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

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

• That is a valid proof but seems more complicated than necessary. We are given the setm5 – user247327 Mar 22 at 10:10

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 $$n \geq 1)$$.
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.
• From de definition of supremum you can find that $p = sup A$ if and only if for every $\epsilon > 0$ there is an $x \in A$ with $x > p − ε$, and $x \leq p$ for every$x \in A$ – The Student Mar 25 at 22:45