Let $S_0=[0,1]$ and define every $S_k$ for $k\geq 1$ \begin{align*} S_1&=\left[0,\frac{1}{3}\right]\cup\left[\frac{2}{3}, 1\right],\\ S_2&=\left[0,\frac{1}{9}\right]\cup\left[\frac{2}{9}, \frac{3}{9}\right]\cup\left[\frac{6}{9}, \frac{7}{9}\right]\cup\left[\frac{8}{9},1\right],\\ S_3&=\left[0, \frac{1}{27}\right]\cup\left[ \frac{2}{27}, \frac{3}{27}\right]\cup\left[ \frac{6}{27}, \frac{7}{27}\right]\cup\left[ \frac{8}{27}, \frac{9}{27}\right]\cup\left[ \frac{18}{27}, \frac{19}{27}\right]\cup\left[ \frac{20}{27}, \frac{21}{27}\right]\cup\left[ \frac{24}{25}, \frac{25}{27}\right]\cup\left[ \frac{26}{27}, 1\right]\\ \vdots \end{align*}

Now let $C=\bigcap_{k=0}^\infty S_k$. This is known as a Cantor set. I have to prove that C is non-empty, by showing that 1/4 is in C. I also have find the cardinal of C and prove my answer.

I have some ideas how to prove non-emptiness, but I am stuck as to how to specifically prove that 1/4 is in C. Also I think that the cardinal is equal to the cardinal of the reals $2^{\aleph_0}$. I probably need to find a bijection to prove this right?

  • $\begingroup$ Notice that the endpoints of each of the intervals that comprise $S_j$ are in the cantor set for all $j$. In particular, $0 \in C$, so the cantor set is non-empty. In fact, its uncountable. $\endgroup$ – Mustafa Said Oct 29 '14 at 8:20
  • $\begingroup$ Notice that my question specifically refers to proving non-emptiness by showing 1/4 is in C. Just generally proving non-emptiness is not hard and I could probably swing a proof myself. $\endgroup$ – prometheus21 Oct 29 '14 at 8:22
  • $\begingroup$ see this $\endgroup$ – John Oct 29 '14 at 8:25
  • $\begingroup$ The proof by induction by John Zhang is pretty neat. Are there any other proofs? How about the cardinal? $\endgroup$ – prometheus21 Oct 29 '14 at 8:30
  • $\begingroup$ See this for an injection of $\mathbb R$ into $C$. The Bernstein-Cantor-Schroeder theorem then gives you a bijection. $\endgroup$ – Andrés E. Caicedo Oct 29 '14 at 13:42

Let's try the coding approach. We will take $0$ for left and $2$ for right. Then

$$C_0:=[0,1/3] \text{ and } C_2:=[2/3,1]$$

and $S_1 = C_0\cup C_2$. Similarly the left and right subintervals of $C_0$ are coded $C_{00}$ and $C_{02}$ and for $C_2$ we have $C_{20}$ and $C_{22}$; for example $C_{02}:=[2/9,3/9]$. This gives us

$$S_2=C_{00}\cup C_{02}\cup C_{20}\cup C_{22}$$

The intervals in $S_3$ are specified for string of length three. In general we can use induction and shows that the intervals of $S_n$ are specified for string of $n$ elements each of which is a chain of $0$ and $2$'s. Now if we have an infinite string of $0$´s and $2$'s, i.e., $a_1a_2a_3a_4\ldots$we have the intervals

$$C_{a_1}\supset C_{a_1a_2}\supset C_{a_1a_2a_3}\supset \ldots C_{a_1\ldots a_n}\supset \ldots$$

The intersection of which is a point in the Cantor set. So any infinite address string defines a point in $C$. Conversely any point $p$ in $C$ defines an infinite address, where its first $n$ symbols are specified by its location in $S_n$. Notice that this addressing is in fact injective, any other point $q$ in the Cantor set has different address, because at some $n$, $p$ and $q$ are in different $C_{\alpha}$ and $C_{\beta}$ in $S_n$. Thus each point in $C$ is in a 1-1 correspondence with the collection of all address.

If each $2$ is replaced by $1$, we have a base two expansion of some $x\in [0,1]$. So as there are uncountable many points in $[0,1]$, uncountable many base two expansion and uncountable many addresses so uncountable many points in $C$.

Now you can interpret the addresses as a base three expansion of points $x\in [0,1]$ as we have shown there is a 1-1 correspondence between the Cantor set and the address and the base three expansion of $1/4_3= .02020202\ldots_3$, so is in a point in $C$.


One way to think of the cantor set is in terms of ternary expansions. We typically write a real number $r \in [0, 1]$ in decimal notation as $$ r = 0.a_1a_2a_3a_4\cdots $$ where $a_i \in \{0, \ldots, 9\}$. But we can pick other bases such as base 2 ($a_i \in \{0, 1\}$) or for our purposes, base 3 ($a_i \in \{0, 1, 2\}$).

Now if you think of numbers in terms of their ternary expansions, you should note that the Cantor set simply removes any number whose ternary expansion includes the digit 1 (that is, essentially, what throwing out the middle interval does). So to show that $1/4$ is in the Cantor set, it would suffice to show that the ternary expansion of $1/4$ does not include a 1.

Furthermore, if you think about this a little more, you should be able to adapt this to showing the uncountability of the cantor set. Remember that every number in the unit interval can be written in a binary expansion, which only involves 0 and 1...

  • $\begingroup$ I think I can sketch a proof for 1/4 based on your answer. Now about the second part. You are saying that by proving C is uncountable, its cardinal is automatically $2^{\aleph_0}$? $\endgroup$ – prometheus21 Oct 29 '14 at 9:07
  • $\begingroup$ Well... that's not quite true. It is true that $2^{\aleph_0}$ is uncountable, but I guess I phrased my answer poorly. There are other uncountable cardinals. In this case though, the cardinality of the Cantor set is exactly what you think it is. $\endgroup$ – Simon Rose Oct 29 '14 at 9:10

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