Bijection from Cartesian Product of Initial Segments to Initial Segment

Theorem

Let $\N_k$ be used to denote the set of the first $k$ non-zero natural numbers:

$\N_k := \set {1, 2, \ldots, k}$

Then a bijection can be established between $\N_k \times \N_l$ and $\N_{k l}$, where $\N_k \times \N_l$ denotes the Cartesian product of $\N_k$ and $\N_l$.

Proof

Let $\phi: \N_k \times \N_l \to \N_{k l}$ be defined as:

$\forall \tuple {m, n} \in \N_k \times \N_l: \map \phi {m, n} = \paren {m - 1} \times l + n$

First it is confirmed that the codomain of $\phi$ is indeed $\N_{k l}$.

This needs considerable tedious hard slog to complete it. In particular: fiddly and tedious, can't think of an elegant way to prove it To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Mappings: $\S 13 \alpha$