Any open subset of $\mathbb{R}$ is a countable union of rational open intervals

Question

There are lots of relevant posts online here and I found some answers with a bit of confusion, I left comments but got no response, please Would anyone be able to help me out with the question?

relevant posts are here

According to @Asinomás's answer:

Let $U$ be an open set in $\mathbb R$ we have to prove there is a countable family of open intervals $(a_i,b_i)$ with $a_i,b_i \in \mathbb R$ so that $U=\bigcup\limits_{i\in\mathbb N}(a_i,b_i)$. To do this for each $x\in U$ we have $(a_x,b_x)$ so that $x\in(a_x,b_x) \subseteq U$. This is because $U$ is open. Of course, there is a rational number $a'_x$ between $a_x$ and $x$ and a rational number $b'_x$ between $b_x$ and $x$. This is because the rational numbers are dense in $\mathbb R$. So we can consider the familiy of open intervals with rational endpoints. $(a'_x,b'_x)$ Why is this family countable? Because the number of intervals with rational coordinates is countable, since its cardinality does not exceed $|\mathbb Q\times \mathbb Q |=|\mathbb N|$.Moreover $\bigcup\limits_{x\in U}(a'_x,b'_x)$ is clearly a subset of $U$ since every interval in the union is contained in $U$. But this union also contains $U$ since every element $x\in U$ is contained in the interval $(a'_x,b'_x)$ and hence is contained in the union. Hence $U$ is a countable union of rational open intervals.

Question 1: It seems that there is an injection for each $x\in U$ to $(a'_x,b'_x)\subset (a_x,b_x)\subset U$, and I agree that there are at most countable rational-endpoint open subsets of $\mathbb{R}$. What really bugs me, isn't $\bigcup_{x\in U}(a'_x,b'x)$ implicitly saying $x\in U$ are countable?

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Accoridng to @DuFong

Let's asume $\mathcal{O}=\bigcup_{i=1}^\infty(a_i,b_i)$,where $a_i,b_i\in \mathbb{R}$, then for each interval $(a_i,b_i)$, since rational number is dense in real number, we can always find rational sequence $\{b_{i,n}\}$ increasingly approaching $b_i$ and $\{a_{i,n}\}$ decreasingly approaching $a_i$, then $(a_i,b_i)=\bigcup_{n=1}^\infty (a_{i,n},b_{i,n})$, then $\mathcal{O}=\bigcup_{i=1}^\infty(a_i,b_i)=\bigcup_{i=1}^\infty\bigcup_{n=1}^\infty(a_{i,n},b_{i,n})$, and countable union of countable union is countable union too.
As for can we find disjoint countable union of rational intervals, I didn't prove it.

Question 2: Why does any open set can be written as countable union $\bigcup^{\infty}_{i=1}(a_i,b_i)$ at the very beginning? What if it is ACTUALLY means uncountable union, namely $\bigcup_{i\in U}(a_i,b_i)$, how do I conclude his countable union of countable union is a countable union, I can see that sequence expansion approach is countable union, but I don't understand the first countable union

Appreciate any answer and please enlighten me.

Answer 1

Let an open set $U$ be given. Let $\mathcal I=\{(a,b):a,b\in\mathbb Q,\,a\lt b,\text{ and }(a,b)\subseteq U\}$, the set of all rational open invervals contained in $U$. Then $\mathcal I$ is countable, because there are only countably many rational intervals in the world; any union of rational intervals is a countable union. I claim that $\bigcup\mathcal I=U$.

Leaving the proof of $\bigcup\mathcal I\subseteq U$ as an exercise, I will show that $U\subseteq\bigcup\mathcal I$. Consider any point $x\in U$. Since $U$ is open, there is a number $\varepsilon\gt0$ such that $(x-\varepsilon,\,x+\varepsilon)\subseteq U$. Since the rational numbers are dense in $\mathbb R$, there are rational numbers $a$ and $b$ such that $x-\varepsilon\lt a\lt x\lt b\lt x+\varepsilon$. Then $x\in(a,b)\subseteq(x-\varepsilon,\,x+\varepsilon)\subseteq U$, so $(a,b)\in\mathcal I$ and $x\in(a,b)\subseteq\bigcup\mathcal I$.

Answer 2

1. No, it doesn't assume $U$ is countable, which would actually be false as long as $U\neq \emptyset$. Rather, we have a function [subject to the axiom of choice] that maps $\phi : U \to \{(p,q) : p,q\in \mathbb{Q}\}$, where the latter is countable. The union $\bigcup_{x \in U} \phi(x)$ is itself an uncountable union, but by removing any duplicates, it becomes countable.
2. You've left out the link in that answer where it's proven that $U$ is a countable disjoint union of open intervals. How does it assume the union is countable? Follow the link.

Answer 3

1. It doesn't. When we say $\bigcup_{x\in U}(a_x^\prime,b_x^\prime)$ is a countable union, what that means is that for different $x$s, the interval $(a_x^\prime,b_x^\prime)$ will turn out to be the same, so even though it is written as a union over $x\in U$, many of these appear several times, so it is really a countable union. You could formalise this: for every rational interval $(a,b)$, if possible pick $x_{(a,b)}$ such that the corresponding $$(a_{x_{(a,b)}}^\prime,b_{x_{(a,b)}}^\prime)=(a,b)$$ Then, look at $U^\prime=\{x_{(a,b)}\mid x_{(a,b)}\text{ exists}\}$. I will claim $$\bigcup_{x\in U}(a_x^\prime,b_x^\prime)=\bigcup_{x\in U^\prime}(a_x^\prime,b_x^\prime)$$ $\supseteq$ is clear. To show $\subseteq$, take $x\in U$, then look at $y=x_{(a_x^\prime,b_x^\prime)}\in U^\prime$. Then, $(a_x^\prime,b_x^\prime)=(a_y^\prime,b_y^\prime)$, which is contained in the union on the RHS. This shows equality. Note that $U^\prime$ is countable.

2. The first countable union comes from the link in the answer. To conclude the countable union of countable union is countable, you can look it up easily, such as here.

Hope this helps. :)