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:

$\forall x \in \Z: \map f x = 2 x$
\(\ds \map f x\) \(=\) \(\ds \map f y\)
\(\ds \leadsto \ \ \) \(\ds 2 x\) \(=\) \(\ds 2 y\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds y\)

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.

$\blacksquare$


Sources