[proofplan] The strategy is to repeat — under the strengthened hypotheses on $\tau$ that make $(E, \tau)$ "reasonable" with respect to the projective tensor-algebra structure — the Baire-category argument that proved [Non-Polishness of $(\mathcal{C}_1, \chi_{\mathrm{pr}})$](/theorems/2510). The hypotheses give us three structural inputs that propagate to $\chi$: continuity of the signature map $S : C_p \to (E, \tau)$, continuity of every signature-projection $\pi_m \circ S$ together with a sequential criterion converting projection-wise convergence to $\tau$-convergence, and continuity of the tensor-algebra group operations. These three inputs allow us to transfer compactness of length balls (via Arzelà–Ascoli on Lipschitz representatives plus the interpolation $\|h\|_p \le (2\|h\|_\infty)^{1-1/p}\|h\|_1^{1/p}$) and to transfer the Lyons–Xu unboundedness construction from $\chi_{\mathrm{pr}}$ to $\chi$. These give a countable cover of $\mathcal{C}_1$ by closed nowhere-dense sets, contradicting the Baire property; the contrapositive of the [Baire Category Theorem](/theorems/1072) and of [Baire for locally compact Hausdorff spaces](/theorems/1105) then rules out both Polishness and local compactness. [/proofplan] [step:Transfer compactness of length balls from $\chi_{\mathrm{pr}}$ to $\chi$] Fix $r > 0$ and let \begin{align*} B(r) := \bigl\{ [x] \in \mathcal{C}_1 : \|x^*\|_1 \leq r \bigr\} \end{align*} be the length ball of radius $r$, where $x^*$ denotes the tree-reduced representative of $[x]$ (which exists and is unique up to reparametrisation by [Hambly-Lyons Uniqueness](/theorems/2488)). We show that $B(r)$ is compact in the topology $\chi$ on $\mathcal{C}_1$. Let $([x_n])_{n=1}^\infty$ be a sequence in $B(r)$. Following the proof of [Length Balls Are Compact](/theorems/2506), choose tree-reduced representatives $x_n^* : [0, 1] \to V$ parametrised proportionally to arc length and translated so that $x_n^*(0) = 0$. Each $x_n^*$ is then $r$-Lipschitz, and the family $(x_n^*)_{n \geq 1}$ is uniformly bounded by $r$ on $[0,1]$. By the [Arzelà-Ascoli Theorem](/theorems/885), there exists a subsequence $(x_{n_k}^*)_{k \geq 1}$ converging uniformly on $[0,1]$ to a continuous limit $x : [0,1] \to V$ which is itself $r$-Lipschitz, hence has $\|x\|_1 \leq r$. We now upgrade uniform convergence to convergence in the $p$-variation topology of $C_p$. For continuous paths $h : [0,1] \to V$ of finite $1$-variation, the standard interpolation between $L^\infty$ and $1$-variation reads \begin{align*} \|h\|_p \leq \bigl(2\|h\|_\infty\bigr)^{1 - 1/p} \, \|h\|_1^{1/p}, \qquad 1 \leq p < \infty. \end{align*} Applying this with $h = x_{n_k}^* - x_{n_l}^*$ — both $r$-Lipschitz, so $\|h\|_1 \leq 2r$ and $\|h\|_\infty \to 0$ as $k, l \to \infty$ by uniform convergence — gives \begin{align*} \|x_{n_k}^* - x_{n_l}^*\|_p \leq (2 \|x_{n_k}^* - x_{n_l}^*\|_\infty)^{1 - 1/p} (2r)^{1/p} \xrightarrow[k,l \to \infty]{} 0. \end{align*} Hence $(x_{n_k}^*)$ is Cauchy in the $p$-variation seminorm; by completeness of $C_p$ and uniqueness of uniform limits, $x_{n_k}^* \to x$ in $C_p$. By hypothesis (3), the signature map $S : C_p \to (E, \tau)$ is continuous. Hence \begin{align*} S(x_{n_k}^*) \to S(x) \quad \text{in } (E, \tau). \end{align*} Since $x \in C_1 \subset C_p$ with $\|x\|_1 \leq r$, we have $S(x) \in \mathcal{S}_1 \subset \mathcal{S}_p \subset E$ by hypothesis (2). Each $S(x_{n_k}^*)$ likewise lies in $\mathcal{S}_1 \subset E$. So the convergence holds in the subspace topology $\tau|_{\mathcal{S}}$. Since $S : (\mathcal{C}_1, \chi) \to (\mathcal{S}, \tau|_{\mathcal{S}})$ is by definition a homeomorphism, $[x_{n_k}] \to [x]$ in $\chi$. By the length-minimisation property of tree-reduced representatives ([Hambly–Lyons Uniqueness](/theorems/2488), which gives $\|[x]^*\|_1 \leq \|x\|_1$), the limit class $[x]$ has tree-reduced length at most $r$, so $[x] \in B(r)$. We have extracted from an arbitrary sequence in $B(r)$ a $\chi$-convergent subsequence with limit in $B(r)$. By assumption, $\tau$ is metrisable, so the subspace topology $\tau|_{\mathcal{S}}$ is metrisable, and the homeomorphism $S$ transports this metric to $\chi$ — hence $\chi$ is metrisable on $\mathcal{C}_1$. In a metrisable space, sequential compactness implies compactness; thus $B(r)$ is $\chi$-compact. [guided] The goal of this step is to show that the length ball $B(r) := \{[x] \in \mathcal{C}_1 : \|x^*\|_1 \leq r\}$ — a set originally proven compact in the canonical product topology $\chi_{\mathrm{pr}}$ via [Length Balls Are Compact](/theorems/2506) — is also compact in the new topology $\chi$. The original proof of theorem 2506 has four moving parts: a metrisability reduction, an Arzelà-Ascoli step at the path level, an interpolation upgrade from uniform to $q$-variation convergence, and a transfer to $\chi_{\mathrm{pr}}$ via continuity of each signature truncation $S_m : C_q \to V^{\otimes m}$. Three of those four parts go through verbatim; only the **transfer step** at the end has to be re-derived, because the topology is now $\chi$, not $\chi_{\mathrm{pr}}$. **Why metrisability of $\chi$ matters first.** We want compactness of $B(r)$ in $\chi$. By hypothesis, $E$ carries a metrisable topology $\tau$; the subspace topology $\tau|_{\mathcal{S}}$ on $\mathcal{S} = \mathcal{S}_1 \subset E$ is then metrisable (subspaces of metrisable spaces are metrisable). The homeomorphism $S : (\mathcal{C}_1, \chi) \to (\mathcal{S}, \tau|_{\mathcal{S}})$ pulls back this metric to $\chi$, so $(\mathcal{C}_1, \chi)$ is itself metrisable. In a metrisable space, a subset is compact if and only if it is sequentially compact, so it suffices to verify that every sequence in $B(r)$ has a $\chi$-convergent subsequence with limit in $B(r)$. **The Arzelà-Ascoli reduction at the path level.** Fix a sequence $([x_n])_{n=1}^\infty$ in $B(r)$. By [Hambly-Lyons Uniqueness](/theorems/2488) each class admits a tree-reduced representative; the natural choice is the one parametrised proportionally to arc length on $[0,1]$ and translated so as to start at $0 \in V$. With these conventions, $x_n^* : [0, 1] \to V$ is Lipschitz with constant equal to its length $\|x_n^*\|_1 \leq r$, and $|x_n^*(t)| \leq r \cdot t \leq r$ for all $t \in [0, 1]$. Hence the family $\mathcal{F} := \{x_n^* : n \geq 1\} \subset C([0,1]; V)$ is uniformly bounded (by $r$) and equicontinuous (by the uniform Lipschitz constant $r$). The codomain $V$ is finite-dimensional, hence a Banach space; the domain $[0,1]$ is a compact metric space. The [Arzelà-Ascoli Theorem](/theorems/885) for $V$-valued continuous functions on a compact metric space then asserts that $\mathcal{F}$ is precompact in the uniform topology on $C([0,1]; V)$ — i.e., there is a subsequence (which we relabel $(x_n^*)$ for brevity) converging uniformly on $[0,1]$ to a continuous limit $x : [0,1] \to V$. The uniform limit inherits the Lipschitz bound: passing to the limit in $|x_n^*(t) - x_n^*(s)| \leq r |t - s|$ gives $|x(t) - x(s)| \leq r |t - s|$, so $x$ is $r$-Lipschitz. By the relation between Lipschitz constants and $1$-variation (Lipschitz constant $\times$ interval length is an upper bound for $1$-variation), $\|x\|_1 \leq r$. **Upgrading to $C_p$ convergence.