Chapter 7.5, Problem 40MS

How much remains? Consider numbers that are not in decimal form (that is, base 10) but instead are represented in base 3. That is, every digit of the number is 0, 1, or 2. The first digit after the “decimal” point tells how many $1/3^1’S$ you have; the next tells how many $1/3^2’S$ you have; the nth place after the decimal point tells how many $1/3^n’S$ you have. So, 0.212 (base 3), for example, represents $2(1/3)+1{(1/3)}^2+2{(1/3)}^3=2/3+1/9+2/27$ or $23/27.$

Show that the points that remain in the Cantor Set are exactly those numbers whose base 3 decimal expansion can be written with only 0’s and

2’s. (Note that, just as

$\mathrm{0.0999...}=0.1$ base $10,\mathrm{0.0222...}=0.1$ base 3, so put all numbers in the ending 2’s form rather than in the ending 0’s form.) Since any sequence of 0’s and 2’s corresponds to a “decimal” number in the Cantor Set, show that there are more numbers in the Cantor Set than there are natural numbers. In fact, the cardinality of the Cantor Set is the same as the cardinality of the real numbers.