[proofplan] We first show unboundedness on every open neighbourhood of the elementary class $[o]$ using the Lyons–Xu sequence $(\Gamma_n)$: each $\Gamma_n$ is tree-reduced with $d([\Gamma_n]) = 2^{n+1}$, yet the sequence converges to $[o]$ in the product topology because $S(\Gamma_n)^{(m)} = 0$ for $m \leq n$. We then transport this local unboundedness from $[o]$ to an arbitrary class $[x]$ via the left-multiplication homeomorphism $L_{[x]}: [y] \mapsto [x] \ast [y]$, which is continuous by the topological group property of $(\mathcal{C}_1, \ast, \chi_{\mathrm{pr}})$ and shifts $d$ by at most the fixed amount $\|x^*\|_1$. [/proofplan] [step:Recall the Lyons–Xu sequence and its key properties] Let $(\Gamma_n)_{n \in \mathbb{N}}$ be the Lyons–Xu sequence in $\mathcal{C}_1$ constructed in the [Lyons–Xu Example](/theorems/???). The construction yields paths $\Gamma_n$ such that: \begin{enumerate} \item[(i)] each $\Gamma_n$ is tree-reduced, so $\Gamma_n^* = \Gamma_n$ and $d([\Gamma_n]) = \|\Gamma_n\|_1 = 2^{n+1}$; \item[(ii)] the truncated signature satisfies $S(\Gamma_n)^{(m)} = 0$ for all $m \leq n$. \end{enumerate} In particular, $d([\Gamma_n]) \to \infty$ as $n \to \infty$. [/step] [step:Show $[\Gamma_n] \to [o]$ in the product topology] Recall that $\chi_{\mathrm{pr}}$ is the product topology on $\mathcal{C}_1$ induced by the level-wise signature projections \begin{align*} \pi_m: \mathcal{C}_1 &\to V^{\otimes m} \\ [x] &\mapsto S(x)^{(m)}, \end{align*} for $m \in \mathbb{N} \cup \{0\}$. Convergence in the product topology is convergence of every coordinate projection. We verify this for $[\Gamma_n] \to [o]$. Fix $m \in \mathbb{N} \cup \{0\}$. By property (ii) of the Lyons–Xu sequence, for every $n \geq m$ we have $S(\Gamma_n)^{(m)} = 0 = S(o)^{(m)}$. Hence the sequence $(\pi_m([\Gamma_n]))_{n \in \mathbb{N}}$ is eventually constantly $0 = \pi_m([o])$, so $\pi_m([\Gamma_n]) \to \pi_m([o])$ in $V^{\otimes m}$. Since each coordinate projection converges, $[\Gamma_n] \to [o]$ in $\chi_{\mathrm{pr}}$. [guided] We must show that $[\Gamma_n] \to [o]$ in the product topology $\chi_{\mathrm{pr}}$. Why is this the right notion of convergence to verify? The topology $\chi_{\mathrm{pr}}$ is by definition the coarsest topology on $\mathcal{C}_1$ making every signature-projection map \begin{align*} \pi_m: \mathcal{C}_1 &\to V^{\otimes m} \\ [x] &\mapsto S(x)^{(m)}, \end{align*} continuous, for all $m \geq 0$. Equivalently, it is the initial topology induced by the family $(\pi_m)_{m \geq 0}$, which makes it a product-style topology. A standard fact about initial topologies: a sequence $([z_n])$ converges to $[z]$ in $\chi_{\mathrm{pr}}$ if and only if $\pi_m([z_n]) \to \pi_m([z])$ in $V^{\otimes m}$ for every $m$. So it suffices to verify level-by-level convergence. Fix $m \in \mathbb{N} \cup \{0\}$ and consider the sequence $(\pi_m([\Gamma_n]))_n$. By property (ii), $S(\Gamma_n)^{(m)} = 0$ for every $n \geq m$. The signature of the elementary path $o$ is zero at every level $\geq 1$ (it is the constant tensor $1$ at level $0$, and $0$ above; for level $0$ the same is true of $\Gamma_n$), so $S(o)^{(m)} = 0$ for every $m \geq 1$ and the level-$0$ entries also match. Either way: \begin{align*} \pi_m([\Gamma_n]) = 0 = \pi_m([o]) \quad \text{for all } n \geq m. \end{align*} A sequence which is eventually constantly equal to its target directly converges to it. Hence $\pi_m([\Gamma_n]) \to \pi_m([o])$ for every $m$, and by the initial-topology characterisation $[\Gamma_n] \to [o]$ in $\chi_{\mathrm{pr}}$. [/guided] [/step] [step:Show every neighbourhood of $[o]$ contains paths of arbitrarily large $d$] Let $U \subseteq \mathcal{C}_1$ be a non-empty open set with $[o] \in U$. By Step 2, $[\Gamma_n] \to [o]$, so there exists $N \in \mathbb{N}$ such that $[\Gamma_n] \in U$ for all $n \geq N$. By property (i) of the Lyons–Xu sequence, \begin{align*} d([\Gamma_n]) = 2^{n+1} \to \infty \text{ as } n \to \infty. \end{align*} Hence $\sup_{[y] \in U} d([y]) \geq \sup_{n \geq N} d([\Gamma_n]) = \infty$, so $d$ is unbounded on $U$. [guided] We have two ingredients from the previous steps: a sequence $([\Gamma_n])$ converging to $[o]$ in $\chi_{\mathrm{pr}}$ (Step 2), and the same sequence having tree-reduced length blowing up, $d([\Gamma_n]) = 2^{n+1} \to \infty$ (property (i) of Step 1). The combination is exactly what we need for unboundedness near $[o]$. Let $U \subseteq \mathcal{C}_1$ be an arbitrary non-empty open neighbourhood of $[o]$. By definition of convergence in a topological space, $[\Gamma_n] \to [o]$ means: for every open neighbourhood $U$ of $[o]$, there exists $N \in \mathbb{N}$ such that $[\Gamma_n] \in U$ for all $n \geq N$. Apply this to our $U$ and get the corresponding index $N$. Now consider $\sup_{[y] \in U} d([y])$. The tail of the sequence $([\Gamma_n])_{n \geq N}$ lies inside $U$, so \begin{align*} \sup_{[y] \in U} d([y]) \geq \sup_{n \geq N} d([\Gamma_n]) = \sup_{n \geq N} 2^{n+1} = +\infty. \end{align*} Hence $d$ is unbounded on $U$. Since $U$ was an arbitrary open neighbourhood of $[o]$, this proves that $d$ is unbounded on every open neighbourhood of $[o]$. This is the **local** version of the unboundedness statement, anchored at $[o]$. The next step transports it to an arbitrary class $[x]$ via the group action. [/guided] [/step] [step:Translate unboundedness from $[o]$ to an arbitrary class via the left-multiplication homeomorphism] Fix an arbitrary class $[x] \in \mathcal{C}_1$ and let $W \subseteq \mathcal{C}_1$ be a non-empty open set with $[x] \in W$. We exhibit elements of $W$ with arbitrarily large tree-reduced length. By the [Topological Group Structure of $(\mathcal{C}_1, \ast, \chi_{\mathrm{pr}})$](/theorems/???), the left-multiplication map \begin{align*} L_{[x]}: \mathcal{C}_1 &\to \mathcal{C}_1 \\ [y] &\mapsto [x] \ast [y] \end{align*} is a homeomorphism, with inverse $L_{[x]^{-1}}: [z] \mapsto [x]^{-1} \ast [z] = [x^{\leftarrow}] \ast [z]$, where $x^{\leftarrow}$ denotes the time-reversal of $x$. Define \begin{align*} V := L_{[x]}^{-1}(W) = \{[y] \in \mathcal{C}_1 : [x] \ast [y] \in W\}. \end{align*} Since $L_{[x]}$ is continuous, $V$ is open. Since $[x] = [x] \ast [o] = L_{[x]}([o]) \in W$ implies $[o] \in V$, the set $V$ is a non-empty open neighbourhood of $[o]$. By Step 3, there exists a sequence $([y_k])_{k \in \mathbb{N}} \subseteq V$ with $d([y_k]) \to \infty$. Set $[z_k] := L_{[x]}([y_k]) = [x] \ast [y_k] \in W$. We claim $d([z_k]) \to \infty$. By the [Sub-Additivity of Tree-Reduced Length under Concatenation](/theorems/???), the tree-reduced length satisfies the reverse triangle inequality \begin{align*} d([y_k]) = d([x]^{-1} \ast [z_k]) \leq d([x]^{-1}) + d([z_k]) = \|x^*\|_1 + d([z_k]), \end{align*} where we used $d([x]^{-1}) = \|(x^{\leftarrow})^*\|_1 = \|x^*\|_1$ since time-reversal preserves tree-reduced length. Rearranging gives \begin{align*} d([z_k]) \geq d([y_k]) - \|x^*\|_1. \end{align*} Since $d([y_k]) \to \infty$ and $\|x^*\|_1$ is a fixed finite