[proofplan] We exhibit $\mathcal{C}_1$ as a countable union of closed sets with empty interior, contradicting the Baire property. The countable cover comes from the length balls $\mathcal{C}_1 = \bigcup_{r=1}^\infty B(r)$. Each $B(r)$ is compact (by the [Compactness of Length Balls](/theorems/???)) and hence closed because $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ is Hausdorff. By [Tree-Reduced Length Is Unbounded on Open Sets](/theorems/2508), each $B(r)$ has empty interior. Failure of the Baire property then rules out complete metrisability — and with it, Polishness — via the [Baire Category Theorem](/theorems/???). [/proofplan]

[step:Cover $\mathcal{C}_1$ by the countable family of length balls $B(r)$]For $r \in \mathbb{N}$, define the length ball \begin{align*} B(r) := \{[x] \in \mathcal{C}_1 : \|x^*\|_1 \leq r\}, \end{align*} where $x^*$ is the tree-reduced representative of $[x]$. By the [Sigma-Compactness of $\mathcal{C}_1$](/theorems/2507), \begin{align*} \mathcal{C}_1 = \bigcup_{r=1}^\infty B(r), \end{align*} exhibiting $\mathcal{C}_1$ as a countable union of length balls.[/step]

[guided]The strategy is a Baire-category argument by contradiction: we will assume $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ is Polish (in particular, completely metrisable, hence Baire), then exhibit a countable cover by closed nowhere-dense sets, contradicting the Baire property. The first ingredient is the cover. Define for $r \in \mathbb{N}$, \begin{align*} B(r) := \{[x] \in \mathcal{C}_1 : \|x^*\|_1 \leq r\}, \end{align*} the closed length ball of radius $r$ as measured by the tree-reduced representative. Why this particular family? Because we already have two structural results about it: - (a) $\mathcal{C}_1 = \bigcup_{r=1}^\infty B(r)$ — countable cover (proved in [Sigma-Compactness of $\mathcal{C}_1$](/theorems/2507)). - (b) Each $B(r)$ is $\chi_{\mathrm{pr}}$-compact (proved in [Compactness of Length Balls](/theorems/???) and reused in /theorems/2507). To verify the hypotheses of [Sigma-Compactness of $\mathcal{C}_1$](/theorems/2507): that theorem requires only that $\mathcal{C}_1$ be the space of tree-equivalence classes of paths of finite one-variation, which is the standing setup. Its conclusion gives the cover above. The cover is **countable** because indexed by $\mathbb{N}$, and the cardinality of $\mathbb{N}$ is $\aleph_0$. The countable cover is the first leg of the Baire-category contradiction; the next two steps establish that each $B(r)$ is closed and has empty interior — i.e., each is nowhere-dense.[/guided]

[step:Show each $B(r)$ is closed in $\chi_{\mathrm{pr}}$]By the [Compactness of Length Balls in $(\mathcal{C}_1, \chi_{\mathrm{pr}})$](/theorems/???), the set $B(r)$ is compact for every $r \in \mathbb{N}$. We claim $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ is Hausdorff: $\chi_{\mathrm{pr}}$ is the initial topology induced by the family of signature-projection maps \begin{align*} \pi_m: \mathcal{C}_1 &\to V^{\otimes m} \\ [x] &\mapsto S(x)^{(m)}, \end{align*} for $m \geq 0$. Each $V^{\otimes m}$ is a finite-dimensional normed space, hence Hausdorff. The family $(\pi_m)_{m \geq 0}$ separates points of $\mathcal{C}_1$ because two paths defining the same equivalence class in $\mathcal{C}_1$ have identical signatures by the [Uniqueness of the Signature for Tree-Reduced Paths](/theorems/???). The initial topology induced by a separating family of continuous maps to Hausdorff spaces is itself Hausdorff. Therefore $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ is Hausdorff. In a Hausdorff space, every compact set is closed. Hence $B(r)$ is closed for every $r \in \mathbb{N}$.[/step]

