A finite coordinate path and a bounded nonrectifiable path

The coordinates are $\mathrm{\Phi }(z)=(1,2)$ and $\mathrm{\Phi }(w)=(3,-1)$. Since $rz=-2-4i$, $\mathrm{\Phi }(rz)=(-2,-4)=-2(1,2)=r\mathrm{\Phi }(z)$ by [F1]. Also $(1-t)z+tw=\frac{2}{3}(1+2i)+\frac{1}{3}(3-i)=\frac{5}{3}+i$, whose coordinates are $(\frac{5}{3},1)=\frac{2}{3}(1,2)+\frac{1}{3}(3,-1)$ by [F1] and [F2].

We have $w-z=2-3i$, so $\mathrm{\mid }w-z\mathrm{\mid }=\sqrt{2^2+(-3{)}^2}=\sqrt{13}$. The Euclidean distance between $(1,2)$ and $(3,-1)$ is the same $\sqrt{13}$, and [F2] gives $\mathrm{\Phi }[B_{\mathbb{C}}(1+i,2)]=B_2((1,1),2)$.

The domain $U=\mathbb{C}$ is open by [F8]. For $\zeta =x+iy$, conjugation has coordinates $f_{\mathbb{R}}(x,y)=(x,-y)$, the matrix map $\mathrm{diag}(1,-1)$. If $h_j=u_j+i{v}_j$ and $r\in \mathbb{R}$, then $\overline{h_1+h_2}=(u_1+u_2)-i(v_1+v_2)=\overline{h_1}+\overline{h_2}$ and $\overline{r{h}_1}=r{u}_1-ir{v}_1=r\overline{h_1}$, so $L(h)=\bar{h}$ is real-linear. For any $h=u+iv$, [F1] and [F7] give $f(\zeta +h)-f(\zeta )=\bar{h}$, so its remainder is zero. The quotient $\mathrm{\mid }r(h)\mathrm{\mid }\mathrm{/}\mathrm{\mid }h\mathrm{\mid }$ is therefore zero for every $h\mathrm{\ne }0$, and [F3] gives $D{f}_{\mathbb{R}}(\mathrm{\Phi }(\zeta ))=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$.

The formulas agree at $s=1$ and $s=2$. On the first and second pieces the distance between values at $s$ and $s_0$ equals $\mathrm{\mid }s-s_0\mathrm{\mid }$, because their differences are respectively real and purely imaginary and the modulus formulas in [F7] and [F10] give this equality; on the constant third piece it is zero. Thus each restriction is continuous. At either breakpoint $s_0\in \{1,2\}$, take $\delta =\min (\epsilon ,1\mathrm{/}2)$. Then every $s\in [0,3]$ with $\mathrm{\mid }s-s_0\mathrm{\mid }<\delta$ lies in the adjacent pieces, where $d_{\mathbb{C}}(\gamma (s),\gamma (s_0))=\mathrm{\mid }s-s_0\mathrm{\mid }$ on a moving piece and is zero on the constant piece; hence the distance is less than $\epsilon$. At 0 and 3 continuity follows from the corresponding endpoint of the continuous first and last restrictions. Thus $\gamma$ is continuous by [F9]. Its coordinate functions are $x(s)=s$ on $[0,1]$ and $1$ on $[1,3]$, and $y(s)=0$ on $[0,1]$, $s-1$ on $[1,2]$, and $1$ on $[2,3]$.

Let $\sigma :[5,5]\to \mathbb{C}$ send $5$ to $2-i$. By [F4], $L_{\mathbb{C}}(\sigma )=0$ and $\sigma$ is rectifiable. The constant surjection $\theta :[7,7]\to [5,5]$ also satisfies [F6], and both sides of its length equality are zero.

For every $n\ge 1$, $0<a_n<m_n<b_n\le 1$, and $b_{n+1}=1\mathrm{/}(2n+1)<1\mathrm{/}(2n)=a_n$ by [F11]. Thus the intervals $I_n$ are disjoint, the two formulas for $\eta$ agree at $m_n$, and their endpoint values are zero. Each interpolation factor lies in $[0,1]$, so $0\le \eta (t)\le 1\mathrm{/}n\le 1$ on $I_n$; outside them $\eta =0$. Hence $\eta$ is well-defined and its trace is contained in $[0,1]$.

Fix $t_0>0$ and choose $N$ with $1\mathrm{/}N<t_0\mathrm{/}2$ by [F11]. For $n\ge N$, $b_n=1\mathrm{/}(2n-1)\le 1\mathrm{/}n\le 1\mathrm{/}N<t_0\mathrm{/}2$, so those intervals lie below $t_0\mathrm{/}2$; only finitely many intervals meet $(t_0\mathrm{/}2,3t_0\mathrm{/}2)$. There $\eta$ is zero or one of finitely many affine pieces; the formulas agree at each midpoint and vanish at the interval endpoints, so their one-sided limits agree at every join. Thus $\eta$ is continuous at $t_0$. At $0$, given $\epsilon >0$ choose $N$ with $2\mathrm{/}N<\epsilon$. If $0<t<1\mathrm{/}N$ lies in $I_n$, then $1\mathrm{/}(2n)=a_n\le t<1\mathrm{/}N$, so $n>N\mathrm{/}2$ and $\mathrm{\mid }\eta (t)\mathrm{\mid }\le 1\mathrm{/}n<2\mathrm{/}N<\epsilon$; outside the intervals $\eta (t)=0$. These real-distance bounds are also the $d_{\mathbb{C}}$ bounds under the canonical embedding by [F10]. Hence $\eta$ is continuous as a complex path at $0$ and at every positive $t_0$.