constant, $d([z_k]) \to \infty$. [guided] The strategy is **transport via the group action**: the topology $\chi_{\mathrm{pr}}$ on $\mathcal{C}_1$ is invariant under left-translation by any class, so a topological property holding at $[o]$ should hold everywhere — provided we can control how $d$ transforms under translation. Fix an arbitrary class $[x] \in \mathcal{C}_1$ and let $W \subseteq \mathcal{C}_1$ be a non-empty open set with $[x] \in W$. Our goal is to exhibit elements of $W$ with arbitrarily large tree-reduced length $d$. Why is left-multiplication the natural tool? The space $(\mathcal{C}_1, \ast, \chi_{\mathrm{pr}})$ is a topological group, so for every fixed $[x]$ the map \begin{align*} L_{[x]}: \mathcal{C}_1 &\to \mathcal{C}_1 \\ [y] &\mapsto [x] \ast [y] \end{align*} is continuous (this is the content of the [Topological Group Structure of $(\mathcal{C}_1, \ast, \chi_{\mathrm{pr}})$](/theorems/???)). Its inverse is $L_{[x]^{-1}}: [z] \mapsto [x]^{-1} \ast [z]$, also continuous; hence $L_{[x]}$ is a homeomorphism. The group inverse $[x]^{-1}$ is the class of the time-reversed path $x^\leftarrow$, since $x \cdot x^\leftarrow$ is tree-equivalent to the elementary path $o$. **Pull $W$ back to a neighbourhood of $[o]$.** Define \begin{align*} V := L_{[x]}^{-1}(W) = \{[y] \in \mathcal{C}_1 : [x] \ast [y] \in W\}. \end{align*} Since $L_{[x]}$ is continuous, $V$ is open. Since $L_{[x]}([o]) = [x] \ast [o] = [x] \in W$, we have $[o] \in V$. So $V$ is a non-empty open neighbourhood of $[o]$, and Step 3 applies to it. **Extract a sequence in $V$ with $d \to \infty$.** By Step 3, there exists a sequence $([y_k])_{k \in \mathbb{N}} \subseteq V$ with $d([y_k]) \to \infty$ as $k \to \infty$. **Push forward to $W$.** Set $[z_k] := L_{[x]}([y_k]) = [x] \ast [y_k]$. Since $[y_k] \in V = L_{[x]}^{-1}(W)$, we have $[z_k] \in W$ by construction. **Control how $d$ transforms under translation.** This is the only substantial step. We claim $d([z_k]) \to \infty$. The relevant fact is the [Sub-Additivity of Tree-Reduced Length under Concatenation](/theorems/???), which gives the reverse-triangle inequality \begin{align*} |d([a] \ast [b]) - d([a])| \leq d([b]) \qquad \text{for all } [a], [b] \in \mathcal{C}_1. \end{align*} We apply this with $[a] = [x]^{-1}$ and $[b] = [z_k]$, noting that $[x]^{-1} \ast [z_k] = [x]^{-1} \ast [x] \ast [y_k] = [y_k]$. The inequality becomes \begin{align*} |d([y_k]) - d([x]^{-1})| \leq d([z_k]). \end{align*} Why is $d([x]^{-1}) = \|x^*\|_1$? The group inverse is realised by time-reversal, and time-reversal preserves the $1$-variation of a path; the tree-reduced representative is also preserved up to time-reversal. Hence $d([x]^{-1}) = \|(x^\leftarrow)^*\|_1 = \|x^*\|_1$, a fixed finite constant. For $k$ large enough that $d([y_k]) > \|x^*\|_1$, the absolute value can be dropped and we get \begin{align*} d([z_k]) \geq d([y_k]) - \|x^*\|_1. \end{align*} Since $d([y_k]) \to \infty$ while $\|x^*\|_1$ is fixed, this forces $d([z_k]) \to \infty$. Therefore $d$ is unbounded on $W$, completing the transport of unboundedness from $[o]$ to the arbitrary class $[x]$. [/guided] [/step] [step:Conclude unboundedness on every non-empty open set] Let $U \subseteq \mathcal{C}_1$ be an arbitrary non-empty open set. Pick any $[x] \in U$. By Step 4 with $W := U$, the function $d$ is unbounded on $U$. Since $U$ was arbitrary, $d$ is unbounded on every non-empty open subset of $(\mathcal{C}_1, \chi_{\mathrm{pr}})$, completing the proof. [/step]