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