For $N\ge 1$, the increasing list $P_N=(0,a_N,m_N,b_N,a_{N-1},m_{N-1},b_{N-1},\dots ,a_1,m_1,b_1)$ is a partition of $[0,1]$, since $b_n<a_{n-1}$ for $n\ge 2$. On each $I_n$, the path values at $a_n,m_n,b_n$ are $0,1\mathrm{/}n,0$, so its two chord lengths sum to $2\mathrm{/}n$ by [F10]; across each gap the chord length is zero. Define $H_N:={\sum }_{j<N}1\mathrm{/}(j+1)$, also for $N=0$. By the prefix recursion, $H_0=0$ and $H_{N+1}=H_N+1\mathrm{/}(N+1)$. Adding the spike contributions gives ${\mathrm{\ell }}_{P_N}(\eta )=2H_N$: this holds at $N=1$, and inserting the next spike adds $2\mathrm{/}(N+1)$, proving it by induction.

By recursion [F13], define natural numbers $q_0=1$ and $q_{k+1}=q_k+q_k$. For every natural $q_k$, the definition of natural order with witness $q_k$, together with $0+q_k=q_k$ [F13], gives $0\le q_k$; compatibility of order with addition then gives $q_k=0+q_k\le q_k+q_k=q_{k+1}$. The base is $q_0=1$. If $1\le q_k$, transitivity with $q_k\le q_{k+1}$ gives $1\le q_{k+1}$, so induction [F13] yields $q_k\ge 1$ for every $k$.

Both coordinates are nondecreasing, with endpoint increments $x(3)-x(0)=1$ and $y(3)-y(0)=1$. For any partition, each absolute coordinate increment equals the nonnegative increment, and the finite sum telescopes to the endpoint increment; the endpoint partition attains $1$. Thus each coordinate has variation exactly $1$. On the three pieces the coordinate velocities are $(1,0)$, $(0,1)$, and $(0,0)$, so the speed integrals are $1$, $1$, and $0$. Therefore the coordinate path has length $2$, and [F5] gives $L_{\mathbb{C}}(\gamma )=2$ and rectifiability.

For each $k$, put $q=q_k$ and $B_k:={\sum }_{j<q}1\mathrm{/}(q+j+1)$. For $j<q$, the positive denominator satisfies $q+j+1\le q+q=q_{k+1}$ in the canonical real images by [F11, F13]. Thus each summand is at least $1\mathrm{/}(q+q)$. Monotonicity and constant-sum scaling in [F12] give $B_k\ge q\mathrm{/}(q+q)=1\mathrm{/}2$, since $q\ge 1$ by step 1.9.

Step 2.1 gives $L_{\mathbb{C}}(\gamma )=2$. The map $\phi (s)=3-s$ on $[0,3]$ is a continuous decreasing surjection, so [F6] gives $L_{\mathbb{C}}(\gamma \circ \phi )=2$. Define $\psi :[0,4]\to [0,3]$ by $\psi (s)=s$ on $[0,1]$, $\psi (s)=1$ on $[1,2]$, and $\psi (s)=s-1$ on $[2,4]$. Its pieces agree at their endpoints; it is continuous and nondecreasing, and its range is $[0,1]\cup \{1\}\cup [1,3]=[0,3]$. Thus it is a surjection with a pause, and [F6] gives $L_{\mathbb{C}}(\gamma \circ \psi )=2$.

Fix a natural $q\ge 1$. The prefix recursions in [F12] and step 1.8 give $H_{q+d}=H_q+{\sum }_{j<d}1\mathrm{/}(q+j+1)$ by induction on $d$: at $d=0$ both sides equal $H_q$, and the successor adds $1\mathrm{/}(q+d+1)$ to both sides. Taking $q=d=q_K$ gives $H_{q_{K+1}}=H_{q_K}+B_K$ with no natural subtraction or lower-index sum. Since $H_{q_0}=H_1=1$, step 2.2 and induction give $H_{q_K}\ge 1+K\mathrm{/}2$ for all $K$. The Archimedean property [F11] makes these lower bounds unbounded, so the harmonic partial sums are unbounded.

The partition sums ${\mathrm{\ell }}_{P_N}(\eta )=2H_N$ are unbounded by steps 1.8 and 3.2. Hence the partition sums of $\eta$ have no real upper bound, so $\eta$ is not rectifiable by [F4], although its trace is bounded by step 1.6.