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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?

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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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 $A$.

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  • $\begingroup$ 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

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