The supremum of the set $X = \{x \in\mathbb{R} : x^2 < 7\}$ is $\sqrt7$.
It's the exact same as with $\sqrt {2}$.
for positive $a,b$ we know $a < b \iff a^2 < b2$ so $X$ is a "cut", i.e. ever element of $X$ is less than any element not in $X$.
So $X$ is not empty and bounded above by ... say $3$ (or anything whose square is greater than $7$). So $x = \sup X$ exists. If $x^2 < 7$ then $x \in X$ and $x = \max X$. If we can prove $X$ has no maximum element we know that is not possible and $x^2 \ge 7$. If we can prove that for any $yy^2 > 7$ then there is a $w < y w^2 > 7$ then that means if $x^2 > 7$ then $x$ is not a least upper bound. The $x^2 = 7$.
Okay, So if $s^2 < 7$ we need to fine $t > s$ and $t^2 < 7$.
The usual trick is to find a $d 0 < d < 7  s^2$ and then $(s+ d)^2 = s^2 + 2ds + d^2 < 7$ so $2ds + d^2 < 7  s^2$. We we can get $d = \frac {d'}s$ and $s > 1$ and $d' \le 1$ then we'd have $2ds + d^2 = 2d' + \frac {d'^2}{s^2} \le 2d' + d'^2 < 2d' + d' = 3d'$. So... heck with it... $2^2 = 4 < 7$ so we can assume $s \ge 2$ and and so $7  s^2 \le 74 = 3$ so if $0< d' = \frac {7s^2} 3< 1$ and $d = \frac {7s^2}{3s}$ we are done.
$(s + d)^2 = s^2 + 2sd + d^2 = s^2 + 2s\frac {d'}s + \frac {d'^2}{s^2}$
$< s^2 + 2d' + d'^2 \le s^2 + 2d' + d' = s^2 + 3d'$
$= s^2 + 3\frac {7s^2}3 = s^2 + 7  s^2 = 7$.
So $X$ has no maximum. And $x= \sup X \not \in X$ so $x^2 \ge 7$.
We can do the same for $t^2 > 7$
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amWhy
Updated on May 19, 2020Comments

amWhy over 3 years
I'm able to show that $\sqrt2$ is the supremum of $X = \{x \in\mathbb{R} : x^2 < 2\},\;$
but I'm having trouble with this set with the set in the title.

Display name over 6 yearsJust prove that the supremum of $X = {\{x ∈ R : x^2 < a}\}$ is $√a$ and then set $a=7$.

Reese Johnston over 6 yearsCan you elaborate on how you showed it for $2$ rather than $7$? For most proof strategies, the $2$ shouldn't have been particularly relevant, so it should work exactly the same for $7$.

copper.hat over 6 yearsThe title is false.

egreg over 6 yearsIf you are able to prove that the supremum is $\sqrt{2}$, how can it be $\sqrt{7}$? A set has at most one supremum.

amWhy over 6 yearsThe asker's question/title, was exactly what we see in the title now; an earlier edit (by @Rafael ) changed the title to ask $\sup x^2 \lt 2$ is $\sqrt 7$.

Display name over 6 yearsLooks like OP changed the question to fix any errors. Anyways, my earlier comment still stands.

fleablood over 6 yearsIf you actually prove that for all $b > 0$ then there exist a unique $a>0$ such that $a^2 = b$ and we call this $a = \sqrt b$ then it is trivial to show that $\sqrt 7$ is an upper bound of $X$ and that if $y < \sqrt{7}$ is not an upper bound (as $y < w < \sqrt{7}$ would imply $w \in X$) so $\sqrt{7} = \sup X$. We usually use the excercise $\sqrt{2}$ exists as means of introducing concepts. Then you prove $\sqrt {b}$ always exist. Then the question is moot.


Vivek Kaushik over 6 yearsWhat's your proof that $\sup(\sqrt{a},\sqrt{a}) =\sqrt{a}$ ?

Mikkel Rev over 6 yearsSuppose that $a >0$, then $\sqrt a$ is an upper bound. To see this, use the defintion 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 $M \in \mathbb R$ 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 suprermum, But if $z \geq \sqrt a$, then $\sqrt a \leq \frac{\sqrt a  M}{\sqrt a} + M \implies \sqrt a \leq M$, a contradiction