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


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

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 $x$?

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

  • $\begingroup$ 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? $\endgroup$ – ParabolicAlcoholic Jul 30 at 17:46
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    $\begingroup$ @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. $\endgroup$ – 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.

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    $\begingroup$ Thanks, that really helped aswell! $\endgroup$ – ParabolicAlcoholic Jul 30 at 15:07

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