Union of Countable Sets of Sets/Proof 1
Theorem
Let $\AA$ and $\BB$ be countable sets of sets.
Then:
- $\set {A \cup B: A \in \AA, B \in \BB}$
is also countable.
Proof
Since $\AA$ is countable, its contents can be arranged in a sequence:
- $\AA = \set {A_1, A_2, \ldots}$
Let $B \in \BB$.
Consider the sequence of sets:
- $\sequence {A_1 \cup B, A_2 \cup B, \ldots}$
We may leave out any possible repetitions, and obtain a countable set:
- $\set {A \cup B: A \in \AA}$
for every $B \in \BB$.
Thus as $B$ varies over all the elements of $\BB$, we obtain the countable family:
- $\sequence {A \cup B: A \in \AA}_{\paren {B \mathop \in \BB} }$
From Countable Union of Countable Sets is Countable, their union is countable.
$\blacksquare$
Axiom of Choice
This proof depends on the Axiom of Choice, by way of Countable Union of Countable Sets is Countable.
Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.
Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.
However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.
Sources
- 1964: Steven A. Gaal: Point Set Topology ... (previous) ... (next): Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability