# What's the idea behind proving $\sup(0,1)=1$?

What's the idea behind proving $$\sup(0,1)=1$$?

I made a plan on how to prove such statements for any $$\sup M = a$$:

(a) $$a$$ is an upper bound of $$M$$

(b) $$\forall \varepsilon$$ $$\exists x\in M: x>a-\varepsilon$$

Proof:

a) is easy to prove. $$1$$ is an upper bound of $$x\in (0,1) \Leftrightarrow 0.

b) Let $$\varepsilon >0$$, then we have to show that there exists an $$x\in M$$ such that $$x>1-\varepsilon$$ is true. How can I do that? What's the reasoning behind choosing a valid $$x$$?

• Try it for a few $\epsilon$'s - say $\epsilon = .1, .01, .001$. Can you see how to do it in general? – user113102 Jul 30 at 14:37
• Note that the set $\mathcal I_\epsilon = (1-\epsilon,1)$ has "some" elements for every $\epsilon \in (0,1)$ – Dominik Kutek Jul 30 at 14:37
• See this post: Let $A = [0,1)$. Then $\sup(A) = 1$ – Jack Jul 30 at 14:40
• @GreyFox: that is rather careless! What if $\varepsilon\ge 2$? – TonyK Jul 30 at 14:40

I always always like to draw a picture. Draw the number line. Draw the points $$1$$ and $$1 - \epsilon$$. You need to show there is a number between $$1- \epsilon$$ and $$\epsilon$$. Can you construct one? Meaning, can you give a formula for such a number, in terms of $$\epsilon$$? How about the average of $$1 - \epsilon$$ and $$1$$? That should be between, shouldn't it? Can you prove it is between?

• What about $M:\{1/n : n\in \mathbb{N}\}$? $0$ is an lower bound. Now, $\exists x\in M: x<0-\varepsilon$. What now, $\varepsilon$ is negative? – ParabolicAlcoholic Jul 30 at 17:46
• @ParabolicAlcoholic your condition $(b)$ is a condition necessary for $a$ to be be the supremum. In order for $a$ to be the infimum, it needs to satisfy $\exists x\in M: x>a+\varepsilon$. I just want to emphasize again that you should try to get really good at thinking of things visually; in these basic examples, if the visual disagrees with the symbolic, it's more likely that you've got a misunderstanding of the symbollic. – Ovi Jul 30 at 18:48

Forget about $$\varepsilon$$ for a minute and prove the following:

Lemma: If $$x \in (0,1)$$ there exists a $$y \in (0,1)$$ such that $$x \lt y$$.

This section is provided to give an argument based on the 'geometry' of the linear ordering.

You can demonstrate that $$\sup(0,1)=1$$ is true without using an $$\varepsilon$$.

Let $$\alpha = \sup(0,1)$$.

The set of real numbers is the disjoint union

$$\tag 1 (-\infty,0] \cup (0,1) \cup [1,+\infty)$$

By the lemma, $$\alpha \notin (0,1)$$.

It is trivial to show that no element in $$(-\infty,0]$$ can be an upper bound.

So it must be true that $$\alpha \in [1,+\infty)$$ (it has to be in one of the three partition blocks).

The number $$1$$ is an upper bound for $$(0,1)$$.

Since $$\alpha$$ is less than or equal to any upper bound, $$\alpha \le 1$$.

Since $$\alpha \in [1,+\infty)$$, $$\alpha \ge 1$$.

We conclude that $$\alpha = 1$$.

The question seems, “Given $$\epsilon>0$$, how can I show that there exists $$x$$ such that $$x > 1-\epsilon$$?”

One strategy is to solve the question, “Given $$\epsilon>0$$ and $$x > 1 - \epsilon$$, what is $$x$$?”. If you can solve for $$x$$ then you’ve implicitly shown that $$x$$ exists (“and here it is”).

When this strategy doesn’t work, we fall back on techniques that establish the existence of $$x$$ but not its value.

• Thanks, that really helped aswell! – ParabolicAlcoholic Jul 30 at 15:07