# Which subsets of $\Bbb R$ are a countable union of open sets and countable sets?

From (1), every open subset of $$\Bbb R$$ is at most a countable union of open intervals. The converse is also true: any countable union of open intervals is an open set. However, I want every subset of $$\Bbb R$$. What if we allow countable sets in our countable union?

For example, any closed interval $$[a,b]$$ can be written as the union:

$$[a, b] = \{ a \} \cup (a, b) \cup \{ b \}$$

For this reason, I think very many subsets of $$\Bbb R$$ can be described this way. On my other hands, I don't believe the set of irrational numbers can be described this way, since it cannot contain any open sets (any open interval contains rationals) not can it be a union of countable sets (it is uncountable).

Edit: I asked only about this set's cardinality, but I am really interested in more specific details and how it compares to the entire power set of $$\Bbb R$$. I know this set has at least the cardinality of the continuum since it contains as a proper subset the collection of all open sets, which also has that cardinality.

• Do you want to know the cardinal of the set of sets that can be written as a countable union of open intervals and countable sets? – svelaz Jul 15 '19 at 16:03
• The Borel heirarchy may be relevant. – Xander Henderson Jul 15 '19 at 16:06
• @XanderHenderson It is true that my sets are Borel sets since every singleton set $\{ x \}$ can be recovered from the relative complement $\{ x \} = (x-\varepsilon,x+\varepsilon) \setminus \left( (x-\varepsilon,x) \cup (x, x+\varepsilon) \right)$ – M. Nestor Jul 15 '19 at 16:11
• $\mathbb R$ itself can be a countable union of countable sets if we deny the axiom of choice. – eyeballfrog Jul 15 '19 at 16:20
• @M.Nestor You may want to change the title. There have now been two deleted answers who have read only the title and answered about the cardinality of the set of such subsets. It would also be helpful to clarify exactly what you want from the answerers. – Theo Bendit Jul 15 '19 at 16:23

The empty set is countable. Therefore, if $$x$$ is any real number then $$I_x:=(x, x+1)$$ is a union of an open interval and a countable set. For distinct real numbers $$x, y$$, $$I_x, I_y$$ are distinct. So the cardinality of the set that you are looking at has cardinality at least that of $$\mathbb{R}$$. It cannot have cardinality greater than that of $$\mathbb{R}$$ because, as you mentioned, open sets are countable unions of open intervals.
• That reference contains the answer to my question. The subset of $\mathcal P(\Bbb R)$ described in my post is the $\sigma$-algebra on the collection of open sets, which is the Borel algebra on $\Bbb R$. Thanks! – M. Nestor Jul 15 '19 at 17:23
• @M.Nestor I don't think that the collection you describe is the Borel $\sigma$-algebra (either that, or your description is incredibly unclear). You describe sets which are a union of an open set and a countable number of singleton points. The Borel $\sigma$-algebra is much larger. For example, the usual ternary Cantor set is a Borel set, but is not the union of an open set and a countable number of points. Your example of the irrationals is another such example---it is Borel, but not of the desired form. – Xander Henderson Jul 15 '19 at 20:18
We want to know what is the cardinal of the set of subsets of $$\mathbb{R}$$ that can be written in the form $$A=\mathcal{U}\cup C,$$ where $$\mathcal{U}$$ is an open set and $$C$$ is countable. Both open sets and countable sets have the cardinality of $$\mathbb{R}$$, so the set in question must have the same cardinality since the function sending $$(U,C)\mapsto U\cup C$$ is surjective. On the other hand, the set in question contains the open sets and therefore its cardinal is equal to the one of $$\mathbb{R}$$.