Ambiguous Ternary Expansions at Cantor Endpoints

Why Endpoint Expansions Need Checking

A real number can have two base-three expansions. The familiar identity \(0.0222\ldots{}_3=0.1000\ldots{}_3\) occurs at the left endpoint \(1/3\) of the first removed middle-third interval. The right endpoint \(2/3\) has a different pattern: \(0.2000\ldots{}_3=0.1222\ldots{}_3\). These differences matter when a function is defined by reading ternary digits. A rule that appears to assign a value to each expansion must give the same answer for both expansions of the same point.

We consider the Cantor function \(C\) on \([0,1]\), using the digit rule from the earlier material on defining the Cantor function by ternary digits. For measure-theoretic context, the ambient space is \(([0,1],\mathcal{B}_{[0,1]},m)\), with finite Lebesgue measure restricted to the Borel sets of \([0,1]\). The endpoint calculations and well-definedness proof below are pointwise statements; they do not rely on an almost-everywhere convention or on any convergence theorem.

The factor \(a_j/2\) converts a ternary digit 0 or 2 into the binary digit 0 or 1. If the first ternary digit 1 occurs at position \(r\), the rule places a binary 1 at that position and ignores later digits. This is the part of the definition that handles points in removed middle-third intervals. The present tutorial checks consistency at ambiguous expansions; it does not establish continuity or monotonicity of \(C\).

How Two Ternary Expansions Can Differ

Proof. Let \(k\) be the first position at which the two sequences differ. Write the digits there as \(d<e\). Equality of the represented numbers requires that the difference \(e-d\), multiplied by \(3^{-k}\), be exactly offset by the largest possible difference between the two tails. That largest difference is $$ \sum_{j=k+1}^{\infty}2\cdot 3^{-j} =2\cdot\frac{3^{-(k+1)}}{1-\frac13} =3^{-k}. $$ Thus \((e-d)3^{-k}\le 3^{-k}\), so \(e-d\le1\). Since the digits differ, \(e-d=1\). Equality in the tail bound is necessary to cancel the digit difference. It requires that the sequence with digit \(d\) have every subsequent digit equal to 2 and the sequence with digit \(d+1\) have every subsequent digit equal to 0: any smaller tail difference would not cancel \(3^{-k}\). As \(d+1\le2\), we have \(d\in\{0,1\}\). Conversely, such a pair represents the same number, because $$ d\,3^{-k}+\sum_{j=k+1}^{\infty}2\cdot3^{-j} =d\,3^{-k}+3^{-k} =(d+1)3^{-k}. $$ The common preceding digits contribute equally, proving the characterization. \(\square\)

For endpoints of removed intervals, one expansion uses only 0s and 2s. Its partner has a 1 followed by zeros at a left endpoint, or a 1 followed by 2s at a right endpoint. For example, at \(1/3\) the partner expansion is \(0.1000\ldots{}_3\); at \(2/3\) it is \(0.1222\ldots{}_3\). The lemma also covers ambiguous expansions whose initial digits already include a 1.

Consistency of the Digit Rule

Proof. If \(x\) has only one ternary expansion, there is nothing to compare. If it has two, apply the lemma and let \(k\) be their first differing position. Write the lower digit as \(d\) and the higher digit as \(d+1\). Both expansions share the same prefix before \(k\).

If that common prefix already contains a 1, the first 1 is at the same position in both expansions. The defining rule uses only the digits before that first 1 and the value \(2^{-r}\) at its position. Those digits agree, so the function values agree.

Now suppose the common prefix contains no 1. There are two cases. If \(d=0\), the lower expansion has digit 0 at position \(k\) and then only 2s. Its contribution from position \(k\) onward is $$ 0\cdot 2^{-k}+\sum_{j=k+1}^{\infty}2^{-j}=2^{-k}. $$ The higher expansion has digit 1 at position \(k\), its first 1, so its contribution from position \(k\) onward is also \(2^{-k}\). If \(d=1\), both expansions have their first 1 at position \(k\): the lower one has digit 1 there and the higher one has digit 2 there. The rule gives \(2^{-k}\) for each, with the common prefix contributing the same amount in both. In either case the values agree. These cases exhaust \(d\in\{0,1\}\), so the digit rule is independent of the expansion. The value at \(1\) was specified separately and is unambiguous under that definition. \(\square\)

The calculation also explains the two endpoint patterns without treating them as interchangeable. At a left endpoint, the all-0-and-2 expansion reaches a tail of 2s after a 0; the other expansion has a 1 followed by zeros. At a right endpoint, the all-0-and-2 expansion reaches a tail of 2s after a 1; the other expansion has a 1 followed by 2s. In both cases the binary value carried by the tail of 2s is exactly the binary contribution of the first 1 in the alternative expansion.

A Common Endpoint Error

A tempting but incorrect statement is that every endpoint of a removed interval has an expansion with a 1 followed by 2s. The calculation at \(1/3\) disproves it: the partner of \(0.0222\ldots{}_3\) is \(0.1000\ldots{}_3\), with zeros after the 1. The corrected description is directional: at a left endpoint the alternate expansion has a 1 followed by zeros; at a right endpoint it has a 1 followed by 2s. The all-0-and-2 expansion is the other representation in each case.

A second common error is to convert every digit independently and keep going after encountering a 1. For \(0.1222\ldots{}_3\), that mistaken procedure would turn every digit 1 or 2 into a binary 1, yielding \(1\). The defining rule instead stops at the first 1 and assigns \(2^{-1}=1/2\), consistent with \(0.2000\ldots{}_3\). The first 1 is not an ordinary digit to be followed by further converted digits; it marks the point at which the rule fixes the value.

These endpoint checks are essential whenever a digit-defined function is introduced: an expansion is a representation, not an intrinsic feature of the point. The theorem establishes only that the Cantor-function digit rule is well-defined despite that ambiguity. Further properties, including continuity and monotonicity, require separate arguments.