** Uniform convergence is too weak to invoke continuity of $S : C_p \to E$, which requires convergence in the $p$-variation topology. The bridge is the standard $L^\infty$/$1$-variation interpolation: for any continuous $h : [0,1] \to V$ of finite $1$-variation and any $p \in [1, \infty)$, \begin{align*} \|h\|_p \leq \bigl(2\|h\|_\infty\bigr)^{1 - 1/p} \, \|h\|_1^{1/p}. \end{align*} (This is a consequence of $|h(t) - h(s)|^p \leq (2\|h\|_\infty)^{p-1}\,|h(t) - h(s)|$ summed over a partition: $\sum_i |h(t_i) - h(t_{i-1})|^p \le (2\|h\|_\infty)^{p-1} \sum_i |h(t_i) - h(t_{i-1})| \le (2\|h\|_\infty)^{p-1}\|h\|_1$, so $\|h\|_p \le (2\|h\|_\infty)^{1-1/p}\|h\|_1^{1/p}$.) Apply this with $h = x_n^* - x_m^*$. Both $x_n^*$ and $x_m^*$ are $r$-Lipschitz, so $\|h\|_1 \leq 2r$. Hence \begin{align*} \|x_n^* - x_m^*\|_p \leq (2 \|x_n^* - x_m^*\|_\infty)^{1 - 1/p} (2r)^{1/p}. \end{align*} The exponent $1 - 1/p > 0$ (since $p > 1$), so as $n, m \to \infty$ the right-hand side vanishes by uniform convergence. Hence $(x_n^*)$ is Cauchy in $p$-variation, and by completeness of $C_p$ (and uniqueness of uniform limits) it converges to $x$ in $C_p$. **Transferring the convergence to $\chi$.** Now we use the hypothesis. By assumption, the signature map $S : C_p \to (E, \tau)$ is continuous. Since $x_n^* \to x$ in $C_p$, continuity of $S$ gives \begin{align*} S(x_n^*) \to S(x) \quad \text{in } (E, \tau). \end{align*} The signatures $S(x_n^*)$ and $S(x)$ all lie in $\mathcal{S} = \mathcal{S}_1 \subset E$ (the limit lies in $\mathcal{S}_1$ because $x \in C_1$, having $1$-variation at most $r < \infty$). So the convergence holds in the subspace topology $\tau|_{\mathcal{S}}$: \begin{align*} S(x_n^*) \to S(x) \quad \text{in } (\mathcal{S}, \tau|_{\mathcal{S}}). \end{align*} By definition of $\chi$, the bijection $S : (\mathcal{C}_1, \chi) \to (\mathcal{S}, \tau|_{\mathcal{S}})$ is a homeomorphism — in particular, $S^{-1}$ is continuous. Apply $S^{-1}$ to the convergence above: \begin{align*} [x_n] = S^{-1}(S(x_n^*)) \to S^{-1}(S(x)) = [x] \quad \text{in } (\mathcal{C}_1, \chi). \end{align*} Since $\|x\|_1 \leq r$, the equivalence class $[x]$ has tree-reduced representative of length at most $\|x\|_1 \leq r$ (the tree-reduced representative is the minimal-length representative in the class, so $\|x^*\|_1 \leq \|x\|_1 \leq r$). Hence $[x] \in B(r)$. **Conclusion.** From an arbitrary sequence in $B(r)$ we extracted a subsequence converging in $\chi$ to a limit in $B(r)$. So $B(r)$ is sequentially $\chi$-compact, and by metrisability, $B(r)$ is $\chi$-compact. **Why does the original proof go through?** The original proof of theorem 2506 closes by invoking continuity of each signature truncation $S_m : C_p \to V^{\otimes m}$ and observing that level-by-level convergence is exactly convergence in $\chi_{\mathrm{pr}}$. Here, we replace that final transfer by a single appeal to the assumed continuity $S : C_p \to (E, \tau)$ — bypassing levels entirely. Note that hypothesis (4) (continuity of each $\pi_m : (E, \tau) \to V^{\otimes m}$) ensures that the levelwise picture survives — $\pi_m \circ S : C_p \to V^{\otimes m}$ is continuous as a composition of continuous maps — but the route through $S$ alone suffices for this step. [/guided] [/step] [step:Deduce $\sigma$-compactness of $(\mathcal{C}_1, \chi)$ from the cover by length balls] By [Hambly-Lyons Uniqueness](/theorems/2488), every $[x] \in \mathcal{C}_1$ admits a tree-reduced representative $x^*$ with $\|x^*\|_1 < \infty$. Setting $r_0 := \lceil \|x^*\|_1 \rceil \in \mathbb{N}$, we have $[x] \in B(r_0)$. Hence \begin{align*} \mathcal{C}_1 = \bigcup_{r=1}^\infty B(r), \end{align*} exactly as in the proof of [Sigma-Compactness of $\mathcal{C}_1$](/theorems/2507). Each $B(r)$ is $\chi$-compact by the previous step, so $\mathcal{C}_1$ is a countable union of $\chi$-compact sets — i.e., $(\mathcal{C}_1, \chi)$ is $\sigma$-compact. [/step] [step:Show every non-empty $\chi$-open set contains paths of arbitrarily large tree-reduced length] Let $(\Gamma_n)_{n \in \mathbb{N}}$ be the Lyons-Xu sequence in $\mathcal{C}_1$ from the construction of *axis paths with matching truncated signatures* on page 1290. The construction gives: 1. each $\Gamma_n$ is tree-reduced of length $\|\Gamma_n\|_1 = 2^{n+1}$; 2. $S(\Gamma_n)^{(m)} = 0 \in V^{\otimes m}$ for every $m \in \{0, 1, \dots, n\}$ (with $S(\Gamma_n)^{(0)} = 1 = S(o)^{(0)}$). We first show $[\Gamma_n] \to [o]$ in $\chi$, then transport the resulting unboundedness from $[o]$ to an arbitrary class via the topological group structure. **Convergence in $\chi$ via the projection-wise sequential criterion.