Worked solution: The planar case: every Kakeya set in $\mathbb{R}^2$ has dimension 2 (Davies, 1971)

Step 1 of 4: Besicovitch's paradox and Davies' resolution

In plain words

In 1919 Besicovitch showed that a needle of length 1 can be continuously turned through every direction while sweeping out a region of arbitrarily small area — the pieces of the needle's path can be squeezed together like an accordion so that they overlap almost completely. This is startling: a set containing a full unit segment in every one of infinitely many directions can be 'thin' in the sense of area. Davies asked a sharper question: however small the area, must such a set still be 'thick' in a finer sense, namely Hausdorff dimension? He answered yes for the plane, proving no Kakeya set in $\mathbb{R}^2$ can have dimension less than 2 — settling the case $n=2$ of what is now called the Kakeya conjecture and prompting its statement in every dimension.

\dim_H(K) = \dim_M(K) = 2 \quad \text{for every Kakeya set } K \subset \mathbb{R}^2

Detailed analysis

Formally, the (lower) Minkowski dimension of a bounded set $E$ is $\dim_M(E) = 2 - \lim_{\delta\to 0}\frac{\log|E_\delta|}{\log(1/\delta)}$ , where $E_\delta$ is its $\delta$ -neighborhood; Hausdorff dimension is defined through arbitrary (not just uniform-scale) covers and always satisfies $\dim_H(E)\le \dim_M(E)$ . A single line segment already forces $\dim_H(K)\ge 1$ trivially. Davies' theorem is the nontrivial reverse-type statement $\dim_H(K)\ge 2$ (hence $=2$ , since $K\subset\mathbb{R}^2$ gives the trivial upper bound), proved via a purely measure-theoretic argument combining projective duality between points and lines with Marstrand-type theorems bounding how much the Hausdorff dimension of a set can drop under orthogonal projection in almost every direction.

Terms in this step

Kakeya (Besicovitch) set: A set $K \subset \mathbb{R}^n$ containing a unit line segment pointing in every direction $e \in S^{n-1}$ .