Cartesian Product of Countable Sets is Countable/Formal Proof 2/Mistake

Source Work

1964: Steven A. Gaal: Point Set Topology:

Introduction to Set Theory:

$2$. Set Theoretical Equivalence and Denumerability

This mistake can be seen in the 2009 Dover edition: ISBN 0-486-47222-1

Mistake

If $A$ and $B$ are denumerable, then so are $A \cup B$ and $A \times B$.

...

For instance, if $a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3, \ldots$ are enumerations of $A$ and $B$, then the map $f$ given by the rule

$\map f {\tuple {a_k, a_l} } = \dfrac {\paren {k + l - 1} \paren {k + l - 2} } 2 + \dfrac {l + \paren {-1}^{k + 1} } 2 k + \dfrac {1 + \paren {-1}^{k + l - 1} } 2 l$

gives an enumeration of $\tuple {A \times B}$.

Correction

The expression starts incorrectly. It should be:

$\map f {\tuple {a_k, b_l} } = \dfrac {\paren {k + l - 1} \paren {k + l - 2} } 2 + \dfrac {l + \paren {-1}^{k + 1} } 2 k + \dfrac {1 + \paren {-1}^{k + l - 1} } 2 l$

because the second element of the ordered pair comes from $B$, not $A$.

Sources

• 1964: Steven A. Gaal: Point Set Topology ... (previous) ... (next): Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability