Constructing the Cantor Function

From Ternary Intervals to a Function

The Cantor set is built by repeatedly removing open middle thirds from \([0,1]\). The Cantor function is a continuous, nondecreasing function on that interval that is constant on each removed interval, while increasing across the remaining set. We construct it by assigning a range to every retained interval and linearly joining the assigned endpoint values. Refining the construction produces a uniformly convergent sequence of continuous functions.

For \(n\ge 0\), let \(C_n\) be the union of the intervals remaining after \(n\) removal steps, with \(C_0=[0,1]\). Thus \(C_1=[0,1/3]\cup[2/3,1]\), and each interval of \(C_n\) is split into its left and right thirds to form \(C_{n+1}\). Let \(C=\bigcap_{n=0}^{\infty} C_n\), the Cantor set. Each interval of \(C_n\) has a label \(w=(\varepsilon_1,\ldots,\varepsilon_n)\), where every \(\varepsilon_i\) is 0 or 1: at step \(i\), choose the left third if \(\varepsilon_i=0\) and the right third if \(\varepsilon_i=1\). Its left endpoint is \(\sum_{i=1}^n 2\varepsilon_i3^{-i}\), and its length is \(3^{-n}\).

The endpoint assignments fit together: the retained intervals occur in increasing order of rank, and the range assigned to rank \(j\) is \([j/2^n,(j+1)/2^n]\). A gap between consecutive retained intervals therefore has the shared endpoint value between their assigned ranges. In particular, if a gap is removed at step \(k\), after a prefix \((\varepsilon_1,\ldots,\varepsilon_{k-1})\), its constant value is \(\sum_{i=1}^{k-1}\varepsilon_i2^{-i}+2^{-k}\). The empty prefix is used when \(k=1\).

Approximations and Their Limit

Each \(F_n\) is real-valued, continuous, and nondecreasing on \([0,1]\). On a retained interval it is linear with positive slope, and on a removed gap it is constant. The pieces join at the assigned endpoint values, which are ordered by the interval ranks; hence no downward jump occurs at any join. Also \(F_n(0)=0\) and \(F_n(1)=1\).

Proof. Fix \(n\ge0\). On every gap already removed by step \(n\), \(F_n\) and \(F_{n+1}\) have the same constant value. On a retained interval of \(C_n\) labelled \(w\), both functions take values in \([j(w)/2^n,(j(w)+1)/2^n]\). For \(F_{n+1}\), this follows because the two child intervals and the newly removed middle third divide this assigned range into its lower half, midpoint, and upper half; for \(F_n\), it follows from its linear definition on the parent interval. The range has width \(2^{-n}\), so \[ \sup_{x\in[0,1]}|F_{n+1}(x)-F_n(x)|\le 2^{-n}. \] For integers \(q>p\), the triangle inequality gives \[ \sup_{x\in[0,1]}|F_q(x)-F_p(x)| \le\sum_{n=p}^{q-1}2^{-n} \le 2^{1-p}. \] The last bound tends to zero as \(p\to\infty\), independently of \(q>p\). Thus \((F_n)\) is uniformly Cauchy. Completeness of the real numbers gives a pointwise limit \(F\), and the same bound, followed by \(q\to\infty\), gives \(\sup_x|F(x)-F_p(x)|\le2^{1-p}\). Hence convergence is uniform.

To check continuity directly, let \(x_0\in[0,1]\) and let \(\varepsilon>0\). Choose \(p\) so that \(\sup_x|F(x)-F_p(x)|<\varepsilon/3\). Since \(F_p\) is continuous at \(x_0\), there is \(\delta>0\) such that \(|x-x_0|<\delta\) implies \(|F_p(x)-F_p(x_0)|<\varepsilon/3\). Therefore \[ |F(x)-F(x_0)| \le |F(x)-F_p(x)|+|F_p(x)-F_p(x_0)|+|F_p(x_0)-F(x_0)| <\varepsilon. \] This proves continuity, including at the endpoints with the relative metric on \([0,1]\).

If \(x\le y\), then \(F_n(x)\le F_n(y)\) for every \(n\). Taking limits gives \(F(x)\le F(y)\), so \(F\) is nondecreasing. The equalities \(F(0)=0\) and \(F(1)=1\) follow by taking limits of \(F_n(0)=0\) and \(F_n(1)=1\). Finally, any interval removed at a finite step remains in the same gap at every later stage, where all subsequent approximations have the same constant value. Their limit is constant there as well. \(\square\)

Computing Values from the Construction

Proof. First let \(x\in C\) have the stated ternary expansion. For each \(n\), \(x\) belongs to the retained interval labelled by its first \(n\) digits. Write \(j_n=\sum_{i=1}^n\varepsilon_i2^{n-i}\). By construction, \(F_n(x)\) lies in \([j_n/2^n,(j_n+1)/2^n]\). For every \(m\ge n\), \(F_m(x)\) lies in that same range, since later refinements stay within the parent interval's assigned range. Thus its limit \(F(x)\) lies in that range. The binary series in the statement also lies there: its first \(n\) terms sum to \(j_n/2^n\), and its remaining terms lie between zero and \(\sum_{i>n}2^{-i}=2^{-n}\). Both values are therefore in an interval of length \(2^{-n}\), for every \(n\). Their difference is at most \(2^{-n}\) for every \(n\), so it is zero.

Now suppose \(x\) lies in a gap first removed at step \(k\). All later approximations are constant on that gap at its assigned midpoint value. The prefix selects a parent interval whose assigned range has width \(2^{-(k-1)}\). Its midpoint is the prefix's binary value plus half that width, namely \(\sum_{i=1}^{k-1}\varepsilon_i2^{-i}+2^{-k}\). Taking the limit gives the claimed value. \(\square\)

A Common Misreading of the Approximations

The approximations converge uniformly, but they need not increase pointwise as the stage number grows. For example, \(F_0(9/10)=9/10\). Since \(9/10\) lies in the right third, \(F_1(9/10)=\frac12+\frac12(3(9/10)-2)=\frac12+\frac12(7/10)=17/20\). Since \(17/20=0.85<0.9=9/10\), \(F_1(9/10)<F_0(9/10)\). It would therefore be incorrect to justify the limit by claiming \(F_n\uparrow F\). The valid argument is the uniform Cauchy estimate, which proves convergence without any monotonicity in \(n\).

The construction establishes the Cantor function's continuity, monotonicity, endpoint values, and digit-based values. These properties identify how the function is assembled from the ternary construction while leaving its finer differentiation properties for separate study.