What does countable union mean?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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 $X=\bigcup A$.