0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ That is a valid proof but seems more complicated than necessary. We are given the setm5 $\endgroup$ – user247327 Mar 22 at 10:10
1
$\begingroup$

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 $n \geq 1)$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ i wonder if this does proves anything $\endgroup$ – J. Deff Mar 25 at 4:44
  • $\begingroup$ How can we reason that supremum cannot be less than square root of 2 f using this $\endgroup$ – J. Deff Mar 25 at 5:13
  • $\begingroup$ 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$ $\endgroup$ – The Student Mar 25 at 22:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.