** Property (2) of the Lyons–Xu construction states \begin{align*} \pi_m(S(\Gamma_n)) = S(\Gamma_n)^{(m)} = 0 = \pi_m(S(o)) \quad \text{for all } m \in \{0, 1, \dots, n\}, \end{align*} (with $\pi_0(S(\Gamma_n)) = 1 = \pi_0(S(o))$). Hence for each fixed $m \ge 0$, the sequence $(\pi_m(S(\Gamma_n)))_{n \ge 1}$ is eventually constant at $\pi_m(S(o)) = 0$ (or $1$ for $m = 0$), so $\pi_m(S(\Gamma_n)) \to \pi_m(S(o))$ in $V^{\otimes m}$ as $n \to \infty$. By hypothesis (5), this projection-wise convergence implies $S(\Gamma_n) \to S(o)$ in $\tau$. Both $S(\Gamma_n)$ and $S(o) = \mathbf{1}$ lie in $\mathcal{S}_1 \subset \mathcal{S} \subset E$, so the convergence holds in $\tau|_{\mathcal{S}}$; applying the homeomorphism $S^{-1} : (\mathcal{S}, \tau|_{\mathcal{S}}) \to (\mathcal{C}_1, \chi)$ yields $[\Gamma_n] \to [o]$ in $\chi$. **Unboundedness near $[o]$.** Let $U \subseteq \mathcal{C}_1$ be a non-empty $\chi$-open set with $[o] \in U$. Since $[\Gamma_n] \to [o]$ in $\chi$, there exists $N \in \mathbb{N}$ such that $[\Gamma_n] \in U$ for all $n \geq N$. By property (1), $d([\Gamma_n]) := \|\Gamma_n^*\|_1 = \|\Gamma_n\|_1 = 2^{n+1} \to \infty$ (the equality $\Gamma_n^* = \Gamma_n$ holds by the tree-reduced criterion: the Lyons-Xu paths consist of orthogonal axis-aligned segments and are tree-reduced). Hence $\sup_{[y] \in U} d([y]) \geq \sup_{n \geq N} 2^{n+1} = \infty$. **Transport from $[o]$ to an arbitrary class.** Fix $[x] \in \mathcal{C}_1$ and let $W \subseteq \mathcal{C}_1$ be a non-empty $\chi$-open set with $[x] \in W$. We claim $(\mathcal{C}_1, \ast, \chi)$ is a topological group. Indeed, the map $S : (\mathcal{C}_1, \chi) \to (\mathcal{S}, \tau|_{\mathcal{S}})$ is a homeomorphism, and the bijection $S$ intertwines path concatenation $\ast$ on $\mathcal{C}_1$ with tensor-algebra multiplication on $\mathcal{S}$ (Chen's relation: $S(x \ast y) = S(x) \cdot S(y)$), and intertwines path inversion $[x] \mapsto [x^\leftarrow]$ on $\mathcal{C}_1$ with tensor-algebra inversion on $\mathcal{S}$. By hypothesis (6), tensor-algebra multiplication and inversion on $E$ are $\tau$-continuous; restricting to $\mathcal{S} \times \mathcal{S} \to E$ and noting that the image lies in $\mathcal{S}$ (since $\mathcal{S} = \mathcal{S}_1$ is a subgroup of $\mathcal{S}_p$ under tensor multiplication, which is a subgroup of $E$ if $E$ is closed under the operations as required by hypothesis (6)), the group operations on $\mathcal{S}$ are $\tau|_{\mathcal{S}}$-continuous. Pulling back through the homeomorphism $S$, the group operations on $(\mathcal{C}_1, \ast, \chi)$ are $\chi$-continuous, establishing the topological group structure (the same conclusion as the [Topological Group Structure](/theorems/2505) for $\chi_{\mathrm{pr}}$, but now under the strengthened hypothesis (6)). In a topological group, left-multiplication $L_{[x]} : [y] \mapsto [x] \ast [y]$ is a homeomorphism with inverse $L_{[x]^{-1}}$. Define $V := L_{[x]}^{-1}(W) = \{[y] \in \mathcal{C}_1 : [x] \ast [y] \in W\}$; this is $\chi$-open (continuity of $L_{[x]}$) and contains $[o]$ (since $L_{[x]}([o]) = [x] \in W$). By the unboundedness near $[o]$ established in the previous paragraph, there exists a sequence $([y_k]) \subset V$ with $d([y_k]) \to \infty$. Set $[z_k] := [x] \ast [y_k] \in W$. By the sub-additivity of tree-reduced length under concatenation $d([a] \ast [b]) \leq