Ternary Expansions of Cantor Set Points

From Ternary Digits to Cantor-Set Points

The middle-third Cantor set is formed by repeatedly removing open middle thirds from the interval \([0,1]\). At every stage, each surviving interval has a left third and a right third; a point that survives can therefore be recorded by choosing left or right at each stage. In ternary notation, those choices correspond to digits \(0\) and \(2\). The main point to keep in view is that the correspondence is exact when stated as an existence claim: a real number belongs to the Cantor set if and only if it has a ternary expansion using only \(0\) and \(2\). Ternary expansions need not be unique, especially at interval endpoints.

For a finite word \(w=(w_1,\ldots,w_n)\), with each \(w_j\in\{0,2\}\), define its value by \(s(w)=\sum_{j=1}^{n}w_j3^{-j}\). The interval remaining at stage \(n\) with this sequence of left-right choices is \(I_w=[s(w),s(w)+3^{-n}]\). The length is \(3^{-n}\), since each choice retains one third of the previous interval. Write \(C_n\) for the union of all such intervals \(I_w\) with words of length \(n\), and \(C=\bigcap_{n=0}^{\infty}C_n\), where \(C_0=[0,1]\). Thus \(C\) is the middle-third Cantor set.

The value of the first \(n\) digits is \(s_n=\sum_{j=1}^{n}a_j3^{-j}\). The rest of the series is between \(0\) and \(\sum_{j=n+1}^{\infty}2\cdot3^{-j}=3^{-n}\). Consequently the ternary value lies in the stage-\(n\) interval \([s_n,s_n+3^{-n}]\), for every \(n\). This gives one direction of the correspondence immediately. The converse requires choosing the sequence of intervals containing a given point.

The Exact Correspondence

Proof. First suppose \(x=\sum_{j=1}^{\infty}a_j3^{-j}\) for digits \(a_j\in\{0,2\}\). For each \(n\), let \(w=(a_1,\ldots,a_n)\). The first \(n\) terms sum to \(s(w)\), and the remaining terms satisfy \(0\le\sum_{j=n+1}^{\infty}a_j3^{-j}\le\sum_{j=n+1}^{\infty}2\cdot3^{-j}=3^{-n}\). Therefore \(x\in I_w\subseteq C_n\) for every \(n\), and hence \(x\in\bigcap_{n=0}^{\infty}C_n=C\).

Conversely, suppose \(x\in C\). For each \(n\ge1\), \(x\in C_n\), so it lies in one of the intervals \(I_w\) of stage \(n\). These intervals are pairwise disjoint: two distinct words first differ at some position \(k\), and the corresponding left and right choices at that stage lie in separated thirds. Thus there is a unique word \(w^{(n)}\) whose interval contains \(x\). The intervals at stage \(n+1\) are contained in their stage-\(n\) parent intervals, so the first \(n\) digits of \(w^{(n+1)}\) must equal \(w^{(n)}\). These compatible words define an infinite sequence \((a_j)\) with \(a_j\in\{0,2\}\). For every \(n\), \(x\) lies in the interval with left endpoint \(s_n=\sum_{j=1}^{n}a_j3^{-j}\) and length \(3^{-n}\). Hence \(0\le x-s_n\le3^{-n}\). Since \(s_n\) converges to \(\sum_{j=1}^{\infty}a_j3^{-j}\) and \(3^{-n}\to0\), it follows that \(x=\sum_{j=1}^{\infty}a_j3^{-j}\). This proves both directions. \(\square\)

Finite Stage Endpoints and Alternative Expansions

A finite ternary expansion is one that ends in zeros. An infinite expansion can have the same value: the tail of repeated \(2\)s exactly fills one unit in the next ternary place. For every integer \(k\ge1\),

This identity explains the basic endpoint ambiguity. If a finite expansion has digit \(2\) in position \(k\) and zeros afterward, reducing that digit to \(1\) and replacing the later zeros by \(2\)s leaves its value unchanged. Conversely, a digit \(1\) followed by zeros can be replaced by a \(0\) in that position followed by \(2\)s. When this produces an all-zero-and-two expansion, it can demonstrate that the endpoint belongs to \(C\), even though another expansion uses a digit \(1\).

A Common Mistake

A tempting but incorrect claim is: “A point of \(C\) has no ternary expansion containing the digit \(1\).” The calculation \(1/3=0.1_3\) disproves this claim, while \(1/3=0.02222\ldots{}_3\) confirms that the same point has an expansion using only \(0\) and \(2\). The correct characterization is existential: at least one such expansion exists. The endpoint examples show why checking only a preferred finite expansion can give the wrong impression about membership.

There is a separate arithmetic pitfall in reading the digits as if they were ordinary decimal digits. For example, \(0.202020\ldots{}_3\) is not the decimal number \(0.202020\ldots\). Its place values are \(1/3,1/9,1/27,\ldots\), not \(1/10,1/100,1/1000,\ldots\). Computing the geometric series with powers of \(3\), as in the worked example, gives its value \(3/4\).