Proving rationals cannot have dense orbit in Cantor set

[razmtaz]

Homework Statement

Let C be the standard Cantor "middle third" set (ie C_k = {x:0$\leq$T$^{k}_{3/2}(x)$$\leq$1} and C = $\bigcap$$^{inf}_{k=0}$C_k
where T$^{k}_{3/2} = 3x if x<1/2, = 3 - 3x if x \geq$ 1/2)

Show that a rational number x = p/q $\in$ C cannot have dense orbit in C.

Homework Equations

{T$^{k}_{3/2}$(x)}$^{inf}_{k=0}$ is the orbit of x.

Def A subset A of the interval J is dense in J if A intersects every nonempty open subinterval of J.

The Attempt at a Solution

Want to show that the orbit A = {T$^{k}_{3/2}$(x)}$^{inf}_{k=0}$ of any rational number x = p/q is not dense in C. so we want to show that NOT every nonempty, open subinterval of C intersects with the orbit A (negation of definition of being dense). So what it comes down to (I think) is finding some element of C that is not in the orbit. the trouble is, I don't know exactly what will be in the orbit for any given rational number. The orbit will obviously contain all rational numbers, so do I just need to find some irrational element in C?

for example, 2/3 is in C, and 2/3 = 0.2000... in base 3, so if we let y = 0.20200200020000200000... then a) its irrational (right?) and b) its not in the orbit A since its irrational, so the orbit can't be dense.

Im having a lot of difficulty with the concepts here (dense sets, dense orbit, the cantor set in general) so any insight is appreciated. thanks

 

[mighty2000]

your goal is to provide a clear and logical explanation for why a rational number cannot have a dense orbit in the Cantor set. Here is a possible solution:

First, let's define the Cantor set C as the set of numbers that can be written in base 3 using only the digits 0 and 2. This means that any rational number x = p/q in C can be written as a repeating decimal in base 3, such as 0.202020... for x = 2/3.

Next, let's consider the orbit A of x in C. This is the set of all numbers that can be obtained by repeatedly applying the function T^{k}_{3/2} to x. In other words, A = {T^{k}_{3/2}(x)}^{inf}_{k=0}.

Now, let's assume that the orbit A is dense in C. This means that for any nonempty, open subinterval of C, there exists an element of A that is also in that subinterval. In other words, for any interval (a,b) in C, there exists an element of A that is also in (a,b).

However, we can show that this is not the case for the rational number x = p/q. Since x is a rational number, its decimal representation in base 3 will eventually repeat. In the case of x = 2/3, the decimal representation is 0.202020... which repeats every two digits. This means that there are only a finite number of elements of A that are in the interval (0,0.1), and therefore, there must exist some subinterval (a,b) of (0,0.1) that does not contain any element of A. This contradicts our assumption that A is dense in C.

Therefore, we can conclude that a rational number x = p/q in C cannot have a dense orbit in C. This is because the decimal representation of x in base 3 will eventually repeat, leading to a finite number of elements in the orbit A within any given subinterval of C.