[guided]We need each $B(r)$ to be **closed** — a topological condition. The standard argument is "compact + Hausdorff $\Rightarrow$ closed", but neither side is automatic here. Compactness was established by citation; we now need Hausdorffness of $(\mathcal{C}_1, \chi_{\mathrm{pr}})$. **Hausdorffness of $(\mathcal{C}_1, \chi_{\mathrm{pr}})$.** Recall that $\chi_{\mathrm{pr}}$ is the **initial topology** induced by the family of signature-projection maps \begin{align*} \pi_m: \mathcal{C}_1 &\to V^{\otimes m} \\ [x] &\mapsto S(x)^{(m)}, \end{align*} indexed by $m \ge 0$. By the standard characterization of initial topologies, $\chi_{\mathrm{pr}}$ is the coarsest topology making every $\pi_m$ continuous. We use a basic fact: if $(f_i)_{i \in I}$ is a separating family of continuous maps $X \to Y_i$ where each $Y_i$ is Hausdorff, then the initial topology on $X$ induced by $(f_i)$ is Hausdorff. We verify both parts of this hypothesis: - *Each $V^{\otimes m}$ is Hausdorff.* $V$ is finite-dimensional and normed, so $V^{\otimes m}$ — also finite-dimensional and normed (with the projective tensor norm) — inherits the Hausdorff property; every normed space is Hausdorff (distinct points have disjoint open balls). - *The family $(\pi_m)_{m \ge 0}$ separates points.* Suppose $[x] \ne [y]$ in $\mathcal{C}_1$. By [Uniqueness of the Signature for Tree-Reduced Paths](/theorems/???), the signature determines a tree-equivalence class: $S(x^*) = S(y^*)$ implies $[x] = [y]$. Contrapositively, $[x] \ne [y]$ implies $S(x^*) \ne S(y^*)$, i.e. there is some level $m$ at which $S(x^*)^{(m)} \ne S(y^*)^{(m)}$, so $\pi_m([x]) \ne \pi_m([y])$. Thus the family separates points. Both conditions hold, so $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ is Hausdorff. **Compact $\Rightarrow$ closed in Hausdorff.** This is a standard topology fact: in a Hausdorff space, every compact subset is closed. (Sketch: if $K$ is compact and $x \notin K$, Hausdorffness produces, for each $y \in K$, disjoint open neighborhoods $U_y \ni x$ and $V_y \ni y$. The cover $\{V_y\}$ of $K$ admits a finite subcover $\{V_{y_1}, \dots, V_{y_n}\}$; the intersection $U_{y_1} \cap \dots \cap U_{y_n}$ is an open neighborhood of $x$ disjoint from $K$. Hence $K^c$ is open.) Applying this with $K = B(r)$ (compact by Step 3 of /theorems/2507, which we cite) gives $B(r)$ closed for every $r \in \mathbb{N}$. Why does Hausdorffness require this much work? In an arbitrary topological space, compact sets need not be closed (consider the indiscrete topology, where every set is compact but only $\varnothing$ and $X$ are closed). The Hausdorff hypothesis is the gateway to converting the compactness from Step 1 into the closedness needed for the Baire-category argument in Step 4. Why do we need closedness specifically? In the Baire argument, we will write $\mathcal{C}_1$ as a countable union of nowhere-dense sets. "Nowhere-dense" means "the closure has empty interior". For closed sets, nowhere-dense reduces to "empty interior". Closedness here is the simplification that lets us check nowhere-density in Step 3 by checking only emptiness of the interior.[/guided]

[step:Show each $B(r)$ has empty interior]Suppose for contradiction that $\operatorname{int}(B(r))$ is non-empty for some $r \in \mathbb{N}$. Then $\operatorname{int}(B(r))$ is a non-empty open subset of $\mathcal{C}_1$. The function \begin{align*} d: \mathcal{C}_1 &\to [0, \infty) \\ [y] &\mapsto \|y^*\|_1 \end{align*} satisfies $d([y]) \leq r$ for every $[y] \in B(r) \supseteq \operatorname{int}(B(r))$, so $d$ is bounded by $r$ on the non-empty open set $\operatorname{int}(B(r))$. This contradicts the [Tree-Reduced Length Is Unbounded on Open Sets](/theorems/2508), which states that $d$ is unbounded on every non-empty open set. Therefore $\operatorname{int}(B(r)) = \varnothing$ for every $r \in \mathbb{N}$.[/step]