d([a]) + d([b])$ (a geometric fact about tree-reduced representatives, used in the proof of [Tree-Reduced Length Is Unbounded on Open Sets](/theorems/2508): tree-reduction can only shorten a path, so $\|(a \ast b)^*\|_1 \le \|a^* \ast b^*\|_1 = \|a^*\|_1 + \|b^*\|_1$), applied to $[y_k] = [x]^{-1} \ast [z_k]$ with $d([x]^{-1}) = \|x^*\|_1$ (since time-reversal preserves $1$-variation), \begin{align*} d([y_k]) \leq d([x]^{-1}) + d([z_k]) = \|x^*\|_1 + d([z_k]), \end{align*} which rearranges to $d([z_k]) \geq d([y_k]) - \|x^*\|_1$. Since $d([y_k]) \to \infty$ and $\|x^*\|_1$ is a fixed finite constant, $d([z_k]) \to \infty$. Hence $d$ is unbounded on $W$. [guided] The aim is to show that $d : \mathcal{C}_1 \to [0, \infty)$, $[y] \mapsto \|y^*\|_1$ is unbounded on every non-empty $\chi$-open set. The proof is exactly the same as that of [Tree-Reduced Length Is Unbounded on Open Sets](/theorems/2508), once we re-derive the two facts that proof depends on but in the new topology $\chi$: - (a) the Lyons-Xu sequence $(\Gamma_n)$ converges to $[o]$ in $\chi$; - (b) $(\mathcal{C}_1, \ast, \chi)$ is a topological group. **Why (a) holds in $\chi$.** The Lyons–Xu construction (given on page 1290) produces explicit piecewise-linear paths $\Gamma_n \in C_1$ with two key properties: 1. each $\Gamma_n$ is tree-reduced (consecutive segments are orthogonal axis vectors, satisfying the tree-reduced criterion), and $\|\Gamma_n\|_1 = 2^{n+1}$; 2. $S(\Gamma_n)^{(m)} = 0$ for every $m \le n$, while $S(\Gamma_n) \ne \mathbf{1}$. In the canonical topology $\chi_{\mathrm{pr}}$, property (2) is exactly the statement of level-by-level convergence $S(\Gamma_n) \to \mathbf{1} = S(o)$ in the product topology on $T((V))$ — and convergence in $\chi_{\mathrm{pr}}$ is, by definition, convergence in the product topology pulled back through $S$. Hence $[\Gamma_n] \to [o]$ in $\chi_{\mathrm{pr}}$, immediately, for free. For the new topology $\chi$, the situation is one step removed. We need $S(\Gamma_n) \to S(o)$ in $(E, \tau)$, not in the product topology. The bridge is the joint use of hypotheses (4) and (5). *Hypothesis (4)* gives that each $\pi_m : (E, \tau) \to V^{\otimes m}$ is continuous; in particular, $\tau$ refines the subspace topology induced on $E$ from the product topology on $T((V))$. (This is the precise meaning of "$\tau$ is reasonable with respect to the projective product structure": $\tau$ resolves at least as much information as the projection-wise topology.) *Hypothesis (5)* gives the converse direction at the level of sequential convergence: if $\pi_m(z_n) \to \pi_m(z)$ in $V^{\otimes m}$ for every $m \ge 0$, then $z_n \to z$ in $\tau$. (This is the precise replacement for the unjustified "compatible with signature convergence" in informal expositions of this lemma.) Together with metrisability, hypotheses (4) and (5) say that $\tau$-convergent sequences and projection-wise-convergent sequences in $E$ coincide — so $\tau$ has the same convergent sequences as the subspace projective topology on $E$. *Apply both to $S(\Gamma_n)$.* By property (2), for each fixed $m$, $\pi_m(S(\Gamma_n)) = 0 = \pi_m(S(o))$ for all $n \ge m$. Hence $\pi_m(S(\Gamma_n)) \to \pi_m(S(o))$ in $V^{\otimes m}$ for every $m \ge 0$ (the sequence is eventually constant). Hypothesis (5) then gives $S(\Gamma_n) \to S(o)$ in $\tau$, and since $S(\Gamma_n), S(o) \in \mathcal{S} \subset E$, the convergence holds in $\tau|_{\mathcal{S}}$. Pulling back through the homeomorphism $S$: $[\Gamma_n] \to [o]$ in $\chi$. **Why (b) holds in $\chi$.