Real Numbers are Uncountably Infinite

Theorem
The set of real numbers $$\R$$ is uncountably infinite.

Proof
Suppose there were a surjection $$f: \N \to \R$$.

Then $$\forall x \in \R: \exists n \in \N: f \left({n}\right) = x$$ as $$f$$ is surjective.

Let $$d_{n, 0}$$ be the integer before the decimal point of $$f \left({n}\right)$$.

Similarly, for all $$m > 0$$, let $$d_{n, m}$$ be the $$m$$th digit in the decimal expansion of $$f \left({n}\right)$$.

Let $$e_0$$ be an integer different from $$d_{0, 0}$$.

Similarly, for all $$m > 0$$, let $$e_m$$ be an integer different from $$d_{m, m}$$.

Specifically, we can define $$e_0$$ to be $$d_{0, 0} + 1$$, and:
 * $$e_m = \begin{cases}

1 & : d_{m, m} \ne 1 \\ 2 & : d_{m, m} = 1 \end{cases}$$

Now consider the real number $$x = e_0 + \sum_{n=1}^\infty \frac {e_n} {10^n}$$.

Its decimal expansion is:
 * $$x = \left[{e_0 . e_1 e_2 e_3 \ldots}\right]_{10}$$

Since $$e_0 \ne d_{0, 0}$$, $$x \ne f \left({0}\right)$$.

Similarly, for each $$n \in \N$$ such that $$n \ge 1$$, we have that $$e_n \ne d_{n, n}$$ and so $$x \ne f \left({n}\right)$$.

Thus $$x$$ is a real number which is not in the set $$\left\{{f \left({n}\right): n \in \N}\right\}$$.

Hence $$f$$ can not be surjective.