Hausdorff dimension is defined by an infinite covering process, which is awkward to attack directly. The standard move is to discretize: thicken each unit segment of K into a thin tube of width δ, keep one tube per δ-separated direction (about N∼1/δ of them), and ask how small their union can be. If one can show this union always has area at least N2−o(1)δ2 — nearly as large as if the N tubes of area δ each were completely disjoint — then letting δ→0 forces K itself to have Minkowski dimension 2, and with extra care, Hausdorff dimension 2 as well.
T={T1,…,TN},N∼δ−1,Ti:δ×1 tube,∣θi−θj∣≳∣i−j∣δ
Detailed analysis
Concretely, the equivalence used is: dimM(K)=2⟺ for every ε>0 there is δ0 such that for all δ<δ0, the δ-neighborhood of K has area ≥δε; since K contains a unit segment in each of the N∼δ−1 directions, its δ-neighborhood contains the union of the corresponding N tubes T1,…,TN (this is where the segment-version of the conjecture implies the tube-version). So it suffices to lower-bound ∣⋃iTi∣ by any power δo(1), and the next two steps produce exactly δ−2/log(1/δ)⋅δ2=1/log(1/δ), which indeed tends to 0 slower than any positive power of δ.
Common mistake.Discreteness matters: the δ-tube bound only controls Minkowski dimension directly. Getting the (a priori weaker, but here equal) Hausdorff dimension requires an extra argument across scales, which is part of what makes Davies' original 1971 paper — using projective duality and Marstrand-type projection theorems rather than the tube counting below — a genuinely different route to the same conclusion.