Every collection of disjoint non-empty open subsets of $\mathbb{R}$ is countable?

Question

This is my first question on the forum. I'm wondering if the following proof is valid.

Proof: Let $\{A_\lambda\}_{\lambda \in L}$ be an arbitrary collection of disjoint non-empty open subsets of $\mathbb{R}$. Since every non-empty open subset of $\mathbb{R}$ can be written uniquely as a countable union of disjoint open intervals, we can take the union $A = \bigcup\limits_{\lambda \in L}A_\lambda$ and decompose it $A = \bigcup\limits_{n \in \mathbb{N}} I_n$ in disjoint open intervals which forms a countable collection. We can also decompose each $A_\lambda$ as $\bigcup\limits_{m \in \mathbb{N}}J_{\lambda,m}$. For $\lambda \neq \mu \in L$, $A_\lambda \cap A_\mu = \emptyset$ and this is a new representation of $A$:

$$A = \bigcup_{n \in \mathbb{N}} I_n = \bigcup_{\substack{\lambda \in L \\ m \in \mathbb{N}}} J_{\lambda,m}$$

No matter how complicated the union over $L$ is, the $J_{\lambda,m}$ are disjoint open intervals. Thus, by the uniqueness, the two collections are exactly the same. As the final argument, we produce an injection $\varphi:\{A_\lambda\} \mapsto \{J_{\lambda,m}\}$ picking for each $A_\lambda$ some $J_{\lambda,m}$.

I can't see any fault, but the result seems incredibly strong to me.

P.S.: stack exchange has some bug related to \bigcup and \bigcap symbols?

For simpler argument, pick a rational in each open set. Since there are only countably many rationals, we are finished. Comment: You seem to be using the symbol for intersection when you mean union. — André Nicolas, Oct 25 '11 at 16:42
Hi @ArturoMagidin, the bug i mentioned is related to the preview window. I wrote \bigcup and the preview showed me \bigcap at the first time. Maybe was my browser, thank you for the feedback. — juliohm, Oct 25 '11 at 16:47
Welcome to math.SE. There is no need to sign your messages because your user-name always appears to the right. If by "bug" you refer to the fact that the limits occur to the right of the symbol when doing in-line equations, that's how $\LaTeX$ always does it. If you want to get the limits "under" and "under" when doing an in-line equation, you need to use the \limits command. For example, \bigcup_{i\in I}A_i produces $\bigcup_{i\in I}A_i$, but \bigcup\limits_{i\in I}A_i will give you $\bigcup\limits_{i\in I}A_i$. — Arturo Magidin, Oct 25 '11 at 16:48
@juliohm: Hmmm... that sound weird. Could you have typed \bigcap instead by mistake? — Arturo Magidin, Oct 25 '11 at 16:48

Weltschmerz · Accepted Answer · 2011-10-25T16:49:47.770

7

To each of those open disjoint subsets you can associate one and only one rational number (just pick a rational number in the set). Thus you obtain an injection from your family of subsets into the set of rational numbers, which is countable. The conclusion follows that your family must indeed be countable.

edited Oct 25 '11 at 16:49

answered Oct 25 '11 at 16:38

Weltschmerz

6,875

That is true! And the same argument is used when showing the structure of open subsets of $\mathbb{R}$. So much simple. Thank you @Weltschmerz. – juliohm Oct 25 '11 at 16:43
@juliohm Sure. Could you try to find conditions under which the statement is true for non-disjoint families? – Weltschmerz Oct 25 '11 at 17:04
Do you have such a generalization? Would be great to discuss about it. How could i proceed to achieve a similar result with non-disjoint families? – juliohm Oct 25 '11 at 17:39
@juliohm Start with small families. See what happens with the intersections. – Weltschmerz Oct 25 '11 at 17:44
Ok @Weltschmerz, i'll think better. – juliohm Oct 25 '11 at 18:04
Here, it is more clear that where Axiom of Choice has been applied ! – Fardad Pouran Feb 14 '15 at 18:14
1

@Fardad I did use the Axiom of Choice. Namely, when I say, pick a rational number from each set. – Weltschmerz Feb 15 '15 at 02:30
yes yes i knew. thanks. I meant in your way, what you just said, is clear – Fardad Pouran Feb 15 '15 at 06:50

Arturo Magidin · Answer 2 · 2011-10-25T16:52:04.693

It seems to me that:

"...every non-empty open subset of $\mathbb{R}$ can be written uniquely as a countable union of disjoint open intervals..."

is essentially what you are trying to prove (in fact, it's stronger than what you are trying to prove; it includes a uniqueness clause). So I would be very wary of using it.

In addition, any argument that argues by saying "No matter how complicated the union..." is likely to be at least a little bit informal.

The result you are trying to prove can be established without invoking that rather strong result.

Here's a hint: since the rational numbers are dense in $\mathbb{R}$, if $A$ is any open subset of $\mathbb{R}$, then $A\cap\mathbb{Q}\neq\emptyset$. Can you see why this observation yields the desired result?

I see, thank you Arturo. This forum is amazing. I'll try to avoid the "No matter..." phrase next time. — juliohm, Oct 25 '11 at 16:49

score 6 · Answer 3 · answered Oct 25 '11 at 16:45

Your argument seems valid to me, though I think that in assuming that every open subset of $\mathbb{R}$ has a unique representation as a countable disjoint union of open intervals, you're swatting a fly with a sledgehammer.

How about this? In a disjoint collection $\{U_i\}_{i \in I}$ of nonempty open subsets, choose a rational number $x_{i}$ in each $U_i$. then you get an injection $I \hookrightarrow \mathbb{Q}$, so $I$ is countable. (Really this argument works in any separable topological space, i.e., whenever you have a countable dense subset.)

Every collection of disjoint non-empty open subsets of $\mathbb{R}$ is countable?

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