** This is now a direct consequence of hypothesis (6), which postulates exactly that the tensor-algebra multiplication $E \times E \to E$ and inversion $E \to E$ are $\tau$-continuous. Restricting to the subgroup $\mathcal{S} = \mathcal{S}_1$ (closed under multiplication and inversion because $S(x \ast y) = S(x) \cdot S(y)$ by Chen's relation, and $S(x^\leftarrow) = S(x)^{-1}$ by the inverse identity), the operations are $\tau|_{\mathcal{S}}$-continuous on $\mathcal{S} \times \mathcal{S} \to \mathcal{S}$. The bijection $S : (\mathcal{C}_1, \chi) \to (\mathcal{S}, \tau|_{\mathcal{S}})$ is a homeomorphism that intertwines $\ast$ with $\cdot$ and intertwines $[x]^{-1}$ with $S(x)^{-1}$; pulling back, the group operations on $(\mathcal{C}_1, \ast, \chi)$ are $\chi$-continuous. (Without hypothesis (6), this conclusion is genuinely unavailable — refining the topology on the codomain of a continuous bilinear map does not preserve continuity, so multiplication being continuous in the product topology of $T((V))$ does not in general imply continuity in a finer topology $\tau$.) **Putting it together: unboundedness near $[o]$.** Let $U \subseteq \mathcal{C}_1$ be an arbitrary non-empty $\chi$-open neighbourhood of $[o]$. By (a), $[\Gamma_n] \to [o]$ in $\chi$, so by definition of convergence, eventually $[\Gamma_n] \in U$ — say for all $n \geq N$. By property (1), $d([\Gamma_n]) = 2^{n+1} \to \infty$, so $\sup_{[y] \in U} d([y]) \geq \sup_{n \geq N} 2^{n+1} = \infty$. Thus $d$ is unbounded on $U$. **Transport to an arbitrary class via left-multiplication.** Fix $[x] \in \mathcal{C}_1$ and let $W$ be an arbitrary non-empty $\chi$-open set with $[x] \in W$. By (b), the left-multiplication map \begin{align*} L_{[x]} : (\mathcal{C}_1, \chi) &\to (\mathcal{C}_1, \chi) \\ [y] &\mapsto [x] \ast [y] \end{align*} is a homeomorphism, with inverse $L_{[x]^{-1}} = L_{[x^\leftarrow]}$. Pull $W$ back: $V := L_{[x]}^{-1}(W) = \{[y] : [x] \ast [y] \in W\}$ is $\chi$-open (continuity of $L_{[x]}$) and contains $[o]$ (because $L_{[x]}([o]) = [x] \in W$). By the unboundedness-near-$[o]$ statement just proved, $d$ is unbounded on $V$, so there exists $([y_k])_k \subset V$ with $d([y_k]) \to \infty$. Push back: $[z_k] := L_{[x]}([y_k]) = [x] \ast [y_k] \in W$. We need $d([z_k]) \to \infty$. By sub-additivity of $\|\cdot^*\|_1$ under concatenation (a fact internal to the geometry of tree-reduced paths, used in the proof of [Tree-Reduced Length Is Unbounded on Open Sets](/theorems/2508)): \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*} using $d([x]^{-1}) = \|(x^\leftarrow)^*\|_1 = \|x^*\|_1$ (time-reversal preserves $1$-variation and the tree-reduced length is the $1$-variation of the minimal-length representative). Rearranging: \begin{align*} d([z_k]) \geq d([y_k]) - \|x^*\|_1 \to \infty. \end{align*} Hence $d$ is unbounded on $W$, completing the transport from $[o]$ to the arbitrary class $[x]$. [/guided] [/step] [step:Conclude every length ball $B(r)$ has empty $\chi$-interior] Suppose for contradiction that $\operatorname{int}_\chi(B(r)) \neq \varnothing$ for some $r > 0$. Then $U := \operatorname{int}_\chi(B(r))$ is a non-empty $\chi$-open subset of $\mathcal{C}_1$. By definition of interior, $U \subseteq B(r)$, so for every $[y] \in U$, \begin{align*} d([y]) := \|y^*\|_1 \leq r. \end{align*} Thus $d$ is bounded above by $r$ on the non-empty $\chi$-open set $U$ — contradicting the unboundedness step, which establishes that $d$ is unbounded on every non-empty $\chi$-open set. Therefore $\operatorname{int}_\chi(B(r)) = \varnothing$ for every $r > 0$. [/step] [step:Apply the Baire category theorem to rule out complete metrisability and local compactness] We now combine the compactness, $\sigma$-compactness, and empty-interior results into a violation of the Baire property and deduce the conclusions. **$B(r)$ is closed in $\chi$.** $\chi$ is metrisable (by the metrisability transfer in the compactness step) and hence Hausdorff. By the same step, each $B(r)$ is $\chi$-compact. In a Hausdorff space, compact subsets are closed. Hence $B(r)$ is closed in $\chi$ for every $r$. **$\mathcal{C}_1$ is meagre in itself.