Compactness of the Cantor Set via Products

From a Product Space to the Cantor Set

The Cantor set can be described by retaining, at every stage of its construction, the left and right thirds of each remaining interval. Its points therefore have ternary expansions using only the digits 0 and 2. A binary sequence records exactly those choices: 0 means choose the left third, and 1 means choose the right third. We will make this correspondence precise and use it to transfer compactness from a product of compact spaces to the Cantor set.

Write \(\{0,1\}\) for the two-point space with the discrete topology, in which every subset is open. Let \(\Sigma=\{0,1\}^{\mathbb{N}}\) have the product topology. A basic open set in this topology specifies the values of a sequence at only finitely many coordinates. We use the usual topology on \(\mathbb{R}\), and give every subset of \(\mathbb{R}\), including the Cantor set, its subspace topology.

Each factor \(\{0,1\}\) is compact: given an open cover, choose one member containing 0 and one member containing 1; these at most two sets cover the factor. The product compactness result we need is Tychonoff's theorem: a product of compact spaces, with the product topology, is compact. Its general form uses the axiom of choice; the theorem is established in the earlier material on Tychonoff's theorem. Applying it to the compact factors \(\{0,1\}\), indexed by \(\mathbb{N}\), shows that \(\Sigma\) is compact.

The Ternary-Digit Map

For a word \(s=(s_1,\ldots,s_n)\), put \(l(s)=\sum_{i=1}^n 2s_i/3^i\), and let \(I_s=[l(s),l(s)+3^{-n}]\). At stage \(n\) of the middle-thirds construction, the intervals that remain are precisely the intervals \(I_s\) for words of length \(n\). Define \(C_n\) to be their union, and define the Cantor set by \(C=\bigcap_{n\ge1}C_n\).

We first verify the image. If \(a=(a_i)\in\Sigma\), then after its first \(n\) digits the remaining sum lies between 0 and \(\sum_{i=n+1}^{\infty}2/3^i=3^{-n}\). Thus \(\Phi(a)\in I_{(a_1,\ldots,a_n)}\subseteq C_n\) for every \(n\), and \(\Phi(a)\in C\). Conversely, suppose \(x\in C\). At each stage \(n\), \(x\) belongs to an interval \(I_s\) of that stage. These intervals are pairwise disjoint, so the word \(s\) is unique. The stage-\(n\) interval containing \(x\) lies inside the stage-\((n-1)\) interval containing \(x\); hence the words at successive stages agree on their shared coordinates. They determine a sequence \(a\in\Sigma\). Since \(x\in I_{(a_1,\ldots,a_n)}\), we have \(0\le x-\sum_{i=1}^n2a_i/3^i\le3^{-n}\). As \(3^{-n}\to0\), these partial sums converge to \(x\), so \(x=\Phi(a)\). Therefore \(\Phi(\Sigma)=C\).

Why the Correspondence Is a Homeomorphism

Proof. Surjectivity was proved by showing \(\Phi(\Sigma)=C\). To prove injectivity, let \(a,b\in\Sigma\) be distinct, and let \(k\) be their first differing coordinate. Suppose \(a_k=0\) and \(b_k=1\), interchanging \(a,b\) if needed. The difference contributed at coordinate \(k\) is \(2/3^k\). The total possible opposing contribution from all later coordinates is at most \(\sum_{j=k+1}^{\infty}2/3^j=1/3^k\). Therefore \(\Phi(b)-\Phi(a)\ge 2/3^k-1/3^k=1/3^k>0\). Thus \(\Phi(a)\ne\Phi(b)\).

For continuity, fix \(a\in\Sigma\) and an open neighborhood \(V\) of \(\Phi(a)\) in \(C\). Since \(C\) has the subspace topology, there is \(\varepsilon>0\) such that \(C\cap(\Phi(a)-\varepsilon,\Phi(a)+\varepsilon)\subseteq V\). Choose \(n\) with \(3^{-n}<\varepsilon\). If \(b\) agrees with \(a\) in the first \(n\) coordinates, the two series have the same first \(n\) terms, and \(\lvert\Phi(b)-\Phi(a)\rvert\le\sum_{i=n+1}^{\infty}2/3^i=3^{-n}<\varepsilon\). Thus \(\Phi([(a_1,\ldots,a_n)])\subseteq V\). The cylinder is an open neighborhood of \(a\), which proves continuity at every \(a\).

It remains to prove continuity of the inverse. For each word \(s\) of length \(n\), the cylinder \([s]\) maps onto \(C\cap I_s\): the first \(n\) terms give \(l(s)\), the remaining terms range from 0 to \(3^{-n}\), and all image points lie in \(C\). At a fixed stage \(n\), the finitely many intervals \(I_s\) are pairwise separated by positive gaps. Choose a positive \(\delta\) smaller than half the distance from \(I_s\) to every other stage-\(n\) interval. Then \(C\cap(l(s)-\delta,l(s)+3^{-n}+\delta)=C\cap I_s\), so \(C\cap I_s\) is open in \(C\). Hence \(\Phi([s])\) is open in \(C\). Cylinders specifying the first \(n\) coordinates form a basis for \(\Sigma\): any finite coordinate specification is refined by specifying all coordinates up to the largest index used. Therefore the inverse image under \(\Phi^{-1}\) of each basis cylinder is open in \(C\), and \(\Phi^{-1}\) is continuous. This proves the theorem. \(\square\)

Transferring Compactness

Proof. Let \(\mathcal{U}\) be an open cover of \(f(X)\) by sets open in the subspace \(f(X)\). For each \(U\in\mathcal{U}\), continuity makes \(f^{-1}(U)\) open in \(X\), and these preimages cover \(X\): for each \(x\in X\), the point \(f(x)\) belongs to some \(U\in\mathcal{U}\). Compactness of \(X\) gives finitely many members \(U_1,\ldots,U_m\) whose preimages cover \(X\). For every \(y\in f(X)\), choose \(x\in X\) with \(f(x)=y\). Since \(x\) belongs to one of those preimages, \(y\) belongs to the corresponding \(U_i\). Thus \(U_1,\ldots,U_m\) cover \(f(X)\), proving compactness. \(\square\)

Since \(\Sigma\) is compact and \(\Phi:\Sigma\to C\) is continuous and surjective, the proposition proves that \(C\) is compact. In fact, the homeomorphism gives more than compactness: it identifies all open-set information in one space with the corresponding information in the other. The compactness conclusion itself needs only continuity and surjectivity, not continuity of the inverse.

A Common Product-Topology Mistake

A common error is to treat the singleton containing the all-zero sequence as an open set in \(\Sigma\), on the grounds that one can specify every coordinate to be zero. In the product topology, a basic open set restricts only finitely many coordinates. To verify the problem concretely, let \(U\) be any basic neighborhood of the all-zero sequence, and let \(F\subseteq\mathbb{N}\) be the finite set of coordinates it restricts. Choose \(j\in\mathbb{N}\setminus F\). The sequence that is 1 at coordinate \(j\) and 0 elsewhere still belongs to \(U\), but is not the all-zero sequence. Thus no basic neighborhood of that point is contained in its singleton, and the singleton is not open.

This distinction matters when proving continuity: control of the first \(n\) coordinates gives an open cylinder, while control of all infinitely many coordinates at once does not give a basic open set. For the ternary-digit map, the tail estimate makes finite-coordinate control sufficient: agreement through coordinate \(n\) bounds the image difference by \(3^{-n}\), which can be made as small as desired.