The Cantor set is set of points lying on a line segment. It is created by taking some interval, for instance and removing the middle third , then removing the middle third of each of the two remaining sections and , then removing the middle third of the remaining four sections, and so on ad infinitum.
It is a closed set consisting entirely of boundary points, and is an important counterexample in set theory and general topology. When learning about cardinality, one is first shown subintervals of the real numbers, , as examples of uncountably infinite sets. This might suggest that any uncountably infinite subset of must contain an interval; such an assertion is false. A counterexample to this claim is the Cantor set , which is uncountable despite not containing any intervals. In addition, Cantor sets are uncountable, may have 0 or positive Lebesgue measures, and are nowhere dense. Cantor sets are the only disconnected, perfect, compact metric space up to a homeomorphism.
The Cantor set is constructed by removing increasingly small subintervals from . In the first step, remove from . In the second step, remove from what remains after the first step. In general, on the step, remove from what remains after the step. After all steps have been taken, what remains is the Cantor set .
This construction can be formalized as follows. Let . Then, after the first step, what remains is . After the second step, what remains is Continuing in this manner, one obtains an infinite collection of sets such that for all . Then, one defines the Cantor set to be Intuitively, this makes sense; since for all , taking an infinite intersection would provide the "limiting set" in this situation.
There is an alternate characterization that will be useful to prove some properties of the Cantor set:
consists precisely of the real numbers in whose base-3 expansions only contain the digits 0 and 2.
Base-3 expansions, also called ternary expansions, represent decimal numbers on using the digits . For instance, the number 3 in decimal is 10 in base-3. Fractions in base-3 are represented in terms of negative powers of 3, where a fraction in decimal can be represented as where , and the fraction is written in base-3 as .
For instance, .
Similarly, decimal fractions can be converted by multiplying by , taking the result mod 3, then multiplying the remainder by , and then taking mod 3 of that... etc. until the remainder is 0:
In base-3, some points have more than one notation:
. So But we also have the following: So, This sometimes means there is ambiguity in , for if expansions only contain the digits 0 and 2, then contains but not even though they are the same. In this case, it's said to contain which implies it contains because of the multiple notations.
Suppose contains only the digits 0 and 2 in its base-3 expansion. Let be the truncation of at places after the decimal point. For example, if , then , , , etc. Certainly, the sequence converges to as . In particular, for every , we have Note that the numbers in
- whose base-3 expansions go on for exactly digits after the decimal point and
- which use only the digits 0 and 2
are precisely the left endpoints of the intervals whose union is . Thus, the interval is contained in . It follows that for all , and hence .
Conversely, suppose . Then for all . Note that the numbers in are precisely those whose truncation (i.e., the number obtained by taking only the first digits after the decimal point) uses only 0 and 2 as digits in base 3. It follows that every truncation of uses only 0 and 2 as digits. This implies uses only 0 and 2 in its base-3 expansion.
From this theorem, the proofs of the following two properties follow:
does not contain any subintervals of .
Let be an arbitrary interval. Write and If some equals 1, then by the previous theorem, we know . Similarly, if some equals 1, we know . Otherwise, suppose is the smallest index such that . Our casework forces and , so , and it follows that .
Define a function as follows. If is an element in , set In other words, take the ternary expansion of , replace every digit with , and consider the result as a binary number.
This function is surjective, since any element of has .
Now, suppose is countable. Write . For each , let equal the index of an element with (remark: in defining this function , we are implicitly using the axiom of choice). Then, we may construct a sequence such that for all . This constitutes a bijection between and , so it follows that is countable. Contradiction!
Hence, we conclude is uncountable.
- 127 "rect", . Cantor ternary set, in seven iterations. Retrieved September 5th, 2016, from https://commons.wikimedia.org/wiki/File:Cantor_set_in_seven_iterations.svg