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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 $A$:

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

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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?

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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.

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.)

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?

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