Length Zero, Points Uncountable

This scene builds the Cantor set. Keep deleting middle thirds: the length dies to zero, yet uncountably many points survive.

Each deletion shrinks the total length to two-thirds. After infinitely many rounds, the length is zero. But some points are never cut away. There are as many of them as there are real numbers.

For example, cut the middle third out of a 🍰 cake. Then cut the middle of each remaining piece, again and again. The cake almost vanishes. Still, uncountably many spots were never touched by the knife.

∣Cn∣=(23)n→0|C_n|=\left(\frac{2}{3}\right)^n\to 0
∣C∣=∣R∣|C|=|\mathbb{R}|

Part of these learning paths

Examples & Exercises

Count the intervals of stage C3C_3 and compute its total length.

23=8,(23)3=8272^3 = 8, \qquad \left(\tfrac{2}{3}\right)^3 = \tfrac{8}{27}
Stage 3 — 8 pieces, total length 8/27≈0.2968/27 \approx 0.296.
Show proof

  1. Every interval splits into two at each stage, so stage nn has 2n2^n intervals. Hence C3C_3 consists of 23=82^3 = 8 intervals.

  2. Each interval at stage nn has length (13)n\left(\frac13\right)^n, so the total length is 2n⋅(13)n=(23)n2^n \cdot \left(\frac13\right)^n = \left(\frac23\right)^n.

  3. At n=3n = 3 the total length is (23)3=827≈0.296\left(\frac23\right)^3 = \frac{8}{27} \approx 0.296.

Decide whether 14\frac14 belongs to the Cantor set using its ternary expansion.

14=0.020202…3∈C\frac14 = 0.020202\ldots_3 \in C
Stage 4 — 14\frac14 lies in a surviving interval at every stage.
Show proof

  1. In ternary, 0.020202…3=∑j=1∞2⋅3−2j0.020202\ldots_3 = \sum_{j=1}^{\infty} 2\cdot 3^{-2j}. The geometric series sums to 2⋅191−19=2/98/9=14\frac{2\cdot \frac19}{1-\frac19} = \frac{2/9}{8/9} = \frac14.

  2. First prove the ternary characterization by induction on stages.

    Claim: x∈Cnx \in C_n iff xx has a ternary expansion whose first nn digits are all 00 or 22.

    For n=0n = 0 this is trivial (C0=[0,1]C_0 = [0,1]).

    Inductive step: each interval of CnC_n has the form [s,s+3−n][s, s + 3^{-n}] with s=∑k≤nck3−ks = \sum_{k \le n} c_k 3^{-k}, ck∈{0,2}c_k \in \{0,2\}; splitting it in thirds, the left and right closed thirds correspond to next digit 00 and 22, while the removed open middle third corresponds to next digit 11.

    Hence the intervals of Cn+1C_{n+1} are exactly those obtained by appending 00 or 22, and a point using digit 11 at this place (falling into the middle interval that digit opens) is removed at this stage — except that endpoints have the alternative expansion …1000…3=…0222…3\ldots 1000\ldots_3 = \ldots 0222\ldots_3, so the characterization must be stated with an expansion that "exists".

  3. This expansion contains no digit 11. By the ternary characterization just proved (numbers with an expansion avoiding 11 are exactly the survivors), 14∈Cn\frac14 \in C_n for every nn: its first nn digits are only 0,20, 2, so 14\frac14 never falls into a deleted middle third at any stage.

  4. Hence 14\frac14 belongs to every CnC_n, and therefore to C=⋂nCnC = \bigcap_n C_n.

Prove that the Cantor set CC has measure zero, via a covering argument.

∀ ε>0  ∃ n: (23)n<ε\forall\, \varepsilon > 0\ \ \exists\, n:\ \left(\tfrac23\right)^n < \varepsilon
Stage 8 — total length (2/3)8≈0.039(2/3)^8 \approx 0.039, made arbitrarily small by taking nn larger.
Show proof

  1. C⊆CnC \subseteq C_n for every nn, and CnC_n consists of 2n2^n intervals of total length (23)n\left(\frac23\right)^n.

  2. Given ε>0\varepsilon > 0, choose nn with (23)n<ε\left(\frac23\right)^n < \varepsilon (possible since (23)n→0\left(\frac23\right)^n \to 0). Then CC is covered by finitely many intervals of total length below ε\varepsilon.

  3. A set coverable by intervals of arbitrarily small total length has measure zero (this is the definition of Lebesgue outer measure zero). For the record, CC is also uncountable — length zero, yet uncountably many points.

5. Topology