# Set of Even Integers is Equivalent to Set of Integers

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## 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.

## Proof

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

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

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

\(\displaystyle \map f x\) | \(=\) | \(\displaystyle \map f y\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 2 x\) | \(=\) | \(\displaystyle 2 y\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(=\) | \(\displaystyle y\) |

demonstrating injectivity.

\(\displaystyle y\) | \(\in\) | \(\displaystyle 2 \Z\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\, \displaystyle \exists x \in \Z: \, \) | \(\displaystyle y\) | \(=\) | \(\displaystyle 2 x\) | Definition of Even Integer | ||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle y\) | \(\in\) | \(\displaystyle f \sqbrk \Z\) |

demonstrating surjectivity.

Hence by definition $f$ is a bijection.

$\blacksquare$

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): Chapter $3$. Mappings: Exercise $10$