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\sum_{n=1}^\infty \frac{a_n}{2^n}, & x = \sum_{n=1}^\infty
\frac{2a_n}{3^n}\in\mathcal{C}\ \mathrm{for}\ a_n\in\{0,1\};
\\ \sup_{y\leq x,\, y\in\mathcal{C}} c(y), & x\in [0,1]\
</math>
This formula is well-defined, since every member of the Cantor set has a ''unique'' base 3 representation that only contains the digits 0 or 2. (For some members of <math>\mathcal{C}</math>, the ternary expansion is repeating with trailing 2's and there is an alternative non-repeating expansion ending in 1. For example, <math>\tfrac13</math> = 0.1<sub>3</sub> = 0.02222...<sub>3</sub> is a member of the Cantor set). Since <math>c(0)=0</math> and <math>c(1)=1</math>, and <math>c</math> is monotonic on <math>\mathcal{C}</math>, it is clear that <math>0\le c(x)\le 1</math> also holds for all <math>x\in[0,1]\
==Properties==
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