12

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.

Community
  • 1
SwiftMo
  • 1,047
  • 1
  • 13
  • 23
  • 1
    How was union defined? Was indexed union defined? Something like $\bigcup \limits_{i\in I}(A_i)$? – Git Gud Jan 05 '14 at 15:01

4 Answers4

8

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.

Michael Albanese
  • 99,526
  • 1
    Thanks, so basically it's like saying let $A = {(a_1, b_1), (a_2, b_2), (a_3, b_3), \cdots }$ and $\mathbb{N} \sim A$? – SwiftMo Jan 05 '14 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 05 '14 at 15:38
  • @MJD how the rhs of the expression countably many open intervals? Shouldnt this be uncountably open intervals as they are infinite? $$\qquad\bigcup_{i=1}^{\infty}(a_i, b_i).$$ – GENIVI-LEARNER Apr 01 '20 at 22:07
  • Each interval is an uncountable set, but the set of intervals contains only countably many intervals. – MJD Apr 01 '20 at 23:02
  • What does N ~ A mean? @SwiftMo – John D Aug 24 '23 at 18:57
  • @JohnD: This notation is defined in the question. It means that there is a map $f : \mathbb{N} \to A$ which is one-to-one and onto (i.e. a bijection), so $\mathbb{N}$ and $A$ have the same cardinality. That is, $A$ is countable. – Michael Albanese Aug 28 '23 at 14:51
3

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.

user119191
  • 356
  • 1
  • 3
1

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.

Ragnar
  • 6,233
1

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 $X=\bigcup A$.

Hagen von Eitzen
  • 374,180