[proofplan] The strategy is a transfer-and-exhaust argument. The signature map identifies $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ with its image $\mathcal{S}_1$ inside the countable product $\Pi := \prod_{i=0}^\infty V^{\otimes i}$, which is Polish. The Lusin property is preserved by homeomorphism, so it suffices to show $\mathcal{S}_1 \subset \Pi$ is a Borel subset of a Polish space — equivalently, a Lusin subspace. The Borelness of $\mathcal{S}_1$ is exhibited explicitly: the length balls $S(B(r)) \subset \Pi$ are continuous images of compact sets, hence compact, hence closed in the Hausdorff space $\Pi$, hence Borel; and $\mathcal{S}_1$ is the countable union of these balls. [/proofplan]

[step:Reduce the Lusin property of $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ to Borelness of $\mathcal{S}_1$ inside the Polish ambient product]Define the ambient product space \begin{align*} \Pi := \prod_{i=0}^\infty V^{\otimes i}, \end{align*} equipped with the product topology, where each factor $V^{\otimes i}$ carries its unique finite-dimensional Hausdorff vector-space topology. By the proof of [Hausdorff, Separability, and Metrisability](/theorems/2504) (specifically Steps 3 and 4 of that proof), each $V^{\otimes i}$ is a [Polish space](/page/Polish%20Space), and the countable product $\Pi$ of Polish spaces is itself Polish. Let $\mathcal{S}_1 := S(\mathcal{C}_1) \subset \Pi$ denote the signature image. By definition of $\chi_{\mathrm{pr}}$, the signature map \begin{align*} S : (\mathcal{C}_1, \chi_{\mathrm{pr}}) &\to \mathcal{S}_1 \\ [x] &\mapsto S(x) \end{align*} is a homeomorphism onto $\mathcal{S}_1$ equipped with the subspace topology inherited from $\Pi$. The Lusin property is preserved by homeomorphism. We invoke the standard descriptive-set-theoretic fact: *every Borel subset of a Polish space is a Lusin space*. It therefore suffices to show that $\mathcal{S}_1$ is a Borel subset of $\Pi$; given this, $\mathcal{S}_1$ is Lusin in its subspace topology, and the homeomorphism $S$ transfers the property to $(\mathcal{C}_1, \chi_{\mathrm{pr}})$.[/step]

[guided]The Lusin property is a descriptive-set-theoretic regularity condition: a topological space $Y$ is Lusin if there exist a Polish space $P$ and a homeomorphism of $Y$ onto a Borel subset of $P$. The plan is to take $P = \Pi$, the ambient product, and exhibit $\mathcal{C}_1$ as Borel inside it via the signature map. **The ambient space is Polish.** Recall the construction. Each tensor power $V^{\otimes i}$ is a finite-dimensional real vector space (of dimension $(\dim V)^i$). All norms on a finite-dimensional real vector space induce the same topology, and that topology is the Euclidean topology — separable, completely metrisable, hence Polish. The countable product \begin{align*} \Pi := \prod_{i=0}^\infty V^{\otimes i} \end{align*} of Polish spaces is Polish in the product topology: a complete compatible metric is the standard weighted-supremum \begin{align*} d_\Pi\bigl((a_i)_{i\ge 0}, (b_i)_{i\ge 0}\bigr) := \sum_{i=0}^\infty 2^{-i} \min(d_i(a_i, b_i), 1), \end{align*} where $d_i$ is any complete metric on $V^{\otimes i}$, and a countable dense subset is built coordinate-wise. (See [Hausdorff, Separability, and Metrisability](/theorems/2504), Steps 3-4, for the full verification.) **The signature map is a homeomorphism.** By definition of $\chi_{\mathrm{pr}}$, the topology on $\mathcal{C}_1$ is precisely the pullback under $S$ of the subspace topology on $\mathcal{S}_1 \subset \Pi$. So \begin{align*} S : (\mathcal{C}_1, \chi_{\mathrm{pr}}) \to (\mathcal{S}_1, \text{subspace topology}) \end{align*} is a homeomorphism by construction. **Reduction.** The Lusin property is purely topological, so it transfers across homeomorphisms. We may therefore replace the question "is $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ Lusin?" by "is $\mathcal{S}_1$, with its subspace topology in $\Pi$, Lusin?" The standard fact from descriptive set theory is: **Fact.** If $P$ is Polish and $A \subset P$ is Borel, then $A$ in its subspace topology is a Lusin space. (Proof sketch: a Polish space embeds as a $G_\delta$ subset of a compact metric space — for instance the Hilbert cube $[0,1]^\mathbb{N}$. A Borel subset of a Polish space is Borel inside that compact metric ambient. So $A$ is homeomorphic to a Borel subset of a compact metric space, which is the original definition of Lusin.) Applying this with $P = \Pi$ and $A = \mathcal{S}_1$: it suffices to prove that $\mathcal{S}_1$ is a Borel subset of $\Pi$. The remainder of the proof is dedicated to exhibiting this Borel structure explicitly via the length balls. **Why the explicit exhibition matters.** One might be tempted to argue abstractly that $\mathcal{S}_1$ is Borel because it is the continuous image of a Polish space (some other description of $C_{0,1}(V)$). But continuous images of Borel sets need not be Borel — they are *analytic*, a strictly larger class. So we must exhibit $\mathcal{S}_1$ as a Borel set in a constructive way. The next steps do this by writing $\mathcal{S}_1$ as a countable union of closed sets (an $F_\sigma$ set), which is automatically Borel.[/guided]

