bigger than $q$, so $q$ wasn't the number "just below $1$". In short: assuming such a number exists leads to a contradiction. So no such number exists. – John Hughes Mar 22 '18 at 18:05
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    @AwnonBhowmik: That's just wrong. Consider the set $S = {1}$. Its sup is $1$, and $1 \in S$. – John Hughes Mar 22 '18 at 18:05
  • @ Awnon Bhowmik, how is this possible is my question. Please elaborate in an answer – Soumyaneel Manna Mar 22 '18 at 18:06
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    Every finite set (of real numbers) has a maximum and a minimum. $[0,1)$ is not a finite set. – Jair Taylor Mar 22 '18 at 18:07
  • BTW, every finite set of real numbers (except the empty set) has a max and a min. But your set is not finite. It's bounded, but not finite. – John Hughes Mar 22 '18 at 18:07
  • I observe a high degree of redundancy in the above comments. – Michael Mar 22 '18 at 19:10

5 Answers5

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

  2. 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 $\alpha$.

Mike
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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.

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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?

TheSimpliFire
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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.

Ethan Bolker
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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.

Xin Fu
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