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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?

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    $\begingroup$ What's the name of the "the number just below 1"? Or a representation of it; can you write it down? $\endgroup$
    – user296602
    Mar 22, 2018 at 18:02
  • $\begingroup$ 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? $\endgroup$ Mar 22, 2018 at 18:03
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    $\begingroup$ No, there doesn't; if you name a number $r$ below $1$, then $(1 + r)/2$ is a bigger number below $1$. $\endgroup$
    – user296602
    Mar 22, 2018 at 18:04
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    $\begingroup$ @AwnonBhowmik: That's just wrong. Consider the set $S = \{1\}$. Its sup is $1$, and $1 \in S$. $\endgroup$ Mar 22, 2018 at 18:05
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    $\begingroup$ Every finite set (of real numbers) has a maximum and a minimum. $[0,1)$ is not a finite set. $\endgroup$ Mar 22, 2018 at 18:07

5 Answers 5

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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 $\alpha$.

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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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  • $\begingroup$ Does that mean every set of real numbers is infinite, since there are infinite numbers between two numbers? $\endgroup$ Mar 23, 2018 at 12:47
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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?

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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.

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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.

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