# Cantor's Diagonal Argument/Corollary

## Theorem

Let $S$ be a set such that $\card S > 1$, that is, such that $S$ is not a singleton.

Let $\mathbb G$ be the set of all mappings from the integers $\Z$ to $S$:

$\mathbb G = \set {f: \Z \to S}$

Then $\mathbb G$ is uncountably infinite.

## Proof

Let $\mathbb F$ be the set of all mappings from the natural numbers $\N$ to $S$:

$\mathbb F = \set {f: \N \to S}$

From Cantor's Diagonal Argument, $\mathbb F$ is uncountably infinite.

Let $s \in S$ be an arbitrary distinguished element of $S$.

Let $\mathbb H$ be the set of mappings $h: \Z \to S$ defined as:

$\forall f \in \mathbf F: \forall x \in \Z: \map h x = \begin {cases} \map f x & : x \ge \0 \\ s & : x < 0 \end {cases}$

It can be shown that $\mathbf H$ and $\mathbf F$ are equivalent.

Hence as $\mathbb F$ is uncountable, $\mathbf H$ is likewise uncountable.

We also have that $\mathbf H$ is a subset of $\mathbf G$.

Aiming for a contradiction, suppose $\mathbf G$ is countable.

Then from Subset of Countable Set is Countable it follows that $\mathbf H$ is countable.

Hence by Proof by Contradiction it follows that $\mathbf G$ is uncountable.

$\blacksquare$