Non-end points of Cantor ternary set

Question

I have studied the Cantor Ternary Set $C$ . We know that the Cantor ternary set is uncountable. Take the set $A$ as the set of all end points of the open removed segments. That is, $A=\{0,\frac{1}{3},\frac{2 }{3},\frac{1}{9},\frac{2}{9}....\}$. Now the set $C\setminus A$ is still uncountable. So the cardinality of $C$ was entirely determined by the non-end points.

My question is: Can someone write any $5$ (at least) members of $C\setminus A$?

I know $\frac{1}{4}, \frac{3}{10} \in C\setminus A$ but I don't know about any others.

Any help will be appreciated!

score 18 · Accepted Answer · answered Aug 14 '17 at 03:06

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Any real number between $0$ and $1$ whose ternary expansion doesn't have any $1$s is in the Cantor set. So, for example, each of the following ternary expansions corresponds to a point in the Cantor set:

$0.2020202020...$
$0.220022002200...$
$0.202200222000...$

and so forth. It's easy to check that these don't correspond to endpoints of the Cantor set.

answered Aug 14 '17 at 03:06

Noah Schweber

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To be precise, the endpoints of the Cantor set are exactly those whose ternary expansions of this form have either only finitely many $0$s or only finitely many $2$s. – Eric Wofsey Aug 14 '17 at 03:30
@HennoBrandsma What? The set of endpoints is definitely not dense. – Noah Schweber Aug 14 '17 at 04:18
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I would add that if you choose a repeating ternary expansion, you can write the number as a nice fraction. – Joonas Ilmavirta Aug 14 '17 at 07:39
@Noah Schweber: Thank You! – Chinnapparaj R Aug 14 '17 at 10:54
Just wanted to add there is a cool video here that explains this fact well – Anirudh Aug 14 '17 at 23:03

score 15 · Answer 2 · answered Aug 14 '17 at 03:55

Noah Schweber's answer gives a complete characterization of points in the Cantor set. A point is an endpoint if it eventually ends in all 0s or all 2s. Otherwise, it is a non-endpoint.

I'll just add a couple other ideas:

The Cantor set has some nice symmetries, like $x\mapsto \frac13 x$ and $x\mapsto 1-x$. These linear symmetries preserve endpoint-ness. So if you know that $\frac14$ is a non-endpoint, then you also know that $\frac{3}{4}$ and $\frac{1}{12}$ and $\frac{11}{12}$ and so on are non-endpoints. Similarly, $\frac{3}{10}$ gets you $\frac{1}{10},\frac{7}{10},\frac{9}{10},\frac{1}{30},$ and so on.

Finally, as Noah's third example illustrates, you can use ternary expansions to describe concrete irrational elements of $C\setminus A$ as well. For another example, we have $\theta\left(\frac13\right)-1=0.2002000020000002\ldots_3\in C\setminus A$. See my answer here for the meaning of the $\theta$ notation: Which irrationals are contained in the Cantor set?

MathMan · Answer 3 · 2017-08-14T04:24:07.557

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Any element $Z$ in the Cantor Set can be written as :

$Z= \sum_{n=1}^\infty x_n/3^n$ where $x_n=0$ or $2$.

You can substitute any combination of $0,2$ and have your number belonging to the cantor set

edited Aug 14 '17 at 04:24

answered Aug 14 '17 at 04:18

MathMan

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Yep. And if we require that the sequence $x$ doesn't have the property that it eventually equals $2$, then all the elements obtained in this way are distinct. – goblin GONE Aug 14 '17 at 04:33

Non-end points of Cantor ternary set

3 Answers3