Length Zero, Points Uncountable

Deleting middle thirds drives the total length (23)n→0\left(\frac{2}{3}\right)^n\to 0, yet uncountably many points remain. Uncountability is stated at the idea level only.

∣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. This expansion contains no digit 11. By the ternary characterization (numbers whose expansion avoids 11 are exactly the survivors — used here without proof), 14\frac14 never falls into a deleted middle third at any stage.
  3. 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.

4. Topology