Set of Even Integers is Equivalent to Set of Integers

Theorem

Let $\Z$ denote the set of integers.

Let $2 \Z$ denote the set of even integers.

Then:

$2 \Z \sim \Z$

where $\sim$ denotes set equivalence.

To demonstrate set equivalence, it is sufficient to construct a bijection between the two sets.

Let $f: \Z \to 2 \Z$ defined as:

$\forall x \in \Z: \map f x = 2 x$

$\ds \map f x$

$=$

$\ds \map f y$

$\ds \leadsto \ \ $

$\ds 2 x$

$=$

$\ds 2 y$

$\ds \leadsto \ \ $

$\ds x$

$=$

$\ds y$

demonstrating injectivity.

$\ds y$

$\in$

$\ds 2 \Z$

$\ds \leadsto \ \ $

$\ds \exists x \in \Z: \, $

$\ds y$

$=$

$\ds 2 x$

Definition of Even Integer

$\ds \leadsto \ \ $

$\ds y$

$\in$

$\ds f \sqbrk \Z$

demonstrating surjectivity.

Hence by definition $f$ is a bijection.

$\blacksquare$

1951: J.C. Burkill: The Lebesgue Integral ... (previous) ... (next): Chapter $\text {I}$: Sets of Points: $1 \cdot 2$. Infinite sets
1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $3$. Mappings: Exercise $10$