A set $S$ is said to be uncountable if there is no injection $f:S\to\mathbb{N}$. Assuming the Axiom of choice, every set that is not uncountable is either finite or countably infinite. The most common example of an uncountable set is the set of real numbers $\mathbb{R}$.

Proof that $\mathbb{R}$ is uncountable

We give an indirect proof here. This is one of the most famous indirect proofs and was first given by Georg Cantor.

Suppose that the set $A=\{x\in\mathbb{R}:0<x< 1\}$ is countable. Let $\{\omega_1, \omega_2, \omega_3, ...\}$ be any enumeration of the elements of $A$ (in other words, take an injection $f: A \to \mathbb{N}$, and denote $\omega_i = f(i)$).

Consider the decimal expansion of each $\omega_i$, say $\omega_i=0.b_{i1}b_{i2}b_{i3} \ldots$ for all $i$. Now construct a real number $\omega= 0.b_1b_2b_3 \ldots$, by choosing the digit $b_i$ so that it differs from $b_{ii}$ by at least 3 and so that $b_i$ is never equal to 9 or 0. It follows that $\omega$ differs from $\omega_i$ by at least $\frac{2}{10^i}$, so $\omega \neq \omega_i$ for every $i$. Thus, $\omega \not \in A$. However, $\omega$ is clearly a real number between 0 and 1, a contradiction. Thus our assumption that $A$ is countable must be false, and since $\mathbb{R} \supset A$ we have that $\mathbb{R}$ is uncountable.

An alternative proof uses Cantor's Theorem, which says that for all sets $S$, we have $|S|<|\mathcal{P}(S)|$, where $\mathcal{P}(S)$ is the power set of $S$. First, we note that the Cantor set $\mathcal{C}$ has cardinality $2^{\aleph_{0}}>\aleph_{0}$, and since $\mathcal{C}\subset\mathbb{R}$, there is an injection $f:\mathcal{C}\rightarrow\mathbb{R}$ and thus $|\mathbb{R}|\geq 2^{\aleph_{0}}>\aleph_{0}$, so $\mathbb{R}$ is uncountable. In fact, it turns out that $|\mathbb{R}|=2^{\aleph_{0}}$.

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