# How to prove $X = \{x\in \mathbb{R} \ : \ x^2 < a \}$ has supremum?

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?

• 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 $\sqrt a$? Sep 23 at 15:44
• 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..... Sep 23 at 17:02
• 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..... Sep 23 at 17:11
• 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 $\sup A$. Sep 23 at 17:18
• @fleablood would you please post your last comment as an answer? I want to accept that Sep 24 at 6:51

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.

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

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 $$0\le y, then $$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, a contradiction.

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

• Do you mean $\sup X>\sqrt{a}$? Sep 23 at 16:21
• @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. Sep 23 at 16:40