# how show that $b$ is the supremum of an interval $[a,b)$

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 $$\forall x \in [a,b)$$, $$x+\epsilon\leq b but this means that $$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.

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.

• Problem 2 is a typo which I will correct, and concerning problem 1 I already replied to this Sep 29, 2021 at 12:22
• @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.
– 5xum
Sep 29, 2021 at 12:34

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.

• what would be the correct negation? Sep 29, 2021 at 12:23
• 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 $$\neg(a\wedge b) \Leftrightarrow \neg a \vee \neg b$$ Sep 29, 2021 at 12:24
• so I am assuming we could pick $\epsilon=b-x+1$ which wouldn't make us reach a contradiction? Sep 29, 2021 at 15:28
• 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 $x\in [a,b)$. Sep 29, 2021 at 15:55
• thank you! all is clear now Sep 29, 2021 at 16:11

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, ..."

• 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) Sep 29, 2021 at 12:20
• 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. Sep 29, 2021 at 12:23
• Your fallacy was basing your argument on the wrong definition of supremum. Sep 29, 2021 at 12:25