Cantor set
Smith–Volterra–Cantor set

Instead of repeatedly removing the middle third of every piece as in the Cantor set, we could also keep removing any other fixed percentage (other than 0% and 100%) from the middle. In the case where the middle 8 / 10 of the interval is removed, we get a remarkably accessible case — the set consists of all numbers in that can be written as a decimal consisting entirely of 0s and 9s. If a fixed percentage is removed at each stage, then the limiting set will have measure zero, since the length of the remainder (1 − f) n → 0 {\displaystyle (1-f)^{n}\to 0} as n → ∞ {\displaystyle n\to \infty } for any f {\displaystyle f} such that 0 < f ≤ 1 {\displaystyle 0<f\leq 1}.

On the other hand, "fat Cantor sets" of positive measure can be generated by removal of smaller fractions of the middle of the segment in each iteration. Thus, one can construct sets homeomorphic to the Cantor set that have positive Lebesgue measure while still being nowhere dense. If an interval of length r n {\displaystyle r^{n}} (r ≤ 1 / 3 {\displaystyle r\leq 1/3}) is removed from the middle of each segment at the n th iteration, then the total length removed is ∑ n = 1 ∞ 2 n − 1 r n = r / (1 − 2 r) {\textstyle \sum _{n=1}^{\infty }2^{n-1}r^{n}=r/(1-2r)}, and the limiting set will have a Lebesgue measure of λ = (1 − 3 r) / (1 − 2 r) {\displaystyle \lambda =(1-3r)/(1-2r)}. Thus, in a sense, the middle-thirds Cantor set is a limiting case with r = 1 / 3 {\displaystyle r=1/3}. If 0 < r < 1 / 3 {\displaystyle 0<r<1/3}, then the remainder will have positive measure with 0 < λ < 1 {\displaystyle 0<\lambda <1}. The case r = 1 / 4 {\displaystyle r=1/4} is known as the Smith–Volterra–Cantor set, which has a Lebesgue measure of 1 / 2 {\displaystyle 1/2}.
change_case:english_lowercase