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Assume that $a>0$, Suppose we have :
$$X = \{x\in \mathbb{R} \ : \ x^2 < a \}$$
We should prove that this set has a supremum, and that's $\sqrt{a}$ .
I saw this answer on one of the related posts:

Suppose that $a>0$ then $\sqrt{a}$ is an upper bound . To see this, use the definition of an open ball . Also $0 \in (-\sqrt{a},\sqrt{a})$ since $|0|<\sqrt{a}$. Therefore supremum exists. Now assume for contradiction that $\sqrt{a}$ is not the least upper bound. Then there exist $M \in R$ which is the supremum and $M<\sqrt{a}$.Consider $z:=\frac{\sqrt{a}-M}{\sqrt{a}}+M$.By construction $z>M$. it is impossible that $z<\sqrt{a}$ since M is the supremum,But if $\sqrt{a}\leq z$, then $\sqrt{a}\leq\frac{\sqrt{a}-M}{\sqrt{a}}+M \to \sqrt{a}\leq M$ ,contradiction.

My first question:
Is how author recognized that she should use $\frac{\sqrt{a}-M}{\sqrt{a}}+M$ ? Can we determine a logical process to achieve this expression for our needs?

My second question:
I have problem with this part:
$$\sqrt{a}\leq\frac{\sqrt{a}-M}{\sqrt{a}}+M \to \sqrt{a}\leq M$$
Can we conclude from $z>M$ and $\sqrt{a}\leq z$ that $\sqrt{a}\leq M$ ? I think that's not possible!
Last one:
Is there any better way to prove that?

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    $\begingroup$ I don't get this proof. The way $z$ is constructed it lies in between $M$ and $\sqrt a$ so how can it be bigger than $\sqrt a$? $\endgroup$
    – PNDas
    Sep 23 at 15:44
  • $\begingroup$ I'm going to say don't waste time with this and read a real proof. $\sqrt a$ is an upper bound but you don't use an open ball to see this. Just note if $b>\sqrt a$ then $b^2 > a$ and $b\not \in X$ so $b\in X\implies b \le \sqrt a$ so $\sqrt a$ is an upper bound. Then the bit about $0\in(-\sqrt a, \sqrt a)$ so supremum exist is nonsense. Supremum exist because $X$ is bounded above (because $\sqrt a$ is an upper bound) and the reals have the least upper bound property. Then end. Bounded above $=$ supremum exists. Period. To be continued..... $\endgroup$
    – fleablood
    Sep 23 at 17:02
  • $\begingroup$ And the rest is garbage. It's true $z > M$ but there is no reason at all to assume $z< \sqrt{a}$. I think they are trying to say if $M < \sqrt a$ there is a $z$ so that $M < z < \sqrt a$ (which is true, take to average, or any of the infinite points between) but completely whiffed it. They say that's impossible because $M$ is the supremume but there is no argument that $z \in X$ at all. Some attempt must be made (if we are to do this proof, which I wouldn't bother with) of proving $z^2< a$. To be concluded..... $\endgroup$
    – fleablood
    Sep 23 at 17:11
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    $\begingroup$ A far more straightforward proof would be. 1) $\sqrt a$ is an upper bound of $X$. Pf. $b\ge \sqrt a$ then $b^2 > a$ and $b\not \in X$. So for all $b \in X$ we have $b \le \sqrt a$. So $\sqrt a$ is an upper bound of $X$. suppose $M< \sqrt a$. Then let $b$ be such that $\max(0,M) < b < \sqrt a$. Then $b^2 < \sqrt a^2 =a$. So $b \in X$. But $b > M$ so $M$ is no upper bound. So ... by definition... $\sqrt a$ is an upper bound of $X$ and anything less than $\sqrt a$ is not an upper bound so $\sqrt a$ is $\sup A$. $\endgroup$
    – fleablood
    Sep 23 at 17:18
  • $\begingroup$ @fleablood would you please post your last comment as an answer? I want to accept that $\endgroup$ Sep 24 at 6:51

3 Answers 3

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This supposed proof has some flaws. For example, if $a > 1$, then $\frac{\sqrt{a} - M}{\sqrt{a}} < \sqrt{a} - M$, which implies that $\frac{\sqrt{a} - M}{\sqrt{a}} + M < \sqrt{a} - M + M = \sqrt{a}$, contrary to the claim made in the cited argument. So, I'd recommend looking more closely at the accepted answer from the cited post for pointers on this problem.

