Binary Sequences And The Cantor Set

Binary Sequences as Points of the Cantor Set

The middle-thirds Cantor set can be described by the digits that remain after repeatedly removing open middle thirds. At each step, a point that remains chooses one of two closed subintervals. Recording those choices gives an infinite binary sequence. This correspondence is more than a way to label points: with the appropriate topology on the sequence space, it is a homeomorphism, so it preserves the topological structure of the Cantor set.

Let \(\Sigma=\{0,1\}^{\mathbb{N}}\) be the set of all infinite sequences \(a=(a_1,a_2,\ldots)\), where each coordinate belongs to \(\{0,1\}\). Give \(\{0,1\}\) the discrete topology and \(\Sigma\) the product topology. A basic open set specifies the values of finitely many coordinates and leaves all other coordinates unrestricted. In particular, the sets specifying an initial segment form a basis: for a finite word \(w=(w_1,\ldots,w_n)\), write \([w]\) for the set of sequences whose first \(n\) coordinates are \(w_1,\ldots,w_n\).

Let \(C\subseteq\mathbb{R}\), with the subspace topology from the standard topology on \(\mathbb{R}\), be the middle-thirds Cantor set. For each finite binary word \(w=(w_1,\ldots,w_n)\), define the closed interval \(I_w=[\sum_{j=1}^n 2w_j/3^j,\ \sum_{j=1}^n 2w_j/3^j+3^{-n}]\). At level \(n\), the Cantor construction retains exactly the union of these intervals. Thus \(C\) is the intersection, over all \(n\), of the level-\(n\) unions.

We will prove that this map is a homeomorphism. The main points are to verify that every Cantor-set point has exactly one such sequence and to show that closeness in either space corresponds to agreement on a sufficiently long initial segment.

The Ternary-Digit Map Is a Homeomorphism

Proof. We first prove that \(f\) takes values in \(C\) and is onto. For \(a\in\Sigma\), let \(w=(a_1,\ldots,a_n)\). The tail of the series satisfies \(0\leq\sum_{j=n+1}^{\infty}2a_j/3^j\leq\sum_{j=n+1}^{\infty}2/3^j=3^{-n}\). Therefore \(f(a)\in I_w\) at every level \(n\), so \(f(a)\in C\).

Conversely, take \(x\in C\). At each level \(n\), \(x\) belongs to one of the intervals \(I_w\). These intervals are pairwise disjoint: if two words first differ in coordinate \(k\), the intervals lie on opposite sides of the open middle third removed at step \(k\). Consequently there is a unique word \(w^{(n)}\) whose interval contains \(x\). The words are consistent as \(n\) increases, because each level-\((n+1)\) interval lies in its level-\(n\) parent interval. They therefore specify a sequence \(a\in\Sigma\). For every \(n\), both \(x\) and \(\sum_{j=1}^n2a_j/3^j\) lie in \(I_{(a_1,\ldots,a_n)}\), whose length is \(3^{-n}\). Hence \(\left|x-\sum_{j=1}^n2a_j/3^j\right|\leq3^{-n}\). As \(3^{-n}\to0\), this gives \(x=f(a)\), proving surjectivity.

To prove injectivity, suppose \(a,b\in\Sigma\) are distinct, and let \(k\) be their first differing coordinate. Interchanging \(a,b\) if necessary, suppose \(a_k=0\) and \(b_k=1\). The contribution at coordinate \(k\) to \(f(b)-f(a)\) is \(2/3^k\). The later coordinates can reduce this difference by at most \(\sum_{j=k+1}^{\infty}2/3^j=1/3^k\). It follows that \(f(b)-f(a)\geq 2/3^k-1/3^k=1/3^k>0\). Thus \(f(a)\ne f(b)\).