[guided]We need each $B(r)$ to have empty interior in $\chi_{\mathrm{pr}}$. This is the **deep** part of the proof — and the place where the path-space character of $\mathcal{C}_1$ makes itself felt. Geometrically, the claim is: every $\chi_{\mathrm{pr}}$-open set contains paths of arbitrarily large length. Topologically, this is a non-meagreness-style claim about the projective topology. **Setup of the contradiction.** Suppose for contradiction that $\operatorname{int}(B(r)) \ne \varnothing$ for some $r \in \mathbb{N}$. Then $U := \operatorname{int}(B(r))$ is a non-empty open subset of $\mathcal{C}_1$ in the topology $\chi_{\mathrm{pr}}$. By definition of interior, $U \subseteq B(r)$, so every $[y] \in U$ satisfies $\|y^*\|_1 \le r$. **Tree-reduced length as a function.** Define \begin{align*} d: \mathcal{C}_1 &\to [0, \infty) \\ [y] &\mapsto \|y^*\|_1. \end{align*} The map $d$ assigns to each equivalence class the one-variation of its tree-reduced representative. By the previous paragraph, $d \le r$ on $U$, so $d$ is bounded above by $r$ on the non-empty open set $U$. **Contradiction.** This contradicts the [Tree-Reduced Length Is Unbounded on Open Sets](/theorems/2508), whose statement is: for every non-empty open set $U \subseteq \mathcal{C}_1$ (in $\chi_{\mathrm{pr}}$), the function $d: U \to [0, \infty)$ is unbounded — i.e., for every $M > 0$, there exists $[y] \in U$ with $d([y]) > M$. The hypothesis "$U$ non-empty open in $\chi_{\mathrm{pr}}$" is exactly satisfied by our $U = \operatorname{int}(B(r))$. The conclusion then asserts $d$ is unbounded on $U$, contradicting our derivation that $d \le r$ on $U$. The contradiction forces $\operatorname{int}(B(r)) = \varnothing$ for every $r \in \mathbb{N}$. **Why is /theorems/2508 the engine?** It says the projective topology is, in a precise sense, "blind" to length: any open set in $\chi_{\mathrm{pr}}$ — no matter how small in terms of nearby signatures — contains paths of arbitrarily large length. The intuition is that the projective topology controls each signature level $\pi_m([x]) = S(x)^{(m)} \in V^{\otimes m}$ only up to bounded error, and a single signature truncation $(S(x)^{(0)}, \dots, S(x)^{(N)})$ has a well-defined image in $V^{\otimes 0} \times \cdots \times V^{\otimes N}$, but the preimage in $\mathcal{C}_1$ is *huge* — it contains paths of all lengths. So bounding the truncation cannot bound the length. Why-not: One might hope to obtain the empty-interior conclusion by a more direct counting argument (e.g. dimension-type). This fails because $\chi_{\mathrm{pr}}$ is not metrisable in any obvious way and lacks dimension-type invariants. The mechanism in /theorems/2508 — explicit construction of arbitrarily long paths inside any open set, via concatenation with tree-like loops that don't perturb the signature truncation — is essential. Connection to technique: This is the standard "an invariant grows without bound on every open set, hence sublevel sets are nowhere-dense" argument. It is also used to prove the non-meagreness of the rationals as a subset of $\mathbb{R}$, and analogous statements for measure-bounded sets in non-locally-compact spaces.[/guided]

[step:Conclude that $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ fails the Baire property]Recall the [Definition of Baire Space](/page/Baire%20Space): a topological space $X$ is a Baire space if and only if every countable union of closed sets with empty interior has empty interior. Equivalently (taking complements), a countable intersection of dense open sets is dense, or — in the form we use — $X$ cannot be written as a countable union of nowhere-dense closed sets unless one of those sets has non-empty interior. By Steps 1, 2, 3, the family $\{B(r)\}_{r \in \mathbb{N}}$ is a countable cover of $\mathcal{C}_1$ by closed sets each with empty interior. The whole space $\mathcal{C}_1$ has non-empty interior (it equals itself, which is open). Therefore $\mathcal{C}_1$ is the union of countably many closed sets of empty interior, yet has non-empty interior — violating the Baire property.[/step]

