# Set is Infinite iff exist Subsets of all Finite Cardinalities

## Contents

## Theorem

A set $S$ is infinite if and only if for all $n \in \N$, there exists a subset of $S$ whose cardinality is $n$.

## Proof

Let the cardinality of a set $S$ be denoted $\left|{S}\right|$.

### Necessary Condition

Suppose $S$ is infinite.

We use mathematical induction on $n$.

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:

- There exists a subset of $S$ whose cardinality is $n$.

#### Basis for the Induction

From Empty Set is Subset of All Sets:

- $\varnothing \subseteq S$

From Cardinality of Empty Set, its cardinality is $0$.

This is our basis for the induction.

#### Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

- There exists a subset of $S$ whose cardinality is $k$.

Then we need to show:

- There exists a subset of $S$ whose cardinality is $k+1$.

#### Induction Step

This is our induction step:

By the induction hypothesis, there exists $T \subseteq S$ such that $\left|{T}\right| = k$.

First, note that $S \ne T$, otherwise by definition of infinite $S$ would be finite.

So $T$ is therefore a proper subset of $S$.

So:

\(\displaystyle S \setminus T\) | \(\ne\) | \(\displaystyle \varnothing\) | Set Difference with Proper Subset | ||||||||||

\(\displaystyle \exists x\) | \(\in\) | \(\displaystyle S \setminus T\) | Definition of Empty Set | ||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \left\{ {x}\right\}\) | \(\subseteq\) | \(\displaystyle S \setminus T\) | Definition of Singleton | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \left\{ {x}\right\}\) | \(\subseteq\) | \(\displaystyle S\) | Set Difference is Subset: $S \setminus T \subseteq S$ | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle T \cup \left\{ {x}\right\}\) | \(\subseteq\) | \(\displaystyle S\) | $T \subseteq S$ and $\left\{ {x}\right\} \subseteq S$, and Union is Smallest Superset |

As $x \notin T$ it follows that:

- $\left\{ {x}\right\} \cap T = \varnothing$

Thus by Cardinality of Set Union:

- $\left|{T \cup \left\{ {x}\right\}}\right| = k + 1$

That is, $T \cup \left\{ {x}\right\}$ is a subset of $S$ whose cardinality is $k + 1$.

This is the set whose existence was to be be proved.

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

- For all $n \in \N$, there exists a subset of $S$ whose cardinality is $n$.

$\Box$

### Sufficient Condition

Suppose that for all $n \in \N$, there exists a subset of $S$ whose cardinality is $n$.

Assume that $S$ is finite.

Let $N = \left\vert{S}\right\vert$.

As $N \in \N$ it follows that $N + 1 \in \N$.

By hypothesis, there exists a subset $T \subseteq S$ whose cardinality is $N + 1$.

From Cardinality of Subset of Finite Set, $\left\vert{S}\right\vert \ge \left\vert{T}\right\vert$.

But then $\left\vert{S}\right\vert = N \ge N + 1 = \left\vert{T}\right\vert$, which contradicts the fact that $N < N + 1$.

From this contradiction it follows that $S$ can not be finite.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Exercise $18.18$