The supremum of the set $X = \{x \in\mathbb{R} : x^2 < 7\}$ is $\sqrt7$.

Question

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.

Answer 1

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 $y|y^2 > 7$ 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 $7 - s^2 \le 7-4 = 3$ so if $0< d' = \frac {7-s^2} 3< 1$ and $d = \frac {7-s^2}{3s}$ we are done.

$(s + d)^2 = s^2 + 2sd + d^2 = s^2 + 2s\frac {d'}s + \frac {d'^2}{s^2}$

$< s^2 + 2d' + d'^2 \le s^2 + 2d' + d' = s^2 + 3d'$

$= s^2 + 3\frac {7-s^2}3 = s^2 + 7 - s^2 = 7$.

So $X$ has no maximum. And $x= \sup X \not \in X$ so $x^2 \ge 7$.

We can do the same for $t^2 > 7$

Answer 2

If you assume $a > 0$, clearly $$ X = \{x\in \mathbb{R} \ : \ x^2 < a \} = \{x\in \mathbb{R} \ : \ - \sqrt{a} < x < \sqrt{a} \} = (-\sqrt{a},\sqrt{a}) $$ So $\sup X = \sup (-\sqrt{a}, \sqrt{a}) = \sqrt{a}$. Now let $a=7 \qquad \qquad \blacksquare$

Edit: added that $a>0$.