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?

  • $\begingroup$ 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 $2$. $\endgroup$ – Julien May 9 '13 at 16:32
  • $\begingroup$ Okay, I will edit it. Thanks. Is my proof correct? $\endgroup$ – CAF May 9 '13 at 16:35
  • 4
    $\begingroup$ 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, $2\leq \sup A$. $\endgroup$ – Julien May 9 '13 at 16:38
  • $\begingroup$ 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 $n$. $\endgroup$ – CAF May 9 '13 at 16:43
  • 1
    $\begingroup$ 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. $\endgroup$ – Julien May 9 '13 at 16:48

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.


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.

  • $\begingroup$ but what is x representing here? $\endgroup$ – Joseph Rock May 7 '20 at 3:59
  • $\begingroup$ @JosephRock The midpoint between $2$ and $M$. $\endgroup$ – Hasan Saad May 11 '20 at 3:36
  • $\begingroup$ how can the midpoint be larger than the supremum? $\endgroup$ – Joseph Rock May 11 '20 at 3:39

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 $A$.

  • 1
    $\begingroup$ 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. $\endgroup$ – Al.G. Oct 5 '18 at 21:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.