To show that square of the supremum of the set {$t \in \mathbb{R} | t^2 < 2 $} cannot be greater than $2$

Question

Given the set T= {$t \in \mathbb{R} | t^2 < 2 $}.

Take $SupT = \alpha$ and assume that $\alpha ^ 2 > 2$

Now ($\alpha - \frac{1}{n})^2 = \alpha ^ 2 - \frac{2 \alpha}{n} + \frac{1}{n^2}$} > $\alpha ^ 2 - \frac{2 \alpha}{n}$.

Now by archimedian property we have that $\frac{1}{n} < y$ , where y > 0 is real number. So $\frac{-1}{n} < -y$. Now $-y < 0$. Now take $y= \frac {2 - \alpha^2}{2 \alpha}$. Since \alpha is positive so $y < 0$. So we have ($\alpha - \frac{1}{n})^2 > 2$.
Now we have assumed that $\alpha$ is supremum of the set and $\alpha ^ 2 > 2$., but we have found a number less than supremum which is also bigger than all the elements of the set T but less than $\alpha$ which contradicts that alpha is supremum

Is this correct ? Thanks

Answer 1

Your proof is basically correct but you should have justified the statement that $\alpha - {1 \over n}$ is greater than every member of $T$. This isn't hard since if $\beta \geq \alpha - {1 \over n}$ then $\beta^2 \geq (\alpha - {1 \over n})^2 > 2$ and therefore $\beta \notin T$.

(There's no issue about possibly squaring a negative number and having the inequality reversed since $\alpha \geq 1$ and $n \geq 1)$.

Answer 2

Your proof seems to be right but is more complicated than the necessary. I'm going to do a simpler one. The supremum of the set $T = \{t \in \mathbb{R}: t^2 < 2\}$ is $\sqrt{2}$. Suppose that the supremum $r$ is greater than 2, then $\forall \epsilon > 0$ exists $x \in T$ such that $x \geq r - \epsilon$. Fix $0<\epsilon < r - \sqrt{2} \Leftrightarrow r - \epsilon > \sqrt{2}$. Then there is not element $x \in T$ such that $x > r - \epsilon.$ And you have a contradiction.