Real Numbers are Uncountably Infinite

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

Cantor's First Proof
We prove the equivalent result that every sequence $(x_k)_{k \in \N}$ omits at least one $x \in \R$.

Given $(x_k)_{k \in \N}$ of distinct real numbers define a sequence on closed real intervals $I_n$ as follows:

Let $a_1 = \min\{x_1,x_2\}$, $b_1 = \max\{x_1,x_2\}$ and $I_1 = [a_1,b_1]$.

Since $a_1\neq b_1$ this interval is not a singleton.

Now suppose we are given $I_{n-1} = [a_{n-1},b_{n-1}]$.

We may assume that infinitely many of the $x_k$ lie inside $I_{n-1}$ (otherwise we are done).

Let $y$ and $z$ be the first two such numbers, and let


 * $a_n = \min\{y,z\}$, $b_n = \max\{y,z\}$, $I_n = [a_n,b_n]$.

Thus we have sequences $(a_k)_{k \in \N}$ and $(b_k)_{k \in \N}$ with


 * $ a_1 < a_2 < \cdots < b_2 < b_1 $

So $(a_k)_{k \in \N}$ and $(b_k)_{k \in \N}$ are monotone and bounded above and below respectively.

Therefore by the Monotone Convergence Theorem both are convergent. Let


 * $\displaystyle A = \lim_{k \to \infty} a_k$, $\displaystyle B = \lim_{k \to \infty} b_k$.

Clearly we have $A \leq B$, so $[A,B] \neq \emptyset$.

Let $h \in [A,B]$. Then $h \neq a_k,b_k$ for all $k$.

We claim that $h \neq x_k$ for all $k$.

Suppose that $h = x_k$ for some $k$.

Then there are only finitely many points in the sequence before $h$ occurs, and therefore only finitely many of the $a_k$ precede $h$.

Let $a_d$ be the last element of the sequence $(a_k)_{k \in \N}$ preceding $h$.

We defined $a_{d+1}$, $b_{d+1}$ to be interior points of $I_d$, and also $h \in I_{d+1}$ by the definition of $h$.

Therefore $a_{d+1}$ must precede $h$ in the sequence, for the sequence in monotone increasing, a contradiction.

Cantor's Second Proof
By definition, a perfect set is a set $X$ such that every point $x \in X$ is the limit of a sequence of points of $X$ distinct from $x$.

From Real Numbers form a Perfect Set, $\R$ is perfect.

Therefore it is sufficient to show that a perfect subset of $X \subseteq \R^k$ is uncountable.

We prove the equivalent result that every sequence $(x_k)_{k \in \N} \subseteq X$ omits at least one point in $X$.

Let $y_1 \in X$, and let $B_1 := \mathcal B_r(y_1)$ be some closed ball centred at $y_1$.

Given $B_{n-1}$ choose a closed ball $B_{n-1} \supseteq B_n := \mathcal B_{\delta(n)}(y_n)$ such that $\delta(n) \leq \delta(n-1)/2$ (note that $\delta(1) = r$), $y_n \in X$ and $x_n \notin B_n$.

We can satisfy the condition $x_n \notin B_n$ because $X$ is perfect, so every ball centred at a point of $X$ contains infinitely many points of $X$.

Since $\displaystyle \delta(n) \leq \frac r{2^{n-1}}$, $y_n$ is Cauchy.

Therefore since a perfect set is necessarily closed we may let $\displaystyle Y = \lim_{n \to \infty} y_n \in X$.

For $n \in \N$ we have $\{ y_m : m > n\} \subseteq B_n$, so $Y \in B_n$.

But by construction for each $n \in \N$ $x_n \notin B_n$.

Therefore $Y \neq x_n$ for all $n \in \N$, and we are done.

Cantor's Diagonal Argument
We show that the unit interval $[0. . 1]$ is uncountable, from which uncountability of $\R$ follows immediately.

Suppose that $[0. . 1]$ is countable.

Clearly $[0. . 1]$ is not finite because $\displaystyle \frac 1 n$ are distinct for $n \in \N$.

Therefore an injection $[0. . 1]\hookrightarrow \N$ enumerates $[0. . 1]$ with a subset of the natural numbers.

By relabeling, we can associate each $x \in [0. . 1]$ to precisely one natural number to obtain a bijection.

Let $g$ be such a correspondence:

where juxtaposition of digits descibes the decimal expansion of a number.

Note that the decimal expansion is not directly unique, because of 0.999...=1.

This problem is overcome by disallowing infinite strings of $9$'s in the decimal expansion.

For every $k \in \N$ define $f_k = d_{kk} + 1$ taken modulo $10$.

Let $y$ be defined by the decimal expansion:


 * $y = 0.f_1 f_2 f_3 \ldots$

Now


 * $y$ differs from $g(1)$ in the first digit of the decimal expansion
 * $y$ differs from $g(2)$ in the second digit of the decimal expansion

and generally the $n^\text{th}$ digit of the decimal expansion of $g(n)$ and $y$ is different.

By Existence of Base-N Representation, any decimal expansion of a real number is unique (except for the issue from 0.999...=1).

So $y$ can be none of the numbers $g(n)$ for $n \in \N$.

But $g$ is a bijection, a contradiction.

Set-Theoretical Approach: Proof 1
By definition a set $A$ is countable iff there is a surjection from $f: \N \to A$.

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

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.

Set-Theoretical Approach: Proof 2
By Cantor's Theorem there is no surjection $\N \twoheadrightarrow \mathcal P(\N)$.

Additionally, we know that the powerset of the natural numbers is not countable.

Therefore, if we can show that $\mathcal P(\N)$ injects into $\R$ then there is no injection $\R \hookrightarrow \N$ and $\R$ is uncountable.

To prove the theorem we construct an injection, say $f$.

For a subset $S \subseteq \N$, let $\chi_S$ be the characteristic function of $S$, and let $d_i = \chi_S(i)$ for all $i \in \N$.

If the sequence $(d_i)_{i \in \N}$ does not terminate in an infinite sequence of $1$'s, then $f$ sends $S \subseteq \N$ to the binary expansion of a number in $[0,1]$:


 * $ 0.d_1d_2d_3d_4\ldots $

If the sequence $(d_i)_{i \in \N}$ does terminate in an infinite sequence of $1$'s, then $f$ sends $S$ to the binary integer:


 * $d_1d_2d_3\ldots d_k .000\ldots$

where $d_k$ is the last member of the sequence not equal to $1$.

Injectivity of $f$ follows from the uniqueness statement of Existence of Base-N Representation.