### Cantor set

Hi there! This post est divisa in partes tres.

Part 1: This post is divided into three parts, of which this is the first. The Cantor set is a subset of the real numbers (they are also known as the "real line"). More specifically, it is a subset of the interval from 0 to 1 (and, though extremely porous in a strange way, contains just as many numbers as the whole interval). In case you hadn't figured it out, Part 1 was the introductory part, and doesn't ask you to do anything. The Cantor set is named after Mr. Cantor; reference to it occurred in the commentary upon at least one of my previous postings.

The Cantor set can be found this way, starting with the real numbers between 0 and 1: Delete the middle third. That leaves two 1/3s, first and last. From each of these, delete the middle third. That leaves four 1/9s. Exactly, delete right-open intervals, so that after the first step, the two intervals remaining are [0,1/3) and [2/3,1). This persnikkety point can mostly be ignored. Continue forever. What never gets deleted is the Cantor set.

Part 2: This part is segregated in case you are lazy, also, because it concerns a much easier challenge than Part 3: everything in Part 2 can be answered, e.g., in the wikipedia article on the Cantor set, where answers will be found at which those snobby mathematicians would no doubt sniff, "trivial." Nevertheless, if you don't look, it's pretty neat:

The challenge: assuming that you have found some convenient way of *representing* all the numbers between 0 and 1, exhibit a predicate schema P(...) such that if you complete it with a compatible representation of an arbitrary number between 0 and 1, P is true if and only if it is indeed a member of the Cantor set. Equivalently (no philosophers allowed beyond this point), you have found a predicate such that {xP(x)} is the Cantor set. P must be absolute in the sense that, unlike the definition, it does not encompass an infinite series of operations, although an infinite series of *results* is allowed if they can be summarized finitely. (I told you, no philosophers beyond this point! Get the fire hoses!)

Part 3: (You f*********** philosophers won't find this in ************ wikipedia!) Say you have solved Part 2: now you are given the unique expression of a number that is *not* "compatible." Can you determine if it is indeed a member of the Cantor set in a finite number of operations? Obviously, this depends on Part 2's P and its "compatible" representation. For the one I know of (the one you will pick up from wikipedia if you cheat and look), I sure don't, and I seriously doubt anyone else does. Be the first!

1. Andy Stedman1:14 PM

Can I have the number you want to test in, say, base 3?

2. We just discussed this today in math club, as part of a larger discussion on fractal geometry. Cantor dust, right? I shall think on it.

3. Andy, you got it. The Cantor set is the numbers whose ternary representation contains only 0s and 2s, no 1s. Of course, if you then convert all the 2s to 1s and interpret the result as binary, you see at once that the Cantor set is just as numerous as the whole real interval [0,1].