# The set of elements that do not escape the tent map is a Cantor Set

Let $x \in [0,1]$ and $r>2$. Define the Tent map by

$$T(x) = \left\{ \begin{array}{lr} xr & : x \in [0,\frac{1}{2}]\\ r(1-x) & : x \in (\frac{1}{2},1] \end{array} \right.$$

A point $y$ is said to escape the tent map in one iteration if $T(y)>1$. It is said to escape in two iterations if $T(T(y))=T^{(2)}(y)>1$, etc. What I need to show is that the set of points that never escape the tent map, $$C=\{x \in [0,1]:\lim_{n \to +\infty}T^{(n)}(x)\leq 1 \}$$ is a Cantor Set. By "a Cantor Set" I mean a set where we begin by removing the middle $\frac{1}{s}th$ (where $s>2$) portion of the unit interval, then proceed by removing the middle $\frac{1}{s}th$ from the remaining two intervals, then we move to the next four intervals and do the same, etc. In this way it may help to first choose $s$ and then let $r=\frac{2s}{s-1}$ so that $r$ will whittle away the desired portion of the unit interval. In the case of THE Cantor set, we would have $s=3$ for example, since THE Cantor set is built by removing the middle open set of length $\frac{1}{3}$ from $[0,1]$ and each subsequent interval thereafter. I believe proof by induction will lead to the solution. The definition of Cantor Set I am using is taken from the case of $s=3$ $$C_{n} = \frac{C_{n-1}}{3} \cup \left(\frac{2}{3}+\frac{C_{n-1}}{3}\right), \ \ C_{0}=[0,1]$$ where I tweak it to reflect my definition of $r$ from above so that $$C_{n} = \left(C_{n-1}\frac{s-1}{2s}\right) \bigcup \left(\frac{s+1}{2s}+C_{n-1}\frac{s-1}{2s}\right), \ \ C_{0}=[0,1]$$ Again, note that plugging $s=3$ into my result above will take you back to THE Cantor set. At any rate, my approach is showing that $C_{n}$ is the set up points that do not escape the tent map after $n$ iterations. If I succeed in the proof by induction, this should establish that $\lim_{n \rightarrow \infty}C_{n}=C$ as desired. Can anyone provide a proof of this?

• what's a Cantor set ? Commented Sep 3, 2014 at 19:03
• en.wikipedia.org/wiki/Cantor_set Commented Sep 3, 2014 at 19:06
• this gives a definition of "the" Cantor set (and of course yours is not this precise set for $r \neq 3$), but it doesn't say what "a" Cantor set is. (or maybe I can't read) Commented Sep 3, 2014 at 19:37
• I see what you mean.. there isn't really a solid definition there. I guess what I'm going for is a set with the middle $\frac{1}{r}^{th}$ removed from each subsequent interval. And as you pointed out, the case of $r=3$ is THE Cantor set, but a Cantor set is the case of $r \neq 3$. Commented Sep 3, 2014 at 19:43
• what you can do is show there is a homeomorphism of $[0;1]$ that sends $C$ onto the cantor set. That should probably be satisfactory. Commented Sep 3, 2014 at 21:12

## EDIT :

Using the new formulation described in your comments below, we introduce a new parameter $s > 2$ and define the Tent Map parameter $r$ as $$r = \frac{2s}{s-1}.$$ The Cantor-type set thus defined is completely determined by the (internal) endpoints obtained after each set of removals of the inner $s'th$ open intervals. Let us say that an endpoint has rank n if the endpoint is one of those obtained by the n'th set of removals.



To prove that members of this Cantor-type set never escape we would argue using induction on the rank. Key to this is that if $a = \frac pq$ is an endpoint, then the Tent Map will map $a$ to an endpoint of lower rank because $2s$ always divides $q$ and $(s-1)$ always divides $p$, and similarly for $(1-a)$.  The tricky part is obtaining a recursion formula for the endpoints. Each endpoint of rank $n$ determines two endpoints of rank $n+1$. The exact recursion is very fussy and it depends on whether an endpoint's children are to the left or right of the original endpoint. It think each endpoint will looks something like this : $$a = \frac 12 \pm \frac {1}{2s} \pm \frac {1}{4s} \pm \dots$$ where NOT ALL $\pm$ - combinations are permitted.

Unfortunately this is all a bit too time consuming for me presently but it is doable with a bit of patience.  This would get us as far as showing the points in the Cantor-type set do not escape. Proving that those outside of the set do escape is another matter all together since it does not appear that induction would be possible and we would need to rely on $\lt$ and $\gt$ order relations.

