2.5 The Cantor Set

To construct the Cantor set, we proceed as follows:
First denote the closed interval \([0,1]\) by \(F_1\). Next, from \(F_1\) delete the open interval \((\frac {1}{3},\frac {2}{3})\). Denote the remaining closed set by \(F_2=[0,\frac {1}{3}]\cup [\frac {2}{3},1]\).
Next delete from \(F_2\), the open interval \((\frac {1}{9},\frac {2}{9})\) and \((\frac {7}{9},\frac {8}{9})\) which are middle thirds of \(F_2's\) two pieces. Denote the remaining by \[F_3=\Big [0,\frac {1}{9}\Big ]\cup \Big [\frac {2}{9},\frac {1}{3}\Big ]\cup \Big [\frac {2}{3},\frac {7}{9}\Big ]\cup \Big [\frac {8}{9},1\Big ]\] Continuing this process, we get a sequence of sets \(F_n\). Define the Cantor set as \[F=\bigcap ^{\infty }_{i=1}F_i=F_1\cap F_2\cap F_3\cap \dots \] \(F\) is closed, since it is an intersection of closed sets.

\(F\) clearly contains the end points of the closed intervals which make up each \(F_n\). i.e the points \(0,1,\dfrac {1}{3},\dfrac {2}{3},\dfrac {1}{9},\dfrac {2}{9},\dfrac {7}{9},\dfrac {8}{9},\dots \).

It contains a great deal more. The clean way to see this is through ternary expansions: the middle third deleted at each stage is exactly the set of numbers requiring a \(1\) in the next ternary place, so \[F=\left \{\sum _{n=1}^{\infty }\frac {a_n}{3^n}\ :\ a_n\in \{0,2\}\right \},\] the numbers in \([0,1]\) admitting a ternary expansion that uses only the digits \(0\) and \(2\).

This settles the question. The number \(\frac 14\) has ternary expansion \(0.020202\dots \), which uses only \(0\) and \(2\), so \(\frac 14\in F\); but the expansion neither terminates nor ends in repeating \(2\)s, so \(\frac 14\) is not an endpoint of any interval making up any \(F_n\). The same description shows \(F\) is uncountable: sending \(\sum a_n3^{-n}\) to the binary number \(\sum (a_n/2)2^{-n}\) maps \(F\) onto the whole of \([0,1]\).

The removed intervals have total length \(1\)

At stage \(n\) we remove \(2^{n-1}\) intervals, each of length \(3^{-n}\), so the total length removed is \[\sum _{n=1}^{\infty }\frac {2^{n-1}}{3^{n}}=\frac 13\sum _{n=1}^{\infty }\left (\frac 23\right )^{n-1}=\frac 13\cdot \frac {1}{1-\frac 23}=1 .\] So \(F\) takes up none of the length of \([0,1]\) and yet has as many points as \([0,1]\) does. That combination is why the Cantor set is the standard source of counterexamples in analysis.

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