Prove that if $ \sup\{r \in \mathbb{R} | r>0, r^2<2\}=A$ then $A^2\geq 2$

Question

I need to find $$A= \sup\{r \in \mathbb{R} | r>0, r^2<2\}$$ and use a=A, show that $$A^2 \geq 2$$ But this also needs to use a previous result showing that $$a^2<2$$ The recommended way to prove is to use contradiction. Now it seems obvious that $A^2$ would have to be greater than or equal to 2 for it to be the supremum (as the supremum is $\sqrt{2}$). I can't work out how to formalise this argument. It seems that since '$a$' belongs to the interval and $A$ (the supremum) doesn't then that is the contradiction? is that correct? and how would I formalise this as a proof.

Answer 1

From the definition of supremum we know that:

1.$A\geq a$ for every $a$ $\epsilon$ $\{r \in \mathbb{R} | r>0, r^2<2\}$ and

2.for every $\epsilon>0$ there exists an $a$ $\epsilon$ $\{r \in \mathbb{R} | r>0, r^2<2\}$ such that $A\geq a>A-\epsilon$

For the sake of contradiction suppose that $A<2$ and $A= \sup\{r \in \mathbb{R} | r>0, r^2<2\}$ .

So if we can find an $\epsilon>0$ such that $A+\epsilon$ is contained in the set $\{r \in \mathbb{R} | r>0, r^2<2\}$ then we have proved that $A$ can not be a supremum of that set.

Let this $\epsilon=1/n$ , so we need to prove that there exists an $n$ $\epsilon$ $ \mathbb{N}$ such that $A^2<(A+1/n)^2<2$.

Indeed for $n>(2A+1)/(2-A^2)$ the inequality above holds . So we have found an element of the set $\{r \in \mathbb{R} | r>0, r^2<2\}$ that is greater that the supremum of that set, this is a contradiction.

Therefore that $A^2<2$ can't be a supremum of the set $\{r \in \mathbb{R} | r>0, r^2<2\}$

Answer 2

$A$ is not a set, it is a number. It is the supremum of a set. If somebody claims that the set has a sup $B$ such that $B^2 \lt 2$ you can prove them wrong by exhibiting a $C$ that is greater than $B$ and has $C^2 \lt 2$ so $C$ is in the set and $B$ is not the supremum. This means $A^2 \ge 2$

Answer 3

Here is a formal argument: $(\sqrt 2)(1-\frac 1 n)$ $\in \{r \in \mathbb R:r>0,r^{2}<2\}$ for any positive integer n. Hence $A \geq (\sqrt 2)(1-\frac 1 n)$ and $A^{2} \geq 2(1-\frac 1 n)^{2}$. Letting $n \to \infty$ we get $A^{2} \geq 2$.