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I think it is obvious that the supremum of an interval of the form $[a,b)$ is $b$ but I am trying to show it myself.

My try:

$b$ is an upper bound, if it is a supremum, then $\exists x \in [a,b)$ s.t. $x+\epsilon>b\geq x$ for some $\epsilon>0$ (epsilon characterization of the supremum)

Assume it is not, then $\forall x \in [a,b)$, $ x+\epsilon\leq b <x$ but this means that $b$ isn't an upper bound, contradiction. Thus, $b$ must be the supremum of the interval. (also, one can notice that the inequality would imply that $\epsilon<0$ which is another contradiction)

Any help in verifying my error, correcting it, or suggesting better ways to prove this is appreciated.

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3 Answers 3

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Your proof is confusingly written.

Problem 1:

if it is a supremum, then $\exists x \in [a,b)$ s.t. $x+\epsilon>b\geq x$ for some $\epsilon>0$ (epsilon characterization of the supremum)

This is not the epsilon characterization of the supremum. The characterization says that:

$s$ is a supremum of $A$ if, for every $\epsilon > 0$, there exists some $x\in A$ such that $s-\epsilon < a$.

This is completely different from what you wrote. Most importantly, the definition of supremum says something is true for every epsilon, not "for some $\epsilon > 0$ which is what you wrote.

Problem 2:

You wrote "$\forall x \in b$", which makes no sense. $b$ is not a set, so $x\in b$ is nonsensical.

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  • $\begingroup$ Problem 2 is a typo which I will correct, and concerning problem 1 I already replied to this $\endgroup$
    – Sergio
    Sep 29, 2021 at 12:22
  • $\begingroup$ @Sergio I do not know where you replied to this, but the sentence "if it is a supremum, then $\exists x \in [a,b)$ s.t. $x+\epsilon>b\geq x$ for some $\epsilon>0$ (epsilon characterization of the supremum)" is wrong, plain and simple. Maybe you mean "if it is not a supremum"? I don't know, I can't read your mind. $\endgroup$
    – 5xum
    Sep 29, 2021 at 12:34
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You have the right idea, but your formulation of the epsilon chracterization of the supremum isn't correct. It should be

$$\forall \epsilon > 0\quad \exists x\in [a,b) : x+\epsilon > b\quad (+)$$ The negation of that would be $$\exists \epsilon > 0\quad \forall x\in [a,b): x+\epsilon \leq b\quad (*)$$ By choosing $x$ in a suitable way, you can now show that the inequality $(*)$ can't hold for every $x\in [a,b)$. For example, define $x$ by $x:=\text{max}\{a, b-\frac{\epsilon}{2}\}\in [a,b)$. It follows that

$$x+\epsilon \geq b-\frac{\epsilon}{2} + \epsilon = b+\frac{\epsilon}{2} >b,$$ which is a contradiction.

Also, the negation of $x+\epsilon > b\geq x$ isn't $x+\epsilon \leq b<x$.

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  • $\begingroup$ what would be the correct negation? $\endgroup$
    – Sergio
    Sep 29, 2021 at 12:23
  • $\begingroup$ You are looking at two inequalities. $x+\epsilon > b \geq x$ means $x+\epsilon > b$ and $b\geq x$. The negation would be $x+\epsilon \leq b$ or $b<x$, since $$\neg(a\wedge b) \Leftrightarrow \neg a \vee \neg b$$ $\endgroup$
    – Mandelbrot
    Sep 29, 2021 at 12:24
  • $\begingroup$ so I am assuming we could pick $\epsilon=b-x+1$ which wouldn't make us reach a contradiction? $\endgroup$
    – Sergio
    Sep 29, 2021 at 15:28
  • $\begingroup$ The $\epsilon$ in the negation is fixed. We only now that such an $\epsilon$ with the given property exists. We don't know what it looks like and we can't choose it either. What we can choose however is $x$. I have also made a mistake in my answer. In the contradiction part, we don't show that $b$ isn't an upper bound. We simply show that the inequality $x+\epsilon\leq b$ can't hold for every $x\in [a,b)$. $\endgroup$
    – Mandelbrot
    Sep 29, 2021 at 15:55
  • $\begingroup$ thank you! all is clear now $\endgroup$
    – Sergio
    Sep 29, 2021 at 16:11
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The definition of an upper bound $y$ being a supremum of the set $A$ is that for all $\epsilon >0$, there exists $x \in A$ such that $y < x+\epsilon$. Your quantifiers are not correct.

The converse to this statement then (that is if $y$ is not a supremum) is that there exists some $\epsilon >0$ such that for all $x \in A$, $x+\epsilon \leq y$.

So a proof by contradiction would start:

"Suppose that $b$ is an upper bound of $[a,b)$ but $b$ is not the supremum of $[a,b)$. Then there exists an $\epsilon >0$ such that for all $x \in [a,b)$, $x+\epsilon \leq b$. Then, ..."

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  • $\begingroup$ but if 𝜖>0 wouldn't that imply 𝑥+𝜖>𝑥? also, we already know 𝑦>𝑥 as it is an upper bound, so where is the fallacy in what I stated? ( I am not trying to argue, genuinely wondering) $\endgroup$
    – Sergio
    Sep 29, 2021 at 12:20
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    $\begingroup$ Yes, 𝜖>0 implies 𝑥+𝜖>𝑥. 𝑦 being an upper bound only implies 𝑦≥𝑥 for all 𝑥 in your given set. Your mistake is that you haven't used the characterization of the supremum correctly. $\endgroup$
    – Mandelbrot
    Sep 29, 2021 at 12:23
  • $\begingroup$ Your fallacy was basing your argument on the wrong definition of supremum. $\endgroup$
    – Dan Rust
    Sep 29, 2021 at 12:25

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