Cantor's Diagonal Argument

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

Let $\mathbb F$ be the set of all mappings from the natural numbers $\N$ to $S$:
 * $\mathbb F = \set {f: \N \to S}$

Then $\mathbb F$ is uncountably infinite.

Proof
First we note that as $\card S > 1$, there are at least two elements of $S$ which are distinct.

Call these distinct elements $a$ and $b$.

That is:
 * $\exists a, b \in S: a \ne b$

First we show that $\mathbb F$ is infinite, as follows.

For each $m \in \N$, let $\phi_m$ be the mapping defined as:
 * $\map {\phi_m} n = \begin{cases}

a & : n \ne m \\ b & : n = m \end{cases}$

Then:
 * $\forall m_1, m_2 \in \N: m_1 \ne m_2 \implies \phi_{m_1} \ne \phi_{m_2}$

as:
 * $b = \map {\phi_{m_1} } {m_1} \ne \map {\phi_{m_2} } {m_1} = a$

So the mapping $\Psi: \N \to \mathbb F$ defined as:
 * $\map \Psi n = \phi_n$

is an injection.

Thus $\mathbb F$ is infinite from Infinite if Injection from Natural Numbers.

Next we show that $\mathbb F$ is uncountable.

Let $\Phi: \N \to \mathbb F$ be a mapping.

For each $n \in \N$ let $f_n: \N \to S$ be the function $\map \Phi n$.

Let us define $g: \N \to \N$ by:
 * $\map g n = \begin{cases}

a & : \map {f_n} n \ne a \\ b & : \map {f_n} n = a \end{cases}$

Then $g \in \mathbb F$, but:
 * $\forall n \in \N: \map g n \ne \map {f_n} n$

and so $g \ne f_n$.

Since $g$ is an element of $\mathbb F$ which is different from all the values taken by $\Phi$, it follows that $\Phi$ is not a surjection and hence not a bijection.

Thus no bijection exists between $\mathbb F$ and $\N$.

By definition of countably infinite, $\mathbb F$ is countably infinite there is a bijection between $\mathbb F$ and $\N$.

Therefore $\mathbb F$ is uncountable.

Also see

 * Set of Infinite Sequences is Uncountable, which is a basic application of this technique.