Set of Even Integers is Countably Infinite
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Theorem
Let $\Bbb E$ be the set of even integers.
Then $\Bbb E$ is countably infinite.
Proof
Let $f: \Bbb E \to \Z$ be the mapping defined as:
- $\forall x \in \Bbb E: \map f x = \dfrac x 2$
$f$ is well-defined as $x$ is even and so $\dfrac x 2 \in \Z$.
Let $x, y \in \Bbb E$ such that $\map f x = \map f y$.
Then:
\(\ds \map f x\) | \(=\) | \(\ds \map f y\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac x 2\) | \(=\) | \(\ds \dfrac y 2\) | Definition of $f$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds y\) |
Thus $f$ is injective by definition.
Consider the inverse $f^{-1}$.
By inspection:
- $\forall x \in \Z: \map {f^{-1} } x = 2 x$
$f^{-1}$ is well-defined, and $2 x$ is even.
Thus $f^{-1}$ is a mapping from $\Z$ to $\Bbb E$.
Then:
\(\ds \map {f^{-1} } x\) | \(=\) | \(\ds \map {f^{-1} } y\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 x\) | \(=\) | \(\ds 2 y\) | Definition of $f^{-1}$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds y\) |
Thus $f^{-1}$ is injective by definition.
It follows by the Cantor-Bernstein-Schröder Theorem that there exists a bijection between $\Z$ and $\Bbb E$.
$\blacksquare$
Sources
- 1971: Wilfred Kaplan and Donald J. Lewis: Calculus and Linear Algebra ... (previous) ... (next): Introduction: Review of Algebra, Geometry, and Trigonometry: $\text{0-1}$: The Real Numbers