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?

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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


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.



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.


$\sqrt{a}$ 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.

  • $\begingroup$ 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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