= \alpha_{1/2}\bigl(\rho\bigl(s_{0}(x), x\bigr)\bigr) - \rho\bigl(s_{0}(x),x\bigr).$$ The cross section $ \mathcal{C}_{1} $ is then the $ h_{0} $-shift of $ \mathcal{C}_{0} $: $ \mathcal{C}_{1} = \bigl\{ x + h_{0}(x) : x \in \mathcal{C}_{0} \bigr\} $. Geometrically, $ \mathcal{C}_{1} $ is constructed by shifting points by at most $ 1/2 $ within $ \fber[0][0] $-classes relative to the origin provided by the minimal point $ s_{0}(x) $.

The inductive step is very similar, with a notable difference lying in the fact that when moving points within each $ \fber[n][n] $-class, together with any point we move its $ \fber[n-1][n] $-class. More precisely, suppose $ \mathcal{C}_{n}, h_{n-1} $ have been constructed, and let $ s_{n} : \mathcal{C}_{n} \to \mathcal{C}_{n} $ be the Borel selector, which picks $ \prec $-minimal points within $ \fber[n][n] $-classes. Let also $ \tilde{s}_{n-1} : \mathcal{C}_{n} \to \mathcal{C}_{n} $ denote the Borel selector for $ \fber[n-1][n] $-classes. Define $ h_{n} : \mathcal{C}_{n} \to B\bigl(2^{-n-1}\bigr) $ by setting $$h_{n}(x) = \alpha_{2^{-n-1}}\bigl(\rho\bigl(s_{n}(x), \tilde{s}_{n-1}(x)\bigr)\bigr) -
    \rho\bigl(s_{n}(x), \tilde{s}_{n-1}(x)\bigr).$$

It is easy to see that items and are satisfied. Also, $ h_{n} : \mathcal{C}_{n} \to B(2^{-n-1}) $ and Proposition \[prop:lacunarity-of-limit-cross-section\] ensure that $ (\mathcal{C}_{n}, h_{n}) $ converges, and the limit cross section $ \mathcal{D} 