1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
-1
$\begingroup$

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 $C$.

$\endgroup$
1
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.