Borel sets in Cramer's book

Question

I have two following quiestions.

1. On the page 13 (Mathematical methods of statistic) the set R of all rational points x = p/q belonging to the half-open interval (0,1] is considered. It's claimed that every point of [0,1] is a limiting point of R. And the limiting point z is called a limiting point if every neighbourhood of z contains at least one point of set.

Is any proof of that? For me it is not obvious, that there no exists any irrational point, whose tiny neighbourhood does not contain any number p/q.

2. Right after that, in Borel sets section: S is the class of all point sets I such that I is the sum of a finite or enumerable sequence of intervals. ... The set R considered in the preceding paragraph belongs to S.. The difference (0,1) - R, on the other hand, does not contain any non-degenerate interval, and if we try to represent it as a sum of degenerate intervals, a non-enumerable set of such intervals will be required.

"Does not contain any non-degenerate interval" means that it contains only degenerate intervals, doesnt it? But why? If we divide interval (0,1) by enumerable number of points, does we divide it in enumerable number of intervals? For example, 1/2 divides (0,1) into two intervals.

I really appreciate if somebody could explain.

Answer 1

Question 1:

remark: some people define $\mathbb R$ to be the completion of $\mathbb Q$, so the answer to your question is immediate.

Here is a more explicit reason:

Let $\alpha \notin \mathbb Q$ be a real number. Let $\epsilon>0$ and choose a natural number $s$ so that $\frac{1}{s}<\epsilon$, which is possible by the archimedean property (there is a natural number larger than $\frac{1}{\epsilon}$.) Let $r:=\lfloor s \alpha\rfloor$, and take $q=\frac{r}{s}$.

Then $$|\alpha-\frac{r}{s}|=|(\alpha s-r)/s|<1/s<\epsilon.$$

Question 2:

Borel sets are technically elements of the $\sigma$-algebra generated by open sets. In particular, closed under countable intersection and countable union. Recall that $\mathbb Q$ is countable.

Then $$\mathbb Q =\bigcup_{n \in \mathbb N} \bigcap_{k \in \mathbb N} \left(q_n-\frac{1}{k},q_n+\frac{1}{k}\right).$$

In this case, $\mathbb Q \cap [0,1]$ is certainly a Borel set. By taking complements and intersecting, it is not tough to see that $A:=(0,1)-\mathbb Q \cap [0,1]$ is Borel as well.

Now, can $A$ be written as the countable union of non-degenerate intervals? No.

Suppose that $A$ contains a non-degenerate interval $(a,b)$. We can assume that it is open, since $(a,b) \subset[a,b], (a,b], [a,b)$. Pick some $\alpha \in (a,b)$, since $(a,b)$ is open, there is some $\epsilon$-ball around $\alpha$ contained in $(a,b)$. Then by the first point, we can find a rational number $q$ in this $\epsilon$-ball, a contradiction, since $q \notin A$ by construction.

Since every interval in $A$ is degenerate, it just contains points. So, $A$ must be the union of uncountably many "intervals," since they are actually just points (and $|A|$ is uncountable.)

Answer 2

Is any proof of that? For me it is not obvious, that there no exists any irrational point, whose tiny neighbourhood does not contain any number p/q.

If $x \in (0,1]$, let:

$$u_n=\frac{\lfloor 2^nx \rfloor}{2^n}$$

Then $u_n \in (0,1]$ and $u_n \to x$ (if $x=0$, consider $u_n=2^{-n}$ instead). In other words, for all neighbourhood of a real point, there is a rational point inside.

"Does not contain any non-degenerate interval" means that it contains only degenerate intervals, doesnt it? But why? If we divide interval (0,1) by enumerable number of points, does we divide it in enumerable number of intervals? For example, 1/2 divides (0,1) into two intervals.

We are dealing with $(0,1) \setminus R$, not $(0,1)$. $(0,1)\setminus R$ cannot be written as a countable union of intervals (it has a lot of "holes").