# Proving That The Cantor Set is Uncountable Using Base-3

I've been given the following explanation of why the Cantor set is uncountable using base-3 (shortened):

In base-3 we write some arbitrary number $$x \in [0,1]$$ as $$x = 0.b_1b_2b_3..._3 = \frac{b_1}{3}+\frac{b_2}{3^2}+\frac{b_3}{3^3} +\enspace ...$$ with $$b_i \in \left\{0,1,2\right\}$$. So we can write $$x = \sum_{i=1}^{\infty}b_i3^{-i}$$. In this format, the Cantor set has some properties. We divide $$[0,1]$$ into three equal pieces. The first digit, $$b_1$$, tells us whether $$x$$ is in the left, middle or right of a given line segment and so on. Assuming that $$\exists$$ some $$k \in \mathbb{N}$$ such that $$b_k = 1$$ then $$x$$ will be in some middle third of a given interval so $$x \notin \mathcal{C}$$. Conversely, by the definition of the base-3 expansion, if $$b_n \in \left\{0,2\right\}$$ for all $$n \in \mathbb{N}$$, $$x$$ will never be in the middle third of any interval, so $$x \in \mathcal{C}$$

With this formulation of the Cantor set we can easily show that it is uncountable. Suppose instead the opposite, that $$\mathcal{C}$$ is countable, so $$\mathcal{C} = \left\{x^1, x^2, x^3, ...\right\}$$ where each element of $$\mathcal{C}$$ is written like: $$x^1 = 0.b_1^1b_2^1b_3^1..._3$$ $$x^2 = 0.b_1^2b_2^2b_3^2..._3$$ $$x^3 = 0.b_1^3b_2^3b_3^3..._3$$ $$\vdots$$

Here is where I get lost:

Let $$\left(b_1,b_2,b_3,...\right)$$ be the sequence that differs from the diagonal sequence $$\left(b_1^1,b_2^2,b_3^3,...\right)$$. $$\bf\text{In other words if b_i^i = 2 then b_i = 0.}$$ As a result, $$x = 0.b_1b_2b_3..._3$$ never appears in $$\mathcal{C}$$!

Does this mean that because the element composed by this sequence differs at the diagonal every time that it will never equal to any element of $$\mathcal{C}$$? So we can't write every element of $$\mathcal{C}$$? Could we also say "if $$b_i^i = 0$$, then $$b_i = 2$$?".

• Yes, yes, yes. This is the Cantor's diagonal argument – Pedro Apr 9 '20 at 12:18
• Does this apply to all uncountably infinite sets? – yerman Apr 9 '20 at 12:20
• Well, it can be used provided that the elements of the set are characterized by infinite sequences. (Diagonalization arguments also appear in different contexts, as in the proof of the Arzelà-Ascoli theorem) – Pedro Apr 9 '20 at 12:49