** Combining the cover $\mathcal{C}_1 = \bigcup_{r=1}^\infty B(r)$ from the $\sigma$-compactness step with closedness (just shown) and empty interior (from the previous step), each $B(r)$ is nowhere dense, and $\mathcal{C}_1$ is a countable union of nowhere-dense closed subsets of itself. **Failure of the Baire property.** Suppose for contradiction that $(\mathcal{C}_1, \chi)$ were a Baire space. By the category form of the Baire property, the countable union $\bigcup_{r=1}^\infty B(r)$ of nowhere-dense closed sets would have empty interior. But $\bigcup_{r=1}^\infty B(r) = \mathcal{C}_1$ has interior equal to $\mathcal{C}_1$ itself, which is non-empty (since $\mathcal{C}_1$ contains at least the class $[o]$). This contradiction shows $(\mathcal{C}_1, \chi)$ is not a Baire space. **Failure of complete metrisability.** By the [Baire Category Theorem (Complete Metric Spaces)](/theorems/1072), every completely metrisable space is a Baire space. The contrapositive: a non-Baire space is not completely metrisable. Hence $(\mathcal{C}_1, \chi)$ is not completely metrisable. Since a Polish space is, by definition, a separable completely metrisable space, $(\mathcal{C}_1, \chi)$ is not Polish. **Failure of local compactness.** By the [Baire Category Theorem (Locally Compact Hausdorff Spaces)](/theorems/1105), every locally compact Hausdorff space is a Baire space. Since $\chi$ is Hausdorff (being metrisable) but not Baire, $(\mathcal{C}_1, \chi)$ is not locally compact. Combined with the $\sigma$-compactness established earlier, this completes the proof. [guided] We have all the ingredients; this step assembles them via the Baire Category Theorem and its contrapositive consequences. **$B(r)$ is closed in $\chi$.** The compactness step established that $(\mathcal{C}_1, \chi)$ is metrisable, hence Hausdorff (every metric space is Hausdorff: distinct points have disjoint open balls of radius $d(x, y)/3$). The same step also established that each $B(r)$ is $\chi$-compact. The standard fact "compact subsets of a Hausdorff space are closed" then applies, yielding $B(r)$ closed in $\chi$. (Sketch of the standard fact: if $K$ is compact and $z \notin K$, Hausdorffness produces, for each $y \in K$, disjoint open neighbourhoods $U_y \ni z$ and $V_y \ni y$; the cover $\{V_y : y \in K\}$ admits a finite subcover $\{V_{y_1}, \dots, V_{y_N}\}$; the intersection $\bigcap_{i=1}^N U_{y_i}$ is an open neighbourhood of $z$ disjoint from $K$, so $K^c$ is open.) **Each $B(r)$ is nowhere dense in $\chi$.** A set $A \subseteq X$ is **nowhere dense** if $\operatorname{int}(\overline{A}) = \varnothing$. For closed $A$, $\overline{A} = A$, so this reduces to $\operatorname{int}(A) = \varnothing$. We have just shown $B(r)$ is closed (hence $\overline{B(r)} = B(r)$), and the empty-interior step shows $\operatorname{int}_\chi(B(r)) = \varnothing$. Hence $B(r)$ is nowhere dense in $(\mathcal{C}_1, \chi)$ for every $r$. **$\mathcal{C}_1$ is a countable union of nowhere-dense closed sets.** The $\sigma$-compactness step gives $\mathcal{C}_1 = \bigcup_{r=1}^\infty B(r)$, a countable union (indexed by $\mathbb{N}$). Each summand is nowhere-dense closed, by the previous paragraph. So $\mathcal{C}_1$ is **meagre** as a subset of itself (a meagre set is a countable union of nowhere-dense sets). **Failure of the Baire property.