Can an open set in $\mathbb{R}$ be described as the disjoint union of uncountably many open intervals?

The lecture notes I'm working through pose the question of whether any open set in $$\mathbb{R}$$ can be described as the disjoint union of uncountably many open intervals.

The problem doesn't state which topology on $$\mathbb{R}$$ I'm using. I know that any open set can be written as a union of open balls, and an open ball in $$\mathbb{R}$$ with its usual topology is an open interval, so I assume this is using the standard topology. (If it doesn't matter, I'd be interested in knowing.)

Intuitively, I believe the answer is no, but here's my attempt at a proof. It essentially mimics the argument that any collection of non-overlapping discs in $$\mathbb{R}^2$$ is countable.

Let $$U \subset \mathbb{R}$$ be open, and $$\{G_{\alpha}\}_{\alpha \in I}$$ an uncountable collection of disjoint, open intervals such that $$U = \bigcup\limits_{\alpha \in I} G_{\alpha}$$. As each $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$, in each $$G_{\alpha}$$, we can choose a rational $$q_{\alpha}$$. As the $$\{G_{\alpha}\}$$ are disjoint, if $$q_{i}$$ lives in $$G_i$$ for some $$i \in I$$, then $$q_i \not \in G_{\alpha}$$ for every $$\alpha \neq i$$. Therefore, we can write an injection $$f: I \to \mathbb{Q}$$ sending $$\alpha \mapsto q_{\alpha}$$. As $$f$$ is injective, $$I \cong f(I) \subset \mathbb{Q}$$. As $$\mathbb{Q}$$ is countable and a subset of a countable set is countable, $$f(I)$$ is countable, so $$I$$ is countble, and we have a contradiction.

I'd appreciate some advice on how to improve the proof. I don't think I need contradiction, for example. I could just just take $$U$$ to be a union of disjoint open intervals -- which may not even exist; at least, I haven't proved that -- and then prove that $$I$$ must be countable. I also found myself having trouble using appropriate notation to distinguish between the variable $$\alpha$$ and the specific index $$i$$.

• You need to have the usual topology, in general this is not true. For example if you take $\mathbb{R}$ with the discrete topology then $\mathbb{R}=\cup_{x\in\mathbb{R}} \lbrace x\rbrace$. Commented Jan 19, 2022 at 21:18
• About your proof, this looks good to me. Notice that you definition of $f$ implies the axiom of choice, but I don't think that there is a solution without it. Commented Jan 19, 2022 at 21:23
• @Marcos Yes, I definitely used the axiom of choice. Do I need contradiction, though? This is my main uncertainty at this point. I appreciate the feedback. Commented Jan 19, 2022 at 21:25
• @Marcos Is $\{x\}$ an open interval though? Commented Jan 19, 2022 at 21:27
• Your proof is correct, but you’re proving much more than is needed. As @Randall stated any open interval cannot be written as disjoint union of more than 1 open intervals. What you are proving is basically that there cannot be a family of disjoint open intervals, or even sets. Note that you can rephrase your prove: If $F$ is a family of open, disjoint sets then one can find a function that maps $F$ injectively into $\mathbb Q$, thus $F$ is at most countable.
– Lazy
Commented Jan 19, 2022 at 21:30

If $$\Bbb R$$ has the standard order, let $$G_i, i \in I$$ be a pairwise disjoint family of open intervals , say $$G_i = (a_i, b_i)$$ with $$a_i < b_i$$ two reals for each $$i \in I$$.

For each $$i \in I$$ we can pick a rational $$q_i \in \Bbb Q$$ so that $$a_i < q_i < b_i$$. This has nothing to do with any topology but with the construction/definition of $$\Bbb R$$ from $$\Bbb Q$$.

It follows that $$I \to q_i$$ is an injective function from $$I$$ (by disjointness) into a countable set. Hence $$I$$ is at most countable.

There simply doesn't exist any uncountable disjoint family of open intervals in $$\Bbb R$$, answering the question independently of any topology you might put on $$\Bbb R$$...

