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I have a question and I want to know if my solution is correct. If it's not, I would like to know why. Anyway, I would love to see more solutions to this problem. There is the problem:
Let $A \subseteq P(\mathbb{R})$ Be a set of open Intervals, when there are no empty sets involved ($\emptyset \notin A$).
. if $C_1,C_2 \in A$, $C_1 \neq C_2 \rightarrow C_1 \cap C_2=\emptyset$. We need to prove that $|A|\leq \aleph_0$.
At the first time I solved it using the Axiom of Choice, using the fact that we can find a rational number in any non-empty open interval (It helps because $|\mathbb{Q}|=\aleph_0$).
Now I am required to solve it without using the axiom of choice. There is my solution, Is it correct?
For any real number $r$, let $r_n$ be the smallest integer that is bigger then r. For each $(a,b) \in A$, define $d_{(a,b)}$ as $d_{(a,b)}=|b-a|$.
Also, let $n_{(a,b)}$ be the smallest natural number $n$ such that $\frac{1}{10^n}<d_{(a,b)}$.
let $k_{(a,b)}$ be the biggest $x \in \mathbb{N}$ such that $b_n-x \cdot \frac{1}{10^{n_{(a,b)}}} \geq b$.
Now define a function $f: A \rightarrow \mathbb{Q}$ the following way:
$$ f((a,b))= \begin{cases} \frac{a+b}{2},& \text{if } a \in \mathbb{Q},b \in \mathbb{Q} \\ a+\frac{1}{10^{n_{(a,b)}}}, & \text{if } a \in \mathbb{Q},b \notin \mathbb{Q}\\ b-\frac{1}{10^{n_{(a,b)}}}, & \text{if } a \notin \mathbb{Q},b \in \mathbb{Q}\\ b_n-(k_{(a,b)}+1)\cdot \frac{1}{10^{n_{(a,b)}}} & \text{if } a \notin \mathbb{Q},b \notin \mathbb{Q} \end{cases} $$
Now, this is not that hard to show that $f((a,b))$ is a rational number $q \in (a,b)$. also, we never get the same rational number twice because $C_1 \neq C_2 \rightarrow C_1 \cap C_2=\emptyset$. So this is Injective function which is exactly what we wanted. Is this solution correct? I am not using the axiom of choice, right? Thanks!

Asaf Karagila
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Omer
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    https://math.stackexchange.com/questions/98923/open-sets-of-mathbbr1-and-axiom-of-choice/ – Asaf Karagila Aug 07 '18 at 20:42
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    That reminds me of the first time I proved this sort of proposition in front of a classroom full of first-semester freshmen. I later honed and refined this to avoid the very explicit and overly complicated algorithm. – Asaf Karagila Aug 07 '18 at 21:16

1 Answers1

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Yes, this is perfectly correct. Note that you can avoid all your complicated casework by fixing once and for all a bijection $g:\mathbb{N}\to\mathbb{Q}$ and then define $f((a,b))=g(n)$ for the least $n$ such that $g(n)\in (a,b)$. (Of course, defining such a bijection $g$ explicitly would involve some sort of casework similar to what you did, but fixing such a bijection ahead of time makes the argument conceptually a lot simpler.)

Eric Wofsey
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