[guided]We assemble the previous three steps into a violation of the Baire property. **Definitions.** The [Definition of Baire Space](/page/Baire%20Space) is: a topological space $X$ is **Baire** iff every countable union of closed sets with empty interior has empty interior. Equivalent reformulations: - (B1) Every countable union of nowhere-dense sets has empty interior. - (B2) Every countable intersection of dense open sets is dense. - (B3) $X$ is non-meagre as a subset of itself, i.e. $X$ cannot be written as a countable union of nowhere-dense subsets. A set $A \subseteq X$ is **nowhere-dense** if $\operatorname{int}(\overline{A}) = \varnothing$. For closed sets, $\overline{A} = A$, so nowhere-dense reduces to "empty interior". A set is **meagre** (or "of first category") if it is a countable union of nowhere-dense sets. **Assembling the contradiction.** From the previous three steps: - (Step 1) $\mathcal{C}_1 = \bigcup_{r=1}^\infty B(r)$ — countable cover. - (Step 2) Each $B(r)$ is closed. - (Step 3) Each $B(r)$ has empty interior. Combining (Step 2) and (Step 3): each $B(r)$ is closed with empty interior, hence nowhere-dense (its closure equals itself, with empty interior). Combining with (Step 1): $\mathcal{C}_1$ is a countable union of nowhere-dense subsets of itself, hence meagre as a subset of itself. **Why is this incompatible with the Baire property?** Suppose, for contradiction, that $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ were Baire. Apply form (B1) to the countable cover $\bigcup_{r=1}^\infty B(r)$: since each $B(r)$ is closed with empty interior, (B1) gives $\operatorname{int}(\bigcup_{r=1}^\infty B(r)) = \varnothing$. But $\bigcup_{r=1}^\infty B(r) = \mathcal{C}_1$, so $\operatorname{int}(\mathcal{C}_1) = \varnothing$. This contradicts the fact that $\mathcal{C}_1$ is open in itself (the whole space is always open and equals its own interior, so $\operatorname{int}(\mathcal{C}_1) = \mathcal{C}_1 \ne \varnothing$ — recalling we are working in a non-empty space, since $\mathcal{C}_1$ contains at least the constant path). The contradiction forces $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ to fail the Baire property. Equivalently in form (B3): we have written $\mathcal{C}_1$ as a countable union of nowhere-dense subsets of itself, so $\mathcal{C}_1$ is meagre in itself, i.e. not Baire. **Why this style of argument?** The pattern "exhibit an explicit countable cover by closed nowhere-dense sets" is the standard technique for showing a space fails to be Baire. The non-trivial input is Step 3 — exhibiting an invariant ($d = \|x^*\|_1$) that is unbounded on every open set, so its sublevel sets are all nowhere-dense. Why-not: One might attempt to show non-Bairess directly via form (B2) by exhibiting a countable family of dense open sets with non-dense intersection. This is harder in practice because constructing dense open sets requires a positive description, whereas $B(r)$ admits a clean negative description (sublevel sets of an unbounded invariant). The cover-by-nowhere-dense form (B1)/(B3) is the workhorse here.[/guided]

[step:Conclude failure of complete metrisability and Polishness]By the [Baire Category Theorem](/theorems/???), every completely metrisable space is a Baire space. Equivalently, in contrapositive, a non-Baire space cannot be completely metrisable. By Step 4, $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ is not a Baire space. Therefore $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ is not completely metrisable. By the [Definition of Polish Space](/page/Polish%20Space), a Polish space is a separable completely metrisable topological space. Since complete metrisability fails, Polishness fails as well, completing the proof.[/step]

[guided]We invert the chain "Polish $\Rightarrow$ completely metrisable $\Rightarrow$ Baire" via two contrapositives. **Contrapositive of the Baire Category Theorem.** The [Baire Category Theorem](/theorems/???) has several equivalent forms; the one we cite is: \begin{align*} \text{Every completely metrisable space is a Baire space.} \end{align*} The hypothesis is "completely metrisable" — i.e., the topology is induced by a metric under which the space is complete. The conclusion is the Baire property. We do not need the version of Baire for locally compact Hausdorff spaces; the version for completely metrisable spaces is sufficient. The contrapositive reads: \begin{align*} \text{If $X$ is not Baire, then $X$ is not completely metrisable.} \end{align*} Step 4 established that $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ is not Baire. Applying the contrapositive: $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ is not completely metrisable. **Contrapositive of "Polish implies completely metrisable".** By the [Definition of Polish Space](/page/Polish%20Space), a topological space $(X, \tau)$ is **Polish** iff it is (a) separable, (b) metrisable, and (c) completely metrisable — equivalently, iff there exists a separable complete metric inducing $\tau$. In particular, every Polish space is completely metrisable, by definition. The contrapositive reads: \begin{align*} \text{If $X$ is not completely metrisable, then $X$ is not Polish.} \end{align*} We have just shown $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ is not completely metrisable. Therefore $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ is not Polish. This is precisely the conclusion of the theorem. **Why does Polishness fail via this route?** The strategy was a Baire-category argument. The Baire Category Theorem is the bridge: it says complete-metric topology is rigid enough that meagre sets cannot fill up the whole space. We exhibited an explicit decomposition into countably many nowhere-dense pieces — incompatible with any complete metric. The other ingredients of Polishness (separability, metrisability) are not at issue here; the failure is specifically in **complete** metrisability. Why-not: One might attempt to show $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ is not metrisable at all (which would also imply not Polish). This would actually require a different argument and may not even be true — the projective topology can be metrisable without being completely metrisable. The Baire-category approach gives the sharpest available statement: complete metrisability fails, but the proof is silent on whether metrisability per se fails. Connection to technique: The "non-Baire $\Rightarrow$ non-Polish" pattern is the workhorse for proving non-Polishness in functional analysis. Examples include: $C^\infty(U)$ in the topology of pointwise convergence, the space of distributions $\mathcal{D}'(\Omega)$ in the strong topology, and now $(\mathcal{C}_1, \chi_{\mathrm{pr}})$.[/guided]