5
$\begingroup$

Associate to each sequence $a=\{\alpha_n\},$ in which $\alpha_n$ is $0$ or $2$, the real number $$x(a)=\sum \limits_{n=1}^{\infty}\frac{\alpha_n}{3^n}.$$ Prove that the set of all $x(a)$ is precisely the Cantor set.

We know that $\frac{1}{3}\in C$ but also $\frac{1}{3}=\sum \limits_{n=2}^{\infty}\frac{2}{3^n}$ and $\frac{1}{3}=\frac{1}{3}+\sum \limits_{n=2}^{\infty}\frac{0}{3^n}$. And we have some ambiguity.

Can we also add condition that $\alpha_j=1$ for some $j$?

Can anyone explain this moment to me?

$\endgroup$
  • 1
    $\begingroup$ There's no ambiguity. The number 1/3 can be expressed by means of the first formula so it's in C. The fact that 1/3 also has another representation does not contradict anything. $\endgroup$ – DanielWainfleet Aug 30 '15 at 19:05
  • $\begingroup$ What is the question exactly? $\endgroup$ – Did Aug 30 '15 at 19:07
  • $\begingroup$ @Did, $1/3$ has at least two ternary expansion (above I wrote them). Why we can't add also condition that $a_j=1$ for some $j$? $\endgroup$ – ZFR Aug 30 '15 at 19:09
  • 1
    $\begingroup$ Yeah, and one of these two ternary expansions uses only 0s and 2s hence 1/3 is in the Cantor set. Unambiguously. (Repeating a previous comment above.) $\endgroup$ – Did Aug 30 '15 at 19:14
  • $\begingroup$ @Did, where I can find proof to this fact? $\endgroup$ – ZFR Aug 30 '15 at 19:17
5
$\begingroup$

Let $C_1= [0,\frac{1}{3}] \cup [\frac{2}{3}, 1]$, $C_2 = [0, \frac{1}{9}]\cup[\frac{2}{9},\frac{1}{3}]\cup[\frac{2}{3}, \frac{7}{9}]\cup[\frac{8}{9},1]$, $\ldots$ and continue, in this manner, defining the $C_k$ by "eliminating the middle third" of each interval in the previous $C_{k-1}$, then we have the Cantor tertiary set $$C=\bigcap_{k=1}^\infty C_k$$

and our task is to show that $C$ is precisely the set of all real numbers in $[0,1]$ that can be represented with only zeros and twos in its base-three decimal expansion. NOTE: finite decimal expansions, as always, are not formally well-defined; for instance, $\frac{1}{3} = 0.1 = 0.022222 \ldots$ so that $\frac{1}{3} \in C$ because it can be expressed with only zeros and twos in its base-three decimal expansion. Note, also, that $\frac{1}{3}$ is in every $C_k$ as defined above, so that it is in the intersection of all of them.

This is going to be our general idea. If some $x \in C$, we show that $x$ has a base-three expansion consisting of only zeros and twos. But that doesn't let us off the hook yet - there perhaps could be numbers with base-three expansions consisting of only zeros and twos that are NOT in $C$, so we will, as our second component of the proof, show that given a number with a base-three expansion consisting of only zeros and twos, that number must be in $C$.

We have that $x \in C$ if and only if $x \in C_k$ for all $k \in \mathbb{N}$. For $C_1$ note that $x$ can have a $0$ in the "one-thirds" place if $x$ is in $[0, \frac{1}{3}]$, or that $x$ can have a $2$ in the "one-thirds" place if $x$ is in $[\frac{2}{3}, 1]$. Likewise, for the second decimal place, consider which of the two intervals in $C_2$ where $x$ belongs (why do we only examine two, and not all four intervals used to construct $C_2$?) and note a zero or a two can be assigned accordingly for the second decimal place. Continue likewise, and formalize what I have written, to conclude that $x \in C$ if and only if $x$ has a base-three expansion consisting of only zeros and twos.

$\endgroup$
  • $\begingroup$ "(why do we only examine two, and not all four intervals used to construct $C_2$?)" what is the answer to this? $\endgroup$ – IntegrateThis Oct 11 '18 at 21:07
1
$\begingroup$

As has been stated in the comments, the fact that some members of the Cantor set have a second ternary representation which includes 1 is immaterial to the result you are trying to prove. It states that as long as the number has at least one representation without 1s, it is in the Cantor set.

To prove it, assume that $E_n$ consists of all numbers whose ternary expansions consist only of 0 or 2 up to the $n$th trit (I may be off by one, I don't know exactly how Rudin defines his symbols). Then note that $E_{n+1}$ throws out every number whose $n+1$st trit is 1.

$\endgroup$

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.