Prove that, if $S = \{$real numbers $x > 0 : x^2 < 2\}$, then for every real number $\epsilon > 0$, there is an element $x \in S$ such that $x^2 > 2 − \epsilon$.
My Proof
Proposition: If $S = \{$real numbers $x > 0 : x^2 < 2\}$, then for every real number $\epsilon > 0$, there is an element $x \in S$ such that $x^2 > 2 − \epsilon$.
A (Hypothesis): $S = \{$real numbers $x > 0 : x^2 < 2\}$.
B (Conclusion): For every real number $\epsilon > 0$, there is an element $x \in S$ such that $x^2 > 2 − \epsilon$.
A1: Let $\epsilon > 0$ be a real number.
B1: There is an element $x \in S$ such that $x^2 > 2 – \epsilon$.
A2: Let $x \in S$.
B2: $x^2 > 2 – \epsilon$
$\implies x > \sqrt{2 – \epsilon} > 0$ and $x > 0 > -\sqrt{2 – \epsilon}$
A3: $ 2 > x^2 > 0$
$\implies 2 > x > 0$ since $x > 0$
A4: $2 > \epsilon^2 > 0$
$\implies 2 > \epsilon > 0$ since $\epsilon > 0$
$\implies 2 – \epsilon > 0 > -\epsilon$
$\implies \sqrt{2 – \epsilon} > 0 > -\epsilon$
$\implies -\sqrt{2 – \epsilon} < 0 < \epsilon$
A5: $2 > x > 0 > -\sqrt{2 – \epsilon}$
$\implies 4 > x^2 > 0 > 2 – \epsilon$
$\therefore x^2 > 2 – \epsilon$ $Q.E.D.$
$\implies x > \sqrt{2 – \epsilon}$
I would greatly appreciate it if people could please take the time to review my proof to ensure that it is correct.
EDIT
After reading the responses, I have attempted to fix my proof.
A (Hypothesis): $S = \{$real numbers $x > 0 : x^2 < 2\}$.
B (Conclusion): For every real number $\epsilon > 0$, there is an element $x \in S$ such that $x^2 > 2 − \epsilon$.
A1: Let $\epsilon > 0$ be a real number.
B1: There is an element $x \in S$ such that $x^2 > 2 – \epsilon$.
A2: Let $x \in S$.
B2: $x^2 > 2 – \epsilon$
$\Leftarrow x > \sqrt{2 – \epsilon} > 0$ where $2 > \epsilon > 0$
A3: $2 > x^2 > 0$
$\implies 2 > \sqrt{2} > x > 0$ since $x > 0$.
A4: $2 > \epsilon^2 > 0$
$\implies 2 > \sqrt{2} > \epsilon > 0$ since $\epsilon > 0$.
$\implies 2 > 2 – \epsilon > 0$
$\implies \sqrt{2} > \sqrt{2 – \epsilon} > 0$
A5: Let $\sqrt{2} > x > \sqrt{2 – \epsilon} > 0$
Now we have to show that $x^2 > 2 – \epsilon$ (B2).
$\implies 2 > x^2 > 2 – \epsilon > 0$
$\therefore x^2 > 2 – \epsilon$ $Q.E.D.$
Best Answer
The idea lurking behind this question might be to use an iteration procedure $x\to A(x)$ converging monotonically to $\sqrt2$, for example,
That is, define $x_0=1$ and, for every $n$, $$x_{n+1}=A(x_n)$$ Since every $x_n$ is positive, the goal becomes to show that, for every $\epsilon>0$, there exists some $n$ such that $$2-\epsilon<x_n^2<2$$ To prove this, note that the function $A$ is increasing on $[1,\sqrt2]$, with $A(x)>x$ for every $x$ in $[1,\sqrt2)$ and $A(\sqrt2)=\sqrt2$.
Thus, $x_n<\sqrt2$ for every $n$ and $x_n\to\sqrt2$ when $n\to\infty$. In particular, for every $\epsilon>0$, choosing $n$ large enough yields $x_n$ such that $2-\epsilon<x_n^2<2$.
More quantitatively, note that $2-x_0^2=1$ and that, for every $x<\sqrt2$, $$2-A(x)^2=\frac{2(2-x^2)^2}{(2+x^2)^2}<\frac12(2-x^2)$$ Thus, for every positive $n$, $$2-x_n^2<\frac1{2^n}$$ which indicates that every $n\geqslant1-\log_2\epsilon$ yields $x_n$ such that $2-\epsilon<x_n^2<2$.
Remarks: (1) Every $x_n$ is a rational number hence this actually proves the result for the smaller set $S'=\{x\in\mathbb Q\mid x>0,x^2<2\}$.
(2) No square root is involved in the procedure or in the proof, only rational numbers and rational functions.
(3) The convergence $x_n\to\sqrt2$ is much faster than geometric, since $$\lim_{n\to\infty}\frac{2-x_{n+1}^2}{(2-x_n^2)^2}=\frac18$$ hence each iteration roughly doubles the number of exact digits of $\sqrt2$ in $x_n$.
(4) Other iteration schemes are available, for example, the function $$\bar A(x)=\frac{3x+4}{2x+3}$$ starting from $\bar x_0=1$, yields an increasing sequence $(\bar x_n)$ converging to $\sqrt2$, but "only" at a geometric rate since $$\lim_{n\to\infty}\frac{\sqrt2-\bar x_{n+1}}{\sqrt2-\bar x_n}=(3-2\sqrt2)^2$$ For example, $$x_4=\frac{941664}{665857}\approx\mathbf{1.41421356237}15\qquad\bar x_4=\frac{1393}{985}\approx\mathbf{1.414213}1979695$$ and $$\sqrt2\approx1.4142135623731$$