[step:Identify the length balls $B(r)$ as compact subsets of the ambient product space via continuity of the inclusion]For each $r > 0$, recall the length ball in $\mathcal{C}_1$: \begin{align*} B(r) := \bigl\{ [x] \in \mathcal{C}_1 : \|x^*\|_1 \leq r \bigr\}, \end{align*} where $x^*$ is the tree-reduced representative of $[x]$. Define the inclusion map \begin{align*} \iota : (\mathcal{S}_1, \text{subspace topology}) &\hookrightarrow \Pi \\ S(x) &\mapsto S(x). \end{align*} By definition of the subspace topology, $\iota$ is continuous (in fact, a topological embedding). By [Length Balls Are Compact](/theorems/2506) applied with $p = 1$, each $B(r)$ is compact in $(\mathcal{C}_1, \chi_{\mathrm{pr}})$. Continuous images of compact sets are compact, so $S(B(r)) \subset \mathcal{S}_1$ is compact in the subspace topology, and $(\iota \circ S)(B(r)) = S(B(r))$ is compact when viewed as a subset of $\Pi$. Since $\Pi$ is Hausdorff (as a product of Hausdorff spaces, or directly because $\Pi$ is metrisable by the previous step), every compact subset of $\Pi$ is closed in $\Pi$. Therefore $S(B(r))$ is closed in $\Pi$ for every $r > 0$, and in particular Borel.[/step]

[guided]The goal of this step is to show that each set $S(B(r)) \subset \Pi$ is closed in the ambient product, which makes it Borel. The chain of reductions is: \begin{align*} B(r) \text{ compact in } \mathcal{C}_1 &\implies S(B(r)) \text{ compact in } \Pi \\ &\implies S(B(r)) \text{ closed in } \Pi \\ &\implies S(B(r)) \text{ Borel in } \Pi. \end{align*} **First link: compactness of $B(r)$ in $\mathcal{C}_1$.** This is the substantive fact. We invoke [Length Balls Are Compact](/theorems/2506), which states that for $p \in [1,2)$ and $r > 0$, the set $B(r) := \{[x] \in \mathcal{C}_p : \|x^*\|_p \le r\}$ is compact in $(\mathcal{C}_p, \chi_{\mathrm{pr}})$. We verify the hypotheses: we are working with $p = 1 \in [1,2)$ and $r > 0$, so the theorem applies and $B(r)$ is $\chi_{\mathrm{pr}}$-compact in $\mathcal{C}_1$. **Second link: pushing compactness to $\Pi$.** The signature map $S : (\mathcal{C}_1, \chi_{\mathrm{pr}}) \to \mathcal{S}_1$ is a homeomorphism, hence continuous. The inclusion \begin{align*} \iota : \mathcal{S}_1 &\to \Pi \end{align*} is continuous by definition of the subspace topology — $\iota^{-1}(U) = U \cap \mathcal{S}_1$ is open in $\mathcal{S}_1$ for every $U$ open in $\Pi$. The composition \begin{align*} \iota \circ S : (\mathcal{C}_1, \chi_{\mathrm{pr}}) &\to \Pi \end{align*} is therefore continuous. The standard topological fact "continuous images of compact sets are compact" — verified by checking that any open cover of the image pulls back to an open cover of the domain, which has a finite subcover — gives that \begin{align*} (\iota \circ S)(B(r)) = S(B(r)) \subset \Pi \end{align*} is compact in $\Pi$. **Third link: compact implies closed in a Hausdorff space.** This is a standard fact: in a Hausdorff space, points can be separated from compact sets by disjoint open neighbourhoods, which lets one show the complement of a compact set is open. We verify $\Pi$ is Hausdorff: it was shown Polish in Step 1, and Polish spaces are metrisable, and metrisable spaces are Hausdorff. Therefore $S(B(r))$, being compact in the Hausdorff space $\Pi$, is closed in $\Pi$. **Fourth link: closed implies Borel.** The Borel $\sigma$-algebra $\mathcal{B}(\Pi)$ is by definition the $\sigma$-algebra generated by the open sets, which (taking complements) contains every closed set. So $S(B(r)) \in \mathcal{B}(\Pi)$. **Why this argument is delicate.** The compactness of $B(r)$ inside $\mathcal{C}_1$ does not automatically give compactness in any other space — compactness depends on the topology. The argument works because $S$ is a *continuous* map (in fact a homeomorphism onto its image), so it sends compact sets to compact sets. If the topology on $\mathcal{C}_1$ had been finer than $\chi_{\mathrm{pr}}$ — for example, the $1$-variation topology — then $B(r)$ would generally fail to be compact (it is a closed bounded set in an infinite-dimensional Banach space, where Heine-Borel fails). The compactness of $B(r)$ in $\chi_{\mathrm{pr}}$ is what powers the entire argument and is the content of Theorem 2506.[/guided]

