Real Numbers are Uncountably Infinite/Set-Theoretical Approach: Proof 1

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

Proof
By Surjection from Natural Numbers iff Countable, a set $A$ is countable iff there exists a surjection $f: \N \to A$.

Suppose there exists 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 $\displaystyle x = e_0 + \sum_{n \mathop = 1}^\infty \frac {e_n} {10^n}$.

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.