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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.

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    $\begingroup$ How was union defined? Was indexed union defined? Something like $\bigcup \limits_{i\in I}(A_i)$? $\endgroup$ – Git Gud Jan 5 '14 at 15:01
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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.

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    $\begingroup$ Thanks, so basically it's like saying let $A = \{(a_1, b_1), (a_2, b_2), (a_3, b_3), \cdots \}$ and $\mathbb{N} \sim A$? $\endgroup$ – SwiftMo Jan 5 '14 at 15:36
  • $\begingroup$ 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. $\endgroup$ – MJD Jan 5 '14 at 15:38
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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.

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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.

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  • $\begingroup$ Is it correct to say that $\mathbb R^+ = \bigcup_{x\in \mathbb R^+} A_x$? $\endgroup$ – Pixar Aug 22 '16 at 14:40
  • $\begingroup$ @Pixar yes, that's correct $\endgroup$ – Ragnar Aug 22 '16 at 22:43
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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 $X=\bigcup A$.

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    $\begingroup$ What is a foobar? $\endgroup$ – Gustavo Apr 29 '16 at 2:05
  • $\begingroup$ foobar is a placeholder name, which is often used in computing. See en.wikipedia.org/wiki/Foobar. $\endgroup$ – Jørgen Fogh Apr 2 '18 at 9:59

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