Regard Q(rational numbers) as a metric space, with d(p, q) = |p − q|. Let E be the set of all p ∈ Q such that $2 < p^2 < 3$.

Question

So the problem:Regard Q(rational numbers) as a metric space, with d(p, q) = |p − q|. Let E be the set of all p ∈ Q such that $2 < p^2 < 3$. I want to show, that set $E$ is closed.

I did with the help of arguments, that $\sqrt 2$ and $\sqrt 3$ cannot be represented as rational numbers. Then we can see, that all limit points of interval E = Q union [($-\sqrt 3$;$-\sqrt 2$) union ($\sqrt 2$;$\sqrt 3$)] lie in set E. So set is closed.

I just want to ask about specific solution:

But I can't get the alternative approach(which is more rigourous than mine) What does $\sigma$ mean? Why do we need $y = x + h$? Can you explain me this solution?

P.S And is it good, that I used less rigurous proof(but I understood the idea well)? Or should I try to dive in mathematical details?

Answer 1

I will continue the explanation here, since it will be difficult to read in the hint. Your question is: how the author chooses the $\delta$ (note, this is actually a delta, not a sigma! Though they look very similar, delta is often used to denote small numbers).

I will consider the $0\neq x < \sqrt{2}$ case, as in the proof. We aim to show that $x \in Q\setminus E$ has a neighborhood around it which is entirely contained in $Q\setminus E$. That is, does there exist a $\delta$ sufficiently small so that any rational number $y \in (x-\delta,x+\delta)$ is also in $Q\setminus E$?

We can construct the $\delta$ as follows: note that $y\in (x-\delta,x+\delta)$ satisfies

$$y^2 < (x+\delta)^2 \leq x^2 + 2|x|\delta + \delta^2.$$

So, if we want $y\in Q\setminus E$, then we want $x^2+2|x|\delta+\delta^2<2$. A simple way to see what $\delta$ will work is to simply plug in expressions ad hoc. Also, note that the $\delta$ which I think about may be different for the $2|x|\delta$ term and the $\delta^2$ term. To emphasize this, I will use $\delta_1$ and $\delta_2$, respectively, and then set $\delta = \min \{\delta_1,\delta_2\}.$ This is very typical in analysis and explains why $\delta$ in proofs is often expressed as a minimum of several values.

First, I want to use the fact that $2-x^2>0$, so perhaps I guess $\delta_1= \frac{2-x^2}{2|x|}$ to kill off the middle term. Then

$$x^2+2|x|\delta_1+\delta_2^2 = x^2 + (2-x^2) + \delta_2^2 = 2+\delta_2^2.$$

We see that the problem here is that I have overshot my target since $2+\delta_2^2$ is larger than $2$. However, we can simply take a slightly smaller number (as the authors do) $\delta_1 = \frac{2-x^2}{3|x|}$ so that I get

$$x^2+2|x|\delta_1 +\delta_2^2 = \frac{1}{3}x^2 + \frac{4}{3} + \delta_2^2.$$

Now, I want this expression to be equal to $2$, so I can solve

$$\frac{1}{3} x^2 + \frac{4}{3} +\delta_2^2 = 2$$

for $\delta_2$:

$$\delta_2 = \sqrt{\frac{2-x^2}{3}}.$$

So, taking $\delta = \min \{\delta_1,\delta_2\}$ recovers exactly the $\delta$-ball which is used in the proof above.

Let me know if anything is still unclear!