[step:Express $\mathcal{S}_1$ as a countable union of closed sets and conclude it is Borel]By [$\sigma$-Compactness of $\mathcal{C}_1$](/theorems/2507), we have the countable cover \begin{align*} \mathcal{C}_1 = \bigcup_{r=1}^\infty B(r), \end{align*} where the union runs over positive integers $r$. Applying the bijection $S$ to both sides preserves arbitrary unions, so \begin{align*} \mathcal{S}_1 = S(\mathcal{C}_1) = S\Bigl(\bigcup_{r=1}^\infty B(r)\Bigr) = \bigcup_{r=1}^\infty S(B(r)). \end{align*} By Step 2, each $S(B(r))$ is closed in $\Pi$. The countable union of closed subsets of $\Pi$ is an $F_\sigma$ subset of $\Pi$, hence belongs to the Borel $\sigma$-algebra $\mathcal{B}(\Pi)$. Therefore $\mathcal{S}_1 \in \mathcal{B}(\Pi)$.[/step]

[guided]We now combine the previous step (each $S(B(r))$ is closed) with the structural fact that $\mathcal{C}_1$ is $\sigma$-compact to write $\mathcal{S}_1$ as a countable union of closed sets in $\Pi$. **Setting up the union.** By [$\sigma$-Compactness of $\mathcal{C}_1$](/theorems/2507), the space $\mathcal{C}_1$ admits the explicit cover \begin{align*} \mathcal{C}_1 = \bigcup_{r=1}^\infty B(r), \end{align*} indexed by the positive integers. We verify the hypothesis of Theorem 2507: it applies to $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ unconditionally, with conclusion the displayed identity. **Pushing forward through $S$.** The signature map $S : \mathcal{C}_1 \to \mathcal{S}_1$ is a bijection (injective by the [signature uniqueness theorem](/theorems/2502), surjective onto $\mathcal{S}_1 := S(\mathcal{C}_1)$ by definition). Bijections preserve arbitrary unions: \begin{align*} S\Bigl(\bigcup_{r=1}^\infty B(r)\Bigr) = \bigcup_{r=1}^\infty S(B(r)). \end{align*} (The forward inclusion is immediate; the reverse uses surjectivity of $S$ onto its image, which we have.) Therefore \begin{align*} \mathcal{S}_1 = \bigcup_{r=1}^\infty S(B(r)) \subset \Pi. \end{align*} **From closed sets to Borel.** By Step 2, each set $S(B(r))$ is closed in $\Pi$. A countable union of closed sets is, by definition, an *$F_\sigma$ set*. The Borel $\sigma$-algebra $\mathcal{B}(\Pi)$ is closed under countable unions and contains all closed sets, hence contains all $F_\sigma$ sets: \begin{align*} F_\sigma \subset \mathcal{B}(\Pi). \end{align*} We conclude $\mathcal{S}_1 \in \mathcal{B}(\Pi)$, i.e., $\mathcal{S}_1$ is a Borel subset of the Polish space $\Pi$. **Why the countable index matters.** The $\sigma$-algebra generated by the closed sets is closed under *countable* (not arbitrary) unions, so the fact that $\mathcal{C}_1$ is *$\sigma$-compact* — covered by *countably many* compacta — is essential. An uncountable union of closed sets need not be Borel; for example, $\mathbb{R} = \bigcup_{x \in \mathbb{R}} \{x\}$ writes the Borel set $\mathbb{R}$ as an uncountable union of closed sets, but a non-Borel subset like a Vitali set is also an uncountable union of closed singletons, so the structure is preserved trivially in that direction but no analogous bound holds for Borel structure.[/guided]

[step:Transport the Borel embedding back to $\mathcal{C}_1$ to obtain the Lusin property] Combining the previous steps: $\mathcal{S}_1$ is a Borel subset of the Polish space $\Pi$ (Step 3), so $\mathcal{S}_1$ in its subspace topology is a Lusin space (the standard descriptive-set-theoretic fact invoked in Step 1). The signature map \begin{align*} S : (\mathcal{C}_1, \chi_{\mathrm{pr}}) \to (\mathcal{S}_1, \text{subspace topology}) \end{align*} is a homeomorphism by the definition of $\chi_{\mathrm{pr}}$. The Lusin property is preserved by homeomorphism, so $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ is a Lusin space, completing the proof. [/step]