We next prove continuity. Fix \(a\in\Sigma\) and \(\varepsilon>0\). Choose \(n\in\mathbb{N}\) with \(3^{-n}<\varepsilon\). If \(b\) agrees with \(a\) in its first \(n\) coordinates, then \(\left|f(a)-f(b)\right|\leq\sum_{j=n+1}^{\infty}2/3^j=3^{-n}<\varepsilon\). The initial-segment cylinder \([(a_1,\ldots,a_n)]\) is an open neighborhood of \(a\), and its image is contained in the \(\varepsilon\)-neighborhood of \(f(a)\). This is continuity at every \(a\).

Finally, consider continuity of the inverse. For a word \(w\) of length \(n\), the image of its cylinder is \(f([w])=C\cap I_w\). Distinct level-\(n\) intervals are separated by gaps of length at least \(3^{-n}\): if the first differing coordinate is \(k\leq n\), the gap created at that step has length \(3^{-k}\geq3^{-n}\). Thus, for any \(x\in C\cap I_w\), the open real interval of radius \(3^{-n}\) about \(x\), intersected with \(C\), lies in \(C\cap I_w\). So \(f([w])\) is open in \(C\). Initial-segment cylinders form a basis of \(\Sigma\), and any open set in \(\Sigma\) is a union of such cylinders. Its image under the bijection \(f\) is therefore open in \(C\). This says exactly that \(f^{-1}:C\to\Sigma\) is continuous. The map is a continuous bijection with continuous inverse, hence a homeomorphism. \(\square\)

Uncountability and Total Disconnectedness

The sequence description gives two topological consequences. First, there are too many infinite binary sequences to list in a sequence. Second, cylinders distinguish any two different sequences by a set that is both open and closed.

Proof. Suppose, to the contrary, that \(\Sigma\) can be listed as \(a^{(1)},a^{(2)},a^{(3)},\ldots\), where \(a^{(n)}_j\) denotes coordinate \(j\) of sequence \(a^{(n)}\). Define a sequence \(b\in\Sigma\) by \(b_n=1-a^{(n)}_n\) for every \(n\). Since \(b_n\ne a^{(n)}_n\), the sequence \(b\) differs from the \(n\)-th listed sequence at coordinate \(n\). Thus \(b\) is absent from the list, contradicting that the list contains every element of \(\Sigma\). So \(\Sigma\) is uncountable. The homeomorphism \(f\) is, in particular, a bijection, so \(C\) is uncountable as well. \(\square\)

Proof. Let \(S\subseteq C\) contain distinct points \(x\) and \(y\). By injectivity of \(f\), their sequences \(f^{-1}(x)\) and \(f^{-1}(y)\) differ at some coordinate. Choose \(n\) large enough to include a coordinate where they differ, and let \(w\) be the first \(n\) coordinates of \(f^{-1}(x)\). Then \(x\in f([w])\) and \(y\notin f([w])\). The cylinder \([w]\) is open in \(\Sigma\); its complement is the union of the other length-\(n\) cylinders, so that complement is open too. Since \(f\) is a homeomorphism, \(f([w])\) is clopen in \(C\). Consequently \(S\cap f([w])\) and \(S\setminus f([w])\) are disjoint, nonempty, relatively open subsets of \(S\) whose union is \(S\). They form a separation, so \(S\) is not connected. Every subset of \(C\) with at least two points is therefore disconnected, which proves total disconnectedness. \(\square\)

This proof uses clopen sets as witnesses: the two points need not have a large distance between them, but some finite initial segment distinguishes their codes. The corresponding cylinder divides the Cantor set, and also divides any subset containing both points.

Common pitfall. Do not replace the ternary-digit map by the usual binary-expansion formula \(\sum_{j=1}^{\infty}a_j/2^j\) and assume it is one-to-one. For example, the distinct sequences \((1,0,0,\ldots)\) and \((0,1,1,1,\ldots)\) both map to \(1/2\): the first sum is \(1/2\), and the second is \(1/4+1/8+1/16+\cdots=1/2\). The ternary map avoids this ambiguity because at the first differing coordinate its leading difference is larger than the maximum possible contribution of all later coordinates combined.