Finite Stages of the Middle-Third Cantor Set

Building the Finite Stages

The middle-third construction is a recursive procedure on closed intervals. At each step, every interval still present is divided into three equal parts, and the open middle part is removed. The two closed outer parts survive. Keeping track of the intervals at the first few stages makes the general counting and length formulas visible, and the recursion lets us prove them without relying on a picture.

We work in \(\mathbb{R}\) with Lebesgue measure \(m\), and all intervals below are Borel sets, hence Lebesgue measurable. Set \(C_0=[0,1]\). For each integer \(n\geq 0\), obtain \(C_{n+1}\) by removing the open middle third of every interval of \(C_n\). The sets \(C_n\) are the finite stages of the construction; this tutorial concerns these finite stages, not a measure calculation for their limiting set.

Applying the rule to \(C_0\) gives \(C_1=[0,1/3]\cup[2/3,1]\). Applying it again to both intervals gives four intervals at stage two. A third application gives eight intervals. These lists are useful for checking the recursive rule, but the stage count grows, so we also want a formula that describes every interval at once.

A Formula for Every Surviving Interval

Each stage makes a left-or-right choice inside every interval. Encode those choices by a word \((\varepsilon_1,\ldots,\varepsilon_n)\), where each \(\varepsilon_j\) is either \(0\) (choose the left third) or \(1\) (choose the right third). For a word of length \(n\), define its interval by

For \(n=0\), the word is empty, the sum is zero, and the formula gives \(I_{\varnothing}=[0,1]\). The term \(2\varepsilon_j/3^j\) records the shift caused by choosing the right third at step \(j\); the final interval width is \(3^{-n}\).

Proof. We use induction on \(n\). For \(n=0\), there is one empty word and its interval is \([0,1]=C_0\), so the assertion holds. Suppose it holds at stage \(n\). From an interval with left endpoint \(a=\sum_{j=1}^n2\varepsilon_j/3^j\) and length \(3^{-n}\), the left child has left endpoint \(a\) and length \(3^{-(n+1)}\); the right child has left endpoint \(a+2\cdot3^{-(n+1)}\) and the same length. These are exactly the intervals given by appending \(0\) and \(1\), respectively, in the displayed formula. Thus the recursion produces all the specified intervals at stage \(n+1\).

There are two appended words for each length-\(n\) word, so there are \(2\cdot2^n=2^{n+1}\) intervals. The children of one parent are separated by the nonempty open middle third. Children of different parents are disjoint because the parent intervals are disjoint by the induction hypothesis. For their order, two distinct words first differ at some position \(j\). Their common choices before \(j\) place them in the same parent interval, and at position \(j\) the word with entry \(0\) selects the left child while the word with entry \(1\) selects the right child. This proves the lexicographic order and disjointness at stage \(n+1\). Each child has length one third of its parent, namely \(3^{-(n+1)}\). The induction is complete. \(\square\)

Counting Intervals, Lengths, and Removed Gaps

At stage \(n\), the theorem gives \(2^n\) pairwise disjoint closed intervals, each of length \(3^{-n}\). Since this is a finite disjoint union of measurable intervals, finite additivity of Lebesgue measure gives the length of \(C_n\). At step \(k\), there are \(2^{k-1}\) intervals to split, and each removed middle third has length \(3^{-k}\). These facts give both the remaining and removed lengths.

Proof. The interval count and individual lengths follow from the theorem. The intervals are pairwise disjoint and measurable, so finite additivity yields \(m(C_n)=2^n3^{-n}=(2/3)^n\). At step \(k\), each of the \(2^{k-1}\) intervals present at stage \(k-1\) contributes exactly one removed middle third. Its length is one third of \(3^{-(k-1)}\), which is \(3^{-k}\). The removed intervals at that step are disjoint because they lie in distinct stage-\((k-1)\) intervals. Thus the total length removed at step \(k\) is \(2^{k-1}3^{-k}\).

For \(n\geq1\), summing these finite-stage lengths gives

Here the finite geometric-sum identity applies because \(1-2/3=1/3\neq0\). For \(n=0\), no intervals have been removed, and the same total-length formula gives \(1-(2/3)^0=0\). This proves the proposition for every stated case. \(\square\)

A Common Counting Error

A frequent mistake is to remove the middle third as a closed interval. At the first step, removing \([1/3,2/3]\) would leave \([0,1/3)\cup(2/3,1]\), incorrectly discarding both endpoints \(1/3\) and \(2/3\). The construction removes \((1/3,2/3)\), so the correct stage is the closed union \([0,1/3]\cup[2/3,1]\).

Another count error is to regard each surviving interval as producing three surviving intervals. It produces three thirds in the subdivision, but the middle third is removed; only the two outer closed thirds survive. Thus the number of intervals doubles at each step, rather than triples. For instance, stage two has \(2\cdot2=4\) intervals, not \(3\cdot2=6\).

The finite-stage formulas describe exactly how many intervals remain and their total length at any specified stage. Those statements use only a finite union and finite additivity. A conclusion about the measure of the limiting Cantor set requires a separate argument about the limit of these stages; the finite-stage calculations alone should not be silently substituted for that argument.