Is every open subset of $\mathbb{R}$ uncountable?

Is every open subset of $\mathbb{R}$ uncountable? I was crafting a proof for the theorem that states every open subset of $\mathbb{R}$ can be written as the union of a countable number of disjoint intervals when this question came up. I feel like the answer is yes, but I'm not sure how to go about proving it or whether there is a crazy construction (like the Cantor Set) that creates a countable, open subset of $\mathbb{R}$. Any ideas?

• Every open set is the union of open intervals which are uncountable. Also, the Cantor Set is closed and uncountable! Careful! Or you just mean it is crazy, but not open and countable? =) – Pedro Tamaroff Mar 10 '13 at 5:15
• Wise guy answer: depends on your topology. If you have the discrete topology, there are plenty of countable (and finite) opens. – Lepidopterist Mar 10 '13 at 5:25
• @Lepidopterist I guess everyone assumes we're considering the standard topology unless otherwise stated. – Pedro Tamaroff Mar 10 '13 at 6:12
• None of the answers so far mention this, so: Not only are nonempty open sets uncountable; they in fact have the same size as the reals (and the same size as the Cantor set). – Andrés E. Caicedo Mar 10 '13 at 6:14
• – Cameron Buie Aug 13 '13 at 15:33

The empty set.

Otherwise, yes. Every open interval is uncountable, so every nonempty open subset of $\Bbb R$ is uncountable.

I assume you don't want a proof but hints. To prove all intervals are uncountable you could first try proving (0,1) is uncountable. Can you construct a decimal fraction of a number that is in (0,1) but isn't in a supposed sequence of all numbers in (0,1)?

After you've established (0,1) is uncountable you could try constructing a bijection between (a,b) and (0,1) and thus proving (a,b) is uncountable.

Every open subset of $\mathbb{R}$ is a nonempty union of disjoint open intervals, each of which are open. A union of open sets is always open. Every nonempty open interval is uncountable.

• $\emptyset$ is an empty union of disjoint open intervals. – Squirtle Mar 5 '15 at 16:39

For any point $x$ of an open set $S$, $S$ contains a neigbourhood of $x$, and this neighbourhood has uncountably many elements. So as long as $x\in S$ exists (in other words $S\neq\emptyset$) $S$ is uncountable.