** Recall the [Baire Category Theorem (Complete Metric Spaces)](/theorems/1072): every complete metric space is a Baire space, where **Baire space** means: every countable union of nowhere-dense closed sets has empty interior. Equivalent forms: - (B1) The intersection of countably many dense open sets is dense. - (B2) The space is not the countable union of nowhere-dense closed sets, unless it is empty. - (B3) Every meagre set has empty interior. Suppose for contradiction that $(\mathcal{C}_1, \chi)$ is a Baire space. By (B1)–(B3), the countable union $\bigcup_{r=1}^\infty B(r)$ of nowhere-dense closed sets has empty interior. But this union equals $\mathcal{C}_1$, whose interior in itself is $\mathcal{C}_1$ — non-empty (it contains at least the equivalence class $[o]$ of the constant zero path). The contradiction shows the assumption was wrong: $(\mathcal{C}_1, \chi)$ is not a Baire space. **Failure of complete metrisability and Polishness.** [Baire Category Theorem (Complete Metric Spaces)](/theorems/1072) states: every completely metrisable space is a Baire space. The contrapositive reads: \begin{align*} \text{If } X \text{ is not a Baire space, then } X \text{ is not completely metrisable.} \end{align*} By the previous paragraph, $(\mathcal{C}_1, \chi)$ is not a Baire space. Hence $(\mathcal{C}_1, \chi)$ is not completely metrisable. By definition, a Polish space is a separable completely metrisable topological space. Since complete metrisability fails, Polishness fails: $(\mathcal{C}_1, \chi)$ is not Polish. **Failure of local compactness.** A topological space is **locally compact Hausdorff** if it is Hausdorff and every point has a compact neighbourhood. By the [Baire Category Theorem (Locally Compact Hausdorff Spaces)](/theorems/1105): every locally compact Hausdorff space is a Baire space. The contrapositive: \begin{align*} \text{If } X \text{ is Hausdorff but not Baire, then } X \text{ is not locally compact.} \end{align*} $(\mathcal{C}_1, \chi)$ is Hausdorff (metrisable) and not Baire (above). Hence $(\mathcal{C}_1, \chi)$ is not locally compact. **Why is the Baire Category Theorem the right tool?** The Baire property is a kind of topological "size" axiom: a Baire space is too "thick" to be exhausted by countably many nowhere-dense pieces. Both completeness (in a metric sense) and local compactness furnish enough rigidity to enforce this thickness — the first via Cantor's nested-set theorem, the second via finite intersection arguments on compact closures. So both are sufficient conditions for being Baire, and a single counter-witness — the explicit covering $\mathcal{C}_1 = \bigcup_r B(r)$ by closed nowhere-dense sets — rules out both at once. **Why does $\sigma$-compactness coexist with the failure of all these properties?** $\sigma$-compactness is a much weaker covering property: $\mathbb{Q}$ is $\sigma$-compact (it is countable, hence a countable union of singletons, each compact) but is neither completely metrisable, nor locally compact, nor Polish, nor a Baire space. The space $(\mathcal{C}_1, \chi)$ behaves analogously: it has just enough internal structure to be exhausted by compact sets, but not enough to satisfy any of the stronger regularity properties. The cover by compact sets is the same cover that, viewed differently, exhibits $\mathcal{C}_1$ as meagre in itself — the very same length balls that compactify $\mathcal{C}_1$ also have empty interior, and these two facts pull in opposite directions. **Why-not metrisability per se?** The proof is silent on whether $(\mathcal{C}_1, \chi)$ might fail to be metrisable in some pathological case. The hypotheses guarantee metrisability (established in the compactness-transfer step), so this is not at issue. The failure is **specifically** at the level of complete metrisability: there is no complete metric inducing $\chi$. The same holds for $\chi_{\mathrm{pr}}$ (the canonical product topology), as proved in [Non-Polishness of $(\mathcal{C}_1, \chi_{\mathrm{pr}})$](/theorems/2510); the present lemma generalizes that result by showing the obstruction is universal across the family of metrisable topologies on $\mathcal{C}_1$ compatible with signature continuity. [/guided] [/step]