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Suppose $\emptyset \neq A \subset \mathbb{R} $. Let $A = [\,0,2).\,\,$ Prove that $\sup A = 2$

This is my attempt:

$A$ is the half open interval $[\,0,2)$ and so all the $x_i \in A$ look like $0 \leq x_i < 2$ so clearly $2$ is an upper bound.

To show it is the ${\it least}$ upper bound, suppose that $2 \neq \sup A$, that is there exists a number $M < 2$ for some real $M$ qualifying as $\sup A$. Certainly this $M \in [0,2) $ so $ M > 0 \Rightarrow 2 -M > 0$.

By the Archimedean Principle, for all real numbers $r > 0\,\, \exists\,\, n \in \mathbb{N}$ such that $0 < \frac{1}{n} < r $. By the Approximation Property of Suprema, there exists $a \in [0,2)$ such that $\sup A - \epsilon < a \leq \sup A$, where $\epsilon > 0$.

Suppose $\sup A = M < 2$. Then the above gives $M - \epsilon < a < 2\,\,\,\,\forall \epsilon > 0$. Also, by Archimedean, we have $0 < \frac{1}{n} < 2-M$, so choose $\epsilon = 2-M$. Then $M - (2-M) < a < 2 \,\Rightarrow 2(M-1) < a < 2$

We can assume $M - 1>0$ and so $2(M-1) > 2$ This results in a contradiction in the previous inequality. Hence $M < 2$ cannot be the supremum.

I realise there is probably a simpler way, but is what I have written all good?

user315531
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CAF
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  • The first thing would be to say what $A$ is a subset of. Like that, it does not have an upper bound, so no sup, or $+\infty$ if you will. In $[0,2)\cup [3,+\infty)$, the sup is $3$. But in $\mathbb{R}$, of course, the sup is $2$. – Julien May 09 '13 at 16:32
  • Okay, I will edit it. Thanks. Is my proof correct? – CAF May 09 '13 at 16:35
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    So we consider $A=[0,2)\subseteq \mathbb{R}$. Step 1: $A$ is bounded above by $2$. So there is a finite sup and $\sup A\leq 2$. Step 2: if $M$ is a bound of $A$ from above, then $x\leq M$ for every $0\leq x<2$. In particular, $2-\frac{1}{n}\leq M$ for every $n\geq 1$. So $2\leq M$ to the limit. In particular, $2\leq \sup A$. – Julien May 09 '13 at 16:38
  • That is quicker indeed - but is my proof okay? I suppose I could have deleted the bit about Archimedean since I didn't have to use such an $n$. – CAF May 09 '13 at 16:43
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    I see at least one mistake: $M-1>0$ implies $2(M-1)>0$, ont $2(M-1)>2$. And if you wanted the latter to hold, you would need $M-1>1$, that is $M>2$. So I'm afraid you are in a circular reasoning. – Julien May 09 '13 at 16:48
  • I see, that is unfortunate, it was going well up to then. – CAF May 09 '13 at 16:59
  • Is there anywhere I can make a slight change that would make it work? – CAF May 09 '13 at 17:04
  • Somehow, yes. But simly: if $0<\sup A=M<2$, then there exists $M<x<2$, contradicting that $\sup A$ is an upper bound for $[0,2)$. – Julien May 09 '13 at 18:01
  • Maybe I could also use the density of rationals: There exists a rational in the interval $2 - \frac{1}{n} < q < 2$, where $q$ is the possible 'lower' bound < $2$. Take the limit and we get $2 < q < 2$, a contradiction? I think a similar way is done to prove the case if this proposed lower bound was irrational. – CAF May 09 '13 at 22:05
  • No. When you take limits, strict inequalities become large. From the moment you have 1 element between $M$ and $2$, you have your contradiction like I said in my previous comment. There is no limit to take. – Julien May 09 '13 at 22:07
  • Do you mean that < becomes $\leq$? – CAF May 09 '13 at 22:09
  • Yes. That's what I meant. Anyway. It does not make sense at this point to take a limit. One value suffices to get the contradiction. – Julien May 09 '13 at 22:22

3 Answers3

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I think there's a way more simple and intuitive proof.

First, as you observed, it is obvious that 2 is an upper bound. Now, to prove it is the supremum. Assume that $M$ is the supremum and $M<2$.

Of course, $M>2$ is trivially impossible since $2$ is an upper bound as well and thus any $M$ bigger than $2$ cannot be a supremum.

Now, let $x=\frac{2-M}{2}+M$.

As you can obviously see $M<x$, so all that is left is to show that $x<2$.

But now, assume it is not, that is $x\geq2$.

Then, $\frac{2-M}{2}+M\geq2$. Multiplying both sides by $2$, we get

$2-M+2M\geq4$, that is $2+M\geq4$. But $M<2$, so $2+M<4$. Thus, by reductio ad absurdum, $x<2$. This shows that $M<2$ cannot be the supremum.

QED.

Now, as for the simplicity of this proof, I have written a lot for clarity and in case you are a beginner on this subject. This can be summarized in $2$ lines, but this is for clarity. I hope this helps, and you must soon learn to find the shortest and more intuitive way. Good luck.

Hasan Saad
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2

Let $a\lt 2$ is $\sup A$.

Therefore, $a=2-b$ , where $b\gt 0$.

We can get some $n\in \mathbb{N} \,|\, 0 \lt \frac1n \lt b$

So, $2-b\lt2-(\frac1n)\lt2$.

$\exists c\in Q\cap A\,|\,2\gt c\gt 2-(\frac1n)\gt 0$.

So,$\,2\gt c \gt 2-(\frac1n)\gt (2-b)=a$.

$0\lt c \lt 2 \implies c\in A$, and also $c\gt a$

i.e. $a$ is not even an upper bound of $A$.

$\implies$ the set of upper bounds of $A$ is $\{x : x\in \mathbb{R}$ and $x\ge 2\}=S$ and the least member of $S$ is $2$.

So, $2$ is the least upper bound of $A$.

jiten
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ReekMaths
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    In short, this means: "Whatever number you take smaller than 2, there will be a bigger than it number in A, so the number you chose is not an upper bound." Am I right? This answer is more clear to me because Hasan defines x in an unfamiliar way for me. – Al.G. Oct 05 '18 at 21:45
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I find the some of the existing proofs either magic or non sensical, hence I decided to write my own proof and submit to this important question.

Note : I use the definition of supremum as smallest upper bound.

We have that the supremum can't be greater than $2$ very easily by the definition which says that it is the "smallest upper bound". Now, the supremum can be in the interval $(-\infty,2]$.

I claim that it can also not be less than or equal $0$. Before that, we first proof $\frac12$ is an element of $[0,2)$. We have that, $ 0 <1$, dividing by 2 both sides, $0 < \frac12$. Taking $0<1$, add $1$ and dividing both sides again by $2$, we get $ \frac12 < 1$. Hence $0 < \frac12 < 1$.

If it were that the supremum of the set, $s$ were less than or equal $0$, we'd have that:

$$ s \leq 0 < \frac12$$

Which is a contradiction with the part of the definition of the supremum which says that the supremum must be greater than any number inside the set (an upper bound).

Hence the supremum is in the interval $(0,2]$. Let us suppose the supremum $s$ is smaller than $2$, then

$$0<s< s+ \frac{2-s}{2} <2$$

That is, the midpoint between the supremum and two is a number that is in the set (field property of real number) and again bigger than it. But, this is also an element in the given set, which means $s$ can't be a supremum as there exist an element greater than it which is also in the set.

Clemens Bartholdy
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