Cantor set—an infinite set of numbers between 0 and 1, defined by an inductive process. To define this set, start with the closed interval [0,1]. Remove the middle third--the open interval (1/3,2/3). (That is, remove all the points between 1/3 and 2/3, except the points 1/3 and 2/3 themselves.) Next take each of the remaining two intervals, and remove its middle third. This leaves us with four intervals: [0,1/9], [2/9,1/3], [2/3,7/9], and [8/9,1]. Now remove the middle third of each of these intervals. This procedure is repeated ad infinitum. The resulting set is the Cantor set, named for Georg Cantor.
This set has many interesting properties. It is an example of a set of measure 0 (see Measurable and nonmeasurable). This is because the interval [0,1] has length 1, and if we add up the lengths of all the removed middle thirds, we also get 1. The Cantor set nevertheless contains infinitely many points--in fact, it contains many points. Moreover, it can be shown that all of these points are "cluster points"--that is, every point in the Cantor set has other points in the Cantor set that get arbitrarily close to it.
We saw above that every fraction whose denominator is a power of three is a member of the Cantor set. How can we tell whether the point 1/100 is in the Cantor set? We can, of course, simply follow the procedure for constructing the set, and check at each stage to see if 1/100 is in one of the intervals being removed. However, for some points, this can take a long time, and there is no way to know in advance how long it will take--the point in question may not be removed until, say, the billionth stage of the process.
There is an alternative method of viewing the Cantor set that makes it very easy to determine whether any given point is, in fact, in the Cantor set. To do this, we switch into base 3 notation. This means that we only use the numerals 0, 1 and 2--thus the number 3 is written as 10, because what we usually think of as the 10's column is now the 3's column. Then 4 is written 11, 5 is 12, and 6 becomes 20. The number 9 is written 100--just as in base 10, each column represents a power of 10, in base 3, each column represents a power of 3. We can also use base 3 notation to the right of the decimal point--the first column represents 1/3, the second 1/9, and so on. Thus, for example, .12 in base 3 is 1/3 + 2/9 = 5/9 (and so the "decimal" point is really now a "base-3" point).
Now consider the first stage in constructing the Cantor set, removing the interval (1/3,2/3). Written in base 3, we are removing the interval (.1,.2). Numbers in this interval have the form .1 ... , i.e., they all have a 1 in the first column. So we have removed all numbers with a 1 in the first column, except for .1 itself.
In the second stage, we remove the intervals (1/9,2/9) and (7/9,8/9). In base 3 notation, these are written (.01,.02) and (.21,.22) respectively, and numbers in these intervals have the form .01 ... and .21.... So we are eliminating all numbers whose second digit to the right of the "decimal" point is 1, except for .01 and .21.
In the third stage, we eliminate the intervals (.001,.002), (.021,.022), (.201,.202) and (.221,.222). Thus we are eliminating all numbers whose third digit to the right of the decimal point is 1, except for the numbers .001,.021,.201, and .221.
Continuing this pattern, we can see that we eliminate all numbers that have a 1 anywhere in their base 3 expansion, with the exception of the numbers that have precisely one 1, as the final digit. Thus to test if any given number--say, 1/100--is in the Cantor set, we merely write it in base 3 (the procedure is the same as in base 10, i.e. long division) and check to see how many ones it has and where they are placed. In the case of 1/100, we get the base 3 expression .000210220022.... Since there is a 1 in this expression and it is not the final digit, we can see that the number 1/100 is not in the Cantor set.
Finally, one other way of looking at the Cantor set is geometrically. If we draw the Cantor set on a number line, we get a fractal: any small portion of it looks the same as the entire set. For example, the portion of the Cantor set in the interval [0,1/3] looks like the entire Cantor set: the middle third of this portion is missing, and the middle thirds of the two remaining segments are missing, and so on.
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