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$$\sqrt{a}\le\frac{\sqrt{a}-M}{\sqrt{a}}+M\iff\frac{\sqrt{a}-M}{1}\le\frac{\sqrt{a}-M}{\sqrt{a}}$$

This is perfectly plausible if $0<\sqrt{a}<1$. As a direct counterexample, let $a=1/4$ and $M=1/8$. $\sqrt{a}>M$ yet $\frac{\sqrt{a}-M}{\sqrt{a}}+M=\frac{7}{8}>\sqrt{a}$, so we don't get the implication "$\sqrt{a}\le M$, a contradiction." The answer to your first question is: It doesn't matter, because what you've written is wrong! I think this answers your second question too.

As for the last one: I'd prove it like this. I will also completely avoid assuming $\sqrt{a}$ even exists, I will assume only that $a\ge0$ (for, if $a<0$ then $X$ is an empty set with no (or no "meaningful") supremum).

An important point that I'll use without further comment: note that, if $x\in X$ and $0\le y<x$, then $y\in X$ as well (check this yourself).

Let $x_0=\max\{1,a\}.$ Then $x_0^2\ge\max\{1,a^2\}\ge a$ in all cases where $a\le 1$ or $a>1$. Therefore $X$ has an upper bound, $x_0$. $X$ is also nonempty since $0\in X$. Therefore there is indeed a supremum of $X$, call it $\alpha\in\Bbb R$. Necessarily $\alpha\ge0$. We want to show $\alpha^2=a$: that is equivalent to showing $\sqrt{a}$ exists and equals the supremum $\alpha$.

Then we get that, for any $\alpha>\varepsilon>0$, that $(\alpha-\varepsilon)\in X$ hence $\alpha^2-2\varepsilon\alpha+\varepsilon^2\le a$, or: $\alpha^2\le a+2\varepsilon\alpha-\varepsilon^2$. $\alpha$ is a finite number... this inequality holds for all arbitrarily small $\varepsilon$, so it must be that $\alpha^2\le a$. Likewise, for any $\varepsilon>0$, $\alpha+\varepsilon\notin X$ (otherwise $\alpha$ would not be the supremum!) so $\alpha^2+\varepsilon^2+2\varepsilon\alpha>a$. Again, this shall hold for arbitrarily small $\varepsilon$, so it must be that $\alpha^2\ge a$.

It follows that $\alpha^2=a$, so $\alpha=\sqrt{a}$ is the supremum of $X$. You can also show that $-\sqrt{a}$ is $\inf X$. Perhaps this would be a good exercise.

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$\sqrt{a}$ is an upper bound of $X$, hence $\sup X$ exists. Suppose for a contradiction $\sup X < \sqrt{a}$. Then by definition of supremum, there exists $x\in X$ such that $\sqrt{a}< x$, which leads to $a<x^2$, a contradiction.

Thus, $\sup X=\sqrt{a}$ must hold.

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  • $\begingroup$ Do you mean $\sup X>\sqrt{a}$? $\endgroup$
    – Snoop
    Sep 23 at 16:21
  • $\begingroup$ @Snoop No, by definition $\sup X \leqslant \sqrt{a}$, because it is the smallest upper bound. So we see what happens should $\sqrt{a}$ not be the smallest one. $\endgroup$
    – AlvinL
    Sep 23 at 16:40

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