So, I'm afraid that is all the time I can spend on this right now. I hope it is of some help. This is a very clever and interesting problem and whoever formulated it clearly knows their stuff. Good Luck! 

## Following is my answer to the problem as originally formulated, which I leave for the sake of completeness.

The generalised definition of a Cantor set that you are assuming requires $r$ be an odd number, for otherwise there is no middle $r'th$.

If $r > 2$ is odd, note that $$\frac{(r-1)}{2r} \in \mathcal C_r,$$ since it is the rightmost point of the left interval resulting from the first removal of the middle $r'th$, and cannot be excluded by subsequent removals.

Further $$T\left(\frac{(r-1)}{2r}\right) = \frac{(r-1)}{2}.$$ So whenever $r$ is odd and greater than $3$, this generalized Cantor set member escapes on the first iteration.

This leaves the case $r=3$ as a possibility, the Cantor Ternary Set itself. The argument I have given above would not apply here since $T(T(\frac13)) = 0.$ So if your Tent set is a Cantor set, it must be The Cantor set.

• r does not have to be odd.. I could remove the middle $\frac{1}{4}$ of every interval should I choose Commented Sep 7, 2014 at 1:35
• @graydad If you divide [0,1] into 4 subintervals as per Cantor, then you have (e.g.) [0, 0.25), [0.25, 0.5), [0.5, 0.75), [0.75, 1]. Which one is the middle interval? Commented Sep 7, 2014 at 1:53
• @graydad If you mean removing an interval of lenght 0.25 from the center, then you can identify the same point - in this case 3/8, which would again escape on the first iteration. Commented Sep 7, 2014 at 1:57
• @graydad Doing a bit of scribbling, I believe that your Tent set corresponds to the case r=2. I'm not sure why you exclude the case r=2 in your OP. If you sit down and look at the end-points remaining after each removal, you will convince yourself that all of these points map to zero. By the case r=2 I mean the case with the first removal leaving [0,1/4] and [3/4,1] ; the second [0,1/16], [3/16, 1/4], etc.. Commented Sep 7, 2014 at 2:39
• I mean what you said in the second case: After the first iteration, you will be left with $[0,\frac{3}{8}] \cup [\frac{5}{8},1]$ and then proceed by removing the same proportion from the middles of the consequent intervals. The case of $r=2$ is important, but points do not escape the tent map when $r=2$; that only happens for \$r>2. Commented Sep 7, 2014 at 2:48

Succeeded with a proof by induction.

I have a relatively simple proof for all $$r>2$$. There is no need for $$r$$ to be odd, nor integer, rational, etc. We only need $$r$$ to be real and greater than two. I think Epsilon's answer confused $$s$$ with $$r$$, since he thought $$\frac{r-1}{2r}\in C\;(wrong)$$, which it doesn't, the point he describes is $$\frac{1}{r}\in C$$. As stated in the question, the set of $$x$$ with bounded orbits is in fact a Cantor set with $$r=\frac{2s}{s-1}$$, but, in order to avoid the confusion of the previous post, I will provide the definition of the Cantor set in terms of $$r$$. First, $$C_0 =[0,1]$$. Then, $$\forall n\in \mathbb{Z_+}:C_n=\left(\frac{1}{r}C_{n-1} \right)\cup\left(\frac{r-1}{r}+\frac{1}{r}C_{n-1} \right)\;.\;(1)$$ You can check that this definition coincides with the one of the question if $$r=\frac{2s}{s-1}$$. We need $$r>2$$ so that the Cantor set just defined is a proper subset of the $$[0,1]$$ interval. In fact:

• If $$1\le r \le 2$$, then $$\forall n \in \mathbb{N}: C_n =[0,1]$$.
• If $$r<1$$ then $$C_1 \not\subset [0,1]$$.

The sketch of the proof is as follows:

1. $$\forall n \in\mathbb{N}: C_n=1-C_n$$. Easily proved by induction.
2. $$\forall n\in\mathbb{Z}_+ : C_n =T^{-1}(C_{n-1})=\left(\frac{1}{r}C_{n-1} \right)\cup\left(1-\frac{1}{r}C_{n-1} \right)$$. Use the definition of $$T(x)$$, Eq. (1), and the previous step.
3. $$T^{(n)}(x)$$ is bounded iff $$x\in C=\bigcap_{n=0}^{\infty}C_n$$. Because $$T^{(n)}(x)$$ is unbounded iff $$\exists k\in\mathbb{N}: T^{(k)}(x)\notin [0,1]$$. The previous step proved that $$x\in C \iff \forall n\in \mathbb{N}:T^{(n)}(x)\in [0,1]$$. $$\square$$