What are the ways of proving that the Cantor set is uncountable apart from Cantor diagonalization?

Question

What are the ways of proving that the Cantor set is uncountable apart from Cantor diagonalization? Are there any based on dynamical systems?

Answer 1

Exercise: Show that Cantor set $C$ is equal to $$C = \left\{\sum_{n=1}^\infty \frac{a_n}{3^n}: a_n \in \{0, 2\}\right\}$$

Then:

One way to intuitively understand the cardinality of the Cantor set is to think of it as essentially the set of ternary (base 3) representations of all $x\in [0, 1] \subset \mathbb{R}$ which can be expressed using only the digits 0 and 2. That is, it is the set of all "trecimal expansions” of $x\in [0, 1] \subset \mathbb{R}$ that consist of of only zeros and twos: e.g. $0.0 \bar{0}_{\text{ base 3}} = 0 \in C, \; 0.2\bar{2}_{\text{ base 3}} = 1\in C,\;\; 0.2020202020202…_\text{(base 3)} \in C \;$, etc.

The set of all such $x$ is uncountable:
There are $2^{|\mathbb{N}|}$ distinct sequences consisting of only zeros and twos: for each $n\in \mathbb{N}$, there are two possible choices for any given $a_n : 0$ or $2$. Since there are uncountably many distinct sequences $\{a_n\} \in \{0, 2\}^\mathbb{N}$, there are uncountably many series $\large\sum {a_n\over 3^n}$, each of which represents a single point in the Cantor set.

Hence there are uncountably many points in the Cantor Set.

Answer 2

Instead of the diagonal proof directly, we can adapt Cantor's first proof of uncountability. Let $C$ be the cantor set, and let $E=\{u_1, u_2, \dots\}$ be a countable set. We construct a point of $C$ not in $E$. First, $C \subseteq [0,1/3]\cup [2/3,1]$, and point $u_1$ does not belong to both of those intervals, so there is an interval $I_1$ of length $1/3$ (one of the ones in the first stage of the construction of the Cantor set) with $u_1 \not\in I_1$. Now when we remove the middle third of $I_1$ we get two intervals of length $1/3^2$. As before, $u_2$ does not belong to both of these intervals, so there is an interval $I_2 \subset I_1$ of length $1/3^2$ (one of the ones in the second stage of the construction of the Cantor set) with $u_2 \not\in I_2$. Continue in this way to get $I_1 \supset I_2 \supset I_3 \supset \dots \supset I_k \supset \dots$ where $I_k$ has length $3^{-k}$ (one of the intervals in the $k$th stage of the construction of the Cantor set) and $u_k \not\in I_k$. Finally, we get a point $x \in \bigcap_{k=1}^\infty I_k$ where $x \in C$ but $x \ne u_k$ for all $k$.

Answer 3

You can use the ol' Baire Category argument: as a closed subspace of $\mathbb{R}$, the Cantor set is completely metrizable (the usual metric is complete) and is thus a Baire space. Therefore the Cantor set is not a countable union of nowhere-dense subsets. As the Cantor set is perfect, the singletons are nowhere dense (in the Cantor set), and so the Cantor set cannot be countable.

Answer 4

In the spirit of Arthur's answer, let's take some tools from a more advanced part of mathematics.

Theorem 1. Every countable and complete metric space is homeomorphic to a countable ordinal with the order topology.

Theorem 2. Every ordinal space contains isolated points. Furthermore, if the ordinal is infinite then there are infinitely many isolated points.

The Cantor space is compact and therefore complete with the metric induced by $\mathbb R$. If the Cantor space was countable then it would be isomorphic an ordinal and would therefore have isolated points. However the Cantor space does not have any isolated points, and therefore cannot be countable.

Answer 5

$\newcommand{\cl}{\operatorname{cl}}$Let $C$ be a Cantor set, and let $f:\Bbb N\to C$ be any function. For $n\in\Bbb N$ let $a_n=f(n)$, and let $A=\{a_n:n\in\Bbb N\}$.

Claim: $A\ne C$.

Since $A$ is an arbitrary countable subset of $C$, it follows from the claim that $C$ is uncountable.

Proof of Claim: Clearly $C$ is infinite, so there is some $x\in C\setminus\{a_0\}$. $C$ is $T_4$, so there is an open nbhd $V_0$ of $x$ such that $a_0\notin\cl V_0$. Suppose that for some $n\in\Bbb N$ we have a non-empty open set $V_n$ such that $\{a_k:k\le n\}\cap\cl V_n=\varnothing$. If $a_{n+1}\notin\cl V_n$, let $V_{n+1}=V_n$. Otherwise, use the fact that $C$ has no isolated points to fix $x\in V_n\setminus\{a_{n+1}\}$, and choose an open nbhd $V_{n+1}$ of $x$ such that $\cl V_{n+1}\subseteq V_n\setminus\{a_{n+1}\}$. In this way we recursively construct open sets $V_n$ such that for $n\in\Bbb N$, $\cl V_{n+1}\subseteq V_n$ and $\{a_k:k\le n\}\cap V_n=\varnothing$. It follows from the compactness of $C$ that $$\bigcap_{n\in\Bbb N}\cl V_n\ne\varnothing\;,$$ but clearly $$A\cap\bigcap_{n\in\Bbb N}\cl V_n=\varnothing\;,$$ so $A\ne C$. $\dashv$

Answer 6

Use a fat Cantor set, and show that it has positive Lebesgue measure. Then, remember that all countable sets are null.

Answer 7

The canonical proof that the Cantor set is uncountable does not use Cantor's diagonal argument directly. It uses the fact that there exists a bijection with an uncountable set (usually the interval $[0,1]$).

Now, to prove that $[0,1]$ is uncountable, one does use the diagonal argument. I'm personally not aware of a proof that doesn't use it.