What does countable union mean?

The book I am reading contains the following two definitions:

Two sets $$A$$ and $$B$$ have the same cardinality if there exists $$f: A \rightarrow B$$ that is one to one and onto. In this case, we write $$A \sim B$$.

A set $$A$$ is countable if $$\mathbb{N} \sim A$$. An infinite set that is not countable is called an uncountable set.

Following on, I read the following statement:

Every open set is either a finite or countable union of open intervals.

Here, what does countable union mean? Clearly it can't mean that the resultant set formed by the union of open intervals is countable (since open intervals are uncountable). But I am not sure how the use of "countable union" connects with the definition provided earlier.

• How was union defined? Was indexed union defined? Something like $\bigcup \limits_{i\in I}(A_i)$? Jan 5, 2014 at 15:01

It just means that every open set can be written in the form $$\bigcup_{i=1}^n(a_i, b_i)\qquad\text{or}\qquad\bigcup_{i=1}^{\infty}(a_i, b_i).$$ That is, every open set can be written as a union of either finitely many open intervals, or countably many open intervals.

• Thanks, so basically it's like saying let $A = \{(a_1, b_1), (a_2, b_2), (a_3, b_3), \cdots \}$ and $\mathbb{N} \sim A$? Jan 5, 2014 at 15:36
• Yes, exactly so. It might be worth pointing out that although finite intersections of open sets are open, countable intersections are not always; that's why we consider the countable-union thing to be interesting.
– MJD
Jan 5, 2014 at 15:38
• @MJD how the rhs of the expression countably many open intervals? Shouldnt this be uncountably open intervals as they are infinite? $$\qquad\bigcup_{i=1}^{\infty}(a_i, b_i).$$ Apr 1, 2020 at 22:07
• Each interval is an uncountable set, but the set of intervals contains only countably many intervals.
– MJD
Apr 1, 2020 at 23:02
• What does N ~ A mean? @SwiftMo Aug 24 at 18:57

It is a set of the form $\cup_{I \in S} I$ where $S$ is a countable set whose elements are open intervals.

We usually write $\cup_{k \in \mathbf{N}} I_k$, where $I_k$ is a sequence of intervals.

The formulations "union of a countable sequence of sets" and "union of a countable set of sets" are equivalent provided we have the axiom of choice.

In addition to the other answers, here is an example of an uncountable union:
Say that $A_x=(0,x)$ for every $x\in \mathbb R^+$. $$\bigcup_{x\in \mathbb R^+} A_x$$ is an uncountable union.

• Is it correct to say that $\mathbb R^+ = \bigcup_{x\in \mathbb R^+} A_x$? Aug 22, 2016 at 14:40
• @Pixar yes, that's correct Aug 22, 2016 at 22:43

By the Axiom of Union (in ZFC), if $A$ is a set, then there is a set $\bigcup A$ that is characterized by $x\in \bigcup A\iff\exists z\in A\colon x\in z$. We say "$X$ is a finite/countable union of foobar sets" if there exists a finite/countable set $A$ such that all elements of $A$ are foobar sets and $X=\bigcup A$.

• What is a foobar? Apr 29, 2016 at 2:05
• foobar is a placeholder name, which is often used in computing. See en.wikipedia.org/wiki/Foobar. Apr 2, 2018 at 9:59