Compactness of equi-Lipschitz curves with values in a compact Euclidean set

By [F1], choose an integer $R\ge 1$ with $C\subseteq [-R,R{]}^d$. For each natural $r\ge 1$, let ${\mathrm{\Gamma }}_r=\{kT\mathrm{/}{2}^r:0\le k\le 2^r\}$ and put $K_r=(R+1)2^r$. Let $G_r$ consist of the lattice points whose coordinates are $m_k{2}^{-r}$ with integers $-K_r\le m_k\le K_r$. Each coordinate index set is finite: its nonnegative and negative parts are copies of finite natural initial segments, which are finite by [F9]; induction on the finite number $d$ of coordinates shows their Cartesian product $G_r$ is finite. To verify its covering radius, fix $x\in [-R,R{]}^d$. For each $k<d$, [F6] applied to $2^r{x}_k$ gives an integer $m_k$ with $m_k\le 2^r{x}_k<m_k+1$. Thus $m_k\le R{2}^r\le K_r$ and $m_k>-R{2}^r-1\ge -(R+1)2^r=-K_r$, using $1\le 2^r$ for $r\ge 1$ (from the power recursion and induction). The corresponding lattice point $g$ has $\mathrm{\mid }{x}_k-g_k\mathrm{\mid }<2^{-r}$ in every coordinate, so [F8] gives $\mathrm{\parallel }x-g{\mathrm{\parallel }}_2^2={\sum }_{k<d}\mathrm{\mid }{x}_k-g_k{\mathrm{\mid }}^2<d{2}^{-2r}$ and hence $\mathrm{\parallel }x-g{\mathrm{\parallel }}_2<{\delta }_r:=\sqrt{d}{2}^{-r}$. Assign each value $u_j(t)$ at each $t\in {\mathrm{\Gamma }}_r$ to the first lattice point within ${\delta }_r$ in a fixed ordering of $G_r$. This gives one of finitely many patterns.

Starting from the positive index set $I_0=\{j\in \mathbb{N}:j\ge 1\}$, partition each infinite index set $I_{r-1}$ by the finitely many patterns at resolution $r$. At least one class is infinite; choose the first such pattern in the fixed finite ordering and let $I_r$ be its infinite class. Then $I_r\subseteq I_{r-1}$.

Set $n_0=1\in I_0$. For $r\ge 1$, recursively choose $n_r$ to be the least member of $I_r$ greater than $n_{r-1}$. The infinite set $I_r$ is unbounded: if it were contained in some finite initial segment $\{0,\dots ,N\}$, that segment would be finite by [F9], contradicting that $I_r$ is infinite. Hence $\{n\in I_r:n>n_{r-1}\}$ is nonempty, and [F5] supplies its least member. For every fixed $r$, all $n_k$ with $k\ge r$ belong to $I_r$, so the maps with those indices have the same resolution-$r$ pattern.

For $k,l\ge r$ and $t\in [0,T]$, choose $s\in {\mathrm{\Gamma }}_r$ with $\mathrm{\mid }t-s\mathrm{\mid }\le T\mathrm{/}{2}^r$. At each lattice time, the common pattern assigns both values to the same lattice point, so the triangle inequality [F8] gives $\mathrm{\parallel }{u}_{n_k}(s)-u_{n_l}(s){\mathrm{\parallel }}_2\le 2{\delta }_r$. Using the Lipschitz bounds and the triangle inequality again, $\mathrm{\parallel }{u}_{n_k}(t)-u_{n_l}(t){\mathrm{\parallel }}_2\le \mathrm{\parallel }{u}_{n_k}(t)-u_{n_k}(s){\mathrm{\parallel }}_2+\mathrm{\parallel }{u}_{n_k}(s)-u_{n_l}(s){\mathrm{\parallel }}_2+\mathrm{\parallel }{u}_{n_l}(s)-u_{n_l}(t){\mathrm{\parallel }}_2\le 2MT\mathrm{/}{2}^r+2{\delta }_r\mathrm{.}$ By [F7], the right side tends to zero as $r\to \mathrm{\infty }$, uniformly in $t$, so $(u_{n_k})$ is uniformly Cauchy.

At each $t$, step 4.1 and [F8] make every coordinate sequence $u_{n_k}(t{)}_j$ Cauchy; [F4] therefore gives a pointwise limit $u(t)\in {\mathbb{R}}^d$. Given $\epsilon >0$, choose $r$ so ${\rho }_r:=2MT\mathrm{/}{2}^r+2{\delta }_r<\epsilon$, using [F7]. For every $k\ge r$, step 4.1 gives ${\sum }_{j<d}\mathrm{\mid }{u}_{n_k}(t{)}_j-u_{n_l}(t{)}_j{\mathrm{\mid }}^2\le {\rho }_r^2$ for all $l\ge r$ and all $t$. Passing to the coordinatewise limit in this finite sum yields ${\sum }_{j<d}\mathrm{\mid }{u}_{n_k}(t{)}_j-u(t{)}_j{\mathrm{\mid }}^2\le {\rho }_r^2$, hence $\mathrm{\parallel }{u}_{n_k}(t)-u(t){\mathrm{\parallel }}_2\le {\rho }_r<\epsilon$, uniformly in $t$. Thus the subsequence converges uniformly. If $u(t)\mathrm{\notin }C$, openness of the complement [F2] gives a ball about $u(t)$ disjoint from $C$, while convergence [F3] puts $u_{n_k}(t)$ in that ball eventually, a contradiction. Thus $u(t)\in C$. For fixed $s,t$, coordinatewise convergence at those two points gives ${\sum }_{j<d}\mathrm{\mid }u(t{)}_j-u(s{)}_j{\mathrm{\mid }}^2={\lim }_k{\sum }_{j<d}\mathrm{\mid }{u}_{n_k}(t{)}_j-u_{n_k}(s{)}_j{\mathrm{\mid }}^2\le M^2\mathrm{\mid }t-s{\mathrm{\mid }}^2$; taking nonnegative square roots yields $\mathrm{\parallel }u(t)-u(s){\mathrm{\parallel }}_2\le M\mathrm{\mid }t-s\mathrm{\mid }$.