Cantor's Diagonal Argument

Theorem
Let $$S$$ be a set such that $$\left|{S}\right| > 1$$, that is, such that $$S$$ is not a singleton.

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

Then $$\mathbb{F}$$ is uncountably infinite.

Proof
First we note that as $$\left|{S}\right| > 1$$, there are at least two elements of $$S$$ which are unequal. Call them $$a$$ 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:
 * $$\phi_m \left({n}\right) = \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 $$b = \phi_{m_1} \left({m_1}\right) \ne \phi_{m_2} \left({m_1}\right) = a$$.

So the mapping $$\psi: \N \to \mathbb{F}$$ defined as:
 * $$\Psi \left({n}\right) = \phi_{n}$$

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 $$\Phi \left({n}\right)$$.

Let us define $$g: \N \to \N$$ by:
 * $$g \left({n}\right) = \begin{cases}

a & : f_n \left({n}\right) \ne a \\ b & : f_n \left({n}\right) = a \end{cases}$$

Then $$g \in \mathbb{F}$$, but $$\forall n \in \N$$, $$g \left({n}\right) \ne f_n \left({n}\right)$$ 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 $$\N$$ and so $$\mathbb{F}$$ is not equivalent to $$\N$$.

Thus from Countably Infinite Iff Equivalent to Natural Numbers, $$\mathbb{F}$$ is uncountable.