Does f(x) have to be in the ternary Cantor set?

[mkkrnfoo85]

Hey all,

I would really like help on this probably simple proof:

That the map x |--> f(x) = (x+2)/3 on [0,1] is a contraction,
and maps the ternary Cantor set into itself. Also, find it's fixed point.


(1) I can easily show the fixed point (where f(x) = x) is 1.
(2) I can also pretty easily show it is a contraction:
where |f(x) - xo| <= q*|x - xo|, where q < 1, and xo is the fixed point.

(3) However, I can't seem to find a way to tell whether it maps the ternary Cantor set into itself. I kno the definition of the ternary Cantor set is taking the interval [0,1] and deleting the middle-third of the interval, and then repeating the process on each remaining interval, infinitely.

What does it mean by mapping the ternary Cantor set into itself? Does x have to start out being in the ternary Cantor set? If so, how is it possible if x can vary from [0,1]? What am I interpreting wrong?

Thanks,

Mark

 

[gammamcc]

Look at how the function maps certain intervals. What sort of intervals must x be contained in for it to be in the Cantor set? What then happens to the intervals under function f(x)?

 

[mkkrnfoo85]

ah thx. just the boost i needed =)

 

[HallsofIvy]

What does it mean by mapping the ternary Cantor set into itself?

It means that if x is in the subset of [0,1] called the Cantor set, then f(x) is also in the Cantor set. You might use the fact that x is in the Cantor set if and only if its base 3 expansion contains no 1s.

(I've edited to change "contains no 0s" to "contains no 1s"!)

I've also given a little more thought to this. Since dividing by 3 base 3 just shifts the "decimal" point, this is trivial!