• Note that if you use another unrelated order, like a well-order, there can be such a family. Commented Jan 19, 2022 at 21:52
• What is your definition for interval in a general topology? I don't think that the statement of the problem has sense for an arbitrary topology. Commented Jan 19, 2022 at 21:54
• @Marcos interval has nothing to do with a topology, only with an order. There are no "intervals" in a general space. But $\Bbb R$ comes with an order so interval is a defined notion. Commented Jan 19, 2022 at 21:55
• but these are open intervals, so the topology is involved here. For example, in the discrete topology what is defined to be an open interval? Commented Jan 19, 2022 at 21:57
• @Marcos On $\Bbb R$ an open interval is a set of the form $\{x\mid a < x < b\}$ for some $a<b$ in $\Bbb R$. It's called open because it's "open-ended" not because it is open in any topology. Commented Jan 19, 2022 at 21:59

As I stated in a comment you are basically proving that there cannot exist an uncountable family of disjoint open sets (you can easily extend your proof to arbitrary separable spaces). But we can in facht prove this without AOC:

Let F be a family of open disjoint sets. We define for $$q\in\mathbb Q$$ the set $$F_q = \{O\in F:q\in O\}$$. The disjoint property then means that $$F_q$$ has at most one element. But for each $$O\in F$$ there exists at least on $$q\in O$$, so $$F = \bigcup_{q\in\mathbb Q} F_q$$ and is thus at most countable.

The trick here is that we do not in fact need any uniqueness of this $$q$$.

• Areyou sure this doesn't use the AOC? In my oppinion this is equivalent to all other proofs. Commented Jan 19, 2022 at 22:16
• @Marcos Yes, this does not use AOC. The reason why the other versions require the AOC is because they try to find a single $q$ for each $O\in F$ to define an injection into $\mathbb Q$, while this does not care about such things and simply defines a surjection from $\mathbb Q$.
– Lazy
Commented Jan 19, 2022 at 22:18
• @Marcos Basically the other proof goes: We take the system of disjoint and nonempty sets defined by $O\cap \mathbb Q$ where $O\in F$ and then find by AOC a system of representatives $q_O$ to define this injection. Here we do the other direction, so for any $q$ we take the set of all $O\in F$ that contain $q$. Then we get that this can contain at most one set simply from the fact that all sets are disjoint.
– Lazy
Commented Jan 19, 2022 at 22:21
• @Marcos fix a bijection $h:\Bbb N \to \Bbb Q$ (this can be done explicitly) and define (in my version) $m(i)= \min \{n \in \Bbb N: a_i < h(n) < b_i\}$ (non-empty set so the min exists) and set $q_i=h(m(i))$. No AC is required at all. Commented Jan 19, 2022 at 22:26
• Thanks, honestly I'm quite suprised that this can be done withouth the AOC. Nice solution!! Commented Jan 19, 2022 at 22:32

This is the direct proof you want:

Let $$U$$ be an open set in $$\mathbb{R}$$ and let us define an equivalence relation: $$x\sim y\Leftrightarrow\exists I\subset U\text{ open interval such that }x,y\in I$$ This is easily proven to be an equivalence relation and thus we can consider $$U/\sim$$. In this space, the equivalence classes are given by the open intervals of $$U$$. If we denote as $$I_x$$ the open interval containing $$x\in U$$, let us prove that $$|U/\sim|\leq\aleph_0$$:

For every $$x\in U$$ there exists $$a_x\in\mathbb{Q}\cap I_x$$ (because $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$) and $$x\nsim y\Rightarrow I_x\cap I_y=\emptyset$$. Thus, using the axiom of choice, we can define $$\phi:\mathbb{R}/\sim\to\mathbb{Q}$$ by $$\phi(I_x)=a_x$$. Trivially $$x\nsim y\Rightarrow \phi(I_x)\neq \phi(I_y)$$ and then $$|U/\sim|\leq|\mathbb{Q}|\leq\aleph_0$$, as we wanted to prove.

Although this is a direct proof the idea is more or less the same as what you did in your contradiction proof.