# Infinite Set is Equivalent to Proper Subset

## Theorem

A set is infinite if and only if it is equivalent to one of its proper subsets.

## Proof 1

Let $T$ be an infinite set.

By Infinite Set has Countably Infinite Subset, it is possible to construct a countably infinite subset of $T$.

Let $S = \set {a_1, a_2, a_3, \ldots}$ be such a countably infinite subset of $T$.

Create a Partition of $S$ into:

$S_1 = \set {a_1, a_3, a_5, \ldots}, S_2 = \set {a_2, a_4, a_6, \ldots}$

Let a bijection be established between $S$ and $S_1$, by letting $a_n \leftrightarrow a_{2 n - 1}$.

This is extended to a bijection between $S \cup \paren {T \setminus S} = T$ and $S_1 \cup \paren {T \setminus S} = T \setminus S_2$ by assigning each element in $T \setminus S$ to itself.

So a bijection has been demonstrated between $T$ and one of its proper subsets $T \setminus S_2$.

That is, if $T$ is infinite, it is equivalent to one of its proper subsets.

Now, let $T_0 \subsetneq T$ be a proper subset of $T$, and $f: T \to T_0$ be a bijection.

It follows from No Bijection between Finite Set and Proper Subset that $T$ must be infinite.

$\blacksquare$

## Proof 2

Let $S$ be a set.

Suppose $S$ is finite.

From No Bijection between Finite Set and Proper Subset we have that $S$ can not be equivalent to one of its proper subsets.

Suppose $S$ is infinite.

From Infinite Set has Countably Infinite Subset, we can construct $v: \N \to S$ such that $v$ is an injection.

We now construct the mapping $h: S \to S$ as follows.

$\map h x = \begin {cases} \map v {n + 1} & : \exists n \in \N: x = \map v n \\ x & : x \notin \Img v \end {cases}$

It is clear that $h$ is an injection.

But we have that $\map v 0 \notin \Img h$ and so $\Img h \subsetneq S$.

The result follows from Injection to Image is Bijection.

$\blacksquare$

## Proof 3

Let $X$ be a set which has a proper subset $Y$ such that:

$\card X = \card Y$

where $\card X$ denotes the cardinality of $X$.

Then:

$\exists \alpha \in \complement_X \paren Y$

and

$Y \subsetneqq Y \cup \set \alpha \subseteq X$
$i_Y: Y \to X: \forall y \in Y: i \paren y = y$
$i_{Y \cup \set \alpha}: Y \cup \set \alpha \to X: \forall y \in Y: i \paren y = y$

give:

$\card X = \card Y \le \card Y + \mathbf 1 \le \card X$

from which:

$\card X = \card Y + \mathbf 1 = \card X + \mathbf 1$

So by definition $X$ is infinite.

$\Box$

Now suppose $X$ is infinite.

That is:

$\card X = \card X + \mathbf 1$

Let $\alpha$ be any object such that $\alpha \notin X$.

Then there is a bijection $f: X \cup \set \alpha \to X$.

Let $f_{\restriction X}$ be the restriction of $f$ to $X$.

Then from Injection to Image is Bijection:

$\image {f_{\restriction X} } = X \setminus \set {f \paren \alpha}$

which is a proper subset of $X$ which is equivalent to $X$.

$\blacksquare$

## Examples

### Even Integers

Let $\Z$ be the set of integers.

By Integers are Countably Infinite, $\Z$ is infinite.

Let $E$ be the set of all even integers.

We have that, for example, $3 \in \Z$ but $3 \notin E$

Hence $E$ is a proper subset of $\Z$

Let $f: \Z \to E$ be the mapping defined as:

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

Then $f$ is a bijection.

Hence a fortiori $\Z$ and $E$ are equivalent.