[proofplan] The argument splits into two independent parts. For the Hausdorff property we exploit point separation by the family $\mathcal{A}$ of continuous coordinate functionals: given two distinct signature classes, a single continuous functional with different values at them produces disjoint open neighbourhoods. For separability and metrisability we transfer the question to the signature image $\mathcal{S}_p$ inside the [Polish](/theorems/???) countable product $\prod_{i \geq 0} V^{\otimes i}$ — every $V^{\otimes i}$ is finite-dimensional real and hence Polish, and a countable product of Polish spaces is Polish. Subspaces of Polish spaces are separable and metrisable, and $(\mathcal{C}_p, \chi_{\mathrm{pr}})$ is homeomorphic to $\mathcal{S}_p$ via the signature map. [/proofplan] [step:Establish the Hausdorff property via point separation by continuous functionals] Let $[x], [y] \in \mathcal{C}_p$ with $[x] \neq [y]$. By the [point-separation property of $\mathcal{A}$](/theorems/???), there exists $f \in T((V))^*$ such that \begin{align*} \Phi_f([x]) \neq \Phi_f([y]). \end{align*} Set $a := \Phi_f([x])$, $b := \Phi_f([y])$, so $a \neq b$ in $\mathbb{R}$. The Euclidean topology on $\mathbb{R}$ is Hausdorff, so there exist disjoint open sets $U_a, U_b \subseteq \mathbb{R}$ with $a \in U_a$ and $b \in U_b$. By definition of $\chi_{\mathrm{pr}}$, every $\Phi_f$ for $f \in T((V))^*$ is $\chi_{\mathrm{pr}}$-continuous. Define \begin{align*} W_x &:= \Phi_f^{-1}(U_a), \\ W_y &:= \Phi_f^{-1}(U_b). \end{align*} By continuity, $W_x, W_y$ are open in $(\mathcal{C}_p, \chi_{\mathrm{pr}})$. They contain $[x]$ and $[y]$ respectively, and they are disjoint because $U_a \cap U_b = \varnothing$ implies $\Phi_f^{-1}(U_a) \cap \Phi_f^{-1}(U_b) = \Phi_f^{-1}(U_a \cap U_b) = \varnothing$. Hence $(\mathcal{C}_p, \chi_{\mathrm{pr}})$ is Hausdorff. [guided] The Hausdorff property says that any two distinct points sit in disjoint open neighbourhoods. Whenever a topological space $X$ admits a family $\mathcal{F}$ of continuous real-valued functions that *separates points* — meaning that for every $p \neq q$ there is some $\phi \in \mathcal{F}$ with $\phi(p) \neq \phi(q)$ — Hausdorffness comes for free: pick $\phi$, separate $\phi(p)$ and $\phi(q)$ by disjoint open intervals in $\mathbb{R}$, and pull back along $\phi$. Here the family is $\mathcal{A} = \{\Phi_f : f \in T((V))^*\}$. Two facts are needed and we use both: 1. Each $\Phi_f$ is continuous on $(\mathcal{C}_p, \chi_{\mathrm{pr}})$. This is the *defining* property of $\chi_{\mathrm{pr}}$: it is constructed precisely so that all $\Phi_f$ are continuous (it is the initial topology generated by $\mathcal{A}$, or contains the initial topology). 2. The family $\mathcal{A}$ separates points of $\mathcal{C}_p$. This is the substantive fact about signatures: distinct path equivalence classes yield distinct signatures, and the dual elements $T((V))^*$ then distinguish them. Given $[x] \neq [y]$, fact 2 produces $f$ with $\Phi_f([x]) = a \neq b = \Phi_f([y])$. Choose disjoint open intervals $U_a, U_b$ around $a, b$ in $\mathbb{R}$. By fact 1, the preimages $W_x = \Phi_f^{-1}(U_a)$ and $W_y = \Phi_f^{-1}(U_b)$ are open in $\chi_{\mathrm{pr}}$, contain $[x]$ and $[y]$, and are disjoint because preimages preserve disjointness. The Hausdorff axiom is verified. [/guided] [/step] [step:Reduce separability and metrisability to a property of $\mathcal{S}_p$] The signature map \begin{align*} S : \mathcal{C}_p &\to \mathcal{S}_p \subset \prod_{i = 0}^\infty V^{\otimes i}, \\ [x] &\mapsto S(x) \end{align*} is a bijection by construction (paths modulo tree-like equivalence are identified with their signatures, by the [uniqueness theorem for signatures](/theorems/???)). The projective topology $\chi_{\mathrm{pr}}$ on $\mathcal{C}_p$ is, by definition, the initial topology induced by $S$ followed by the level projections; equivalently, $\chi_{\mathrm{pr}}$ is the pullback of the subspace product topology on $\mathcal{S}_p$ via $S$. Therefore $S$ is a homeomorphism between $(\mathcal{C}_p, \chi_{\mathrm{pr}})$ and $\mathcal{S}_p$ equipped with its subspace product topology. Topological properties preserved by homeomorphism include separability and metrisability. So it suffices to show that $\mathcal{S}_p$, as a subspace of the product space $\prod_{i \geq 0} V^{\otimes i}$, is separable and metrisable. [guided] The space $\mathcal{C}_p$ is defined as a quotient — paths in a Banach-space-valued path space modulo tree-like equivalence — and arguing topologically with abstract equivalence classes is cumbersome. The standard manoeuvre is to push everything through the signature map onto a concrete subset of an explicit product. **The signature map and its image.** Define \begin{align*} S : \mathcal{C}_p &\to \prod_{i = 0}^\infty V^{\otimes i}, \\ [x] &\mapsto S(x) = (1, S^{(1)}(x), S^{(2)}(x), \ldots), \end{align*} where $S^{(i)}(x) \in V^{\otimes i}$ is the depth-$i$ iterated integral of (any representative of) $[x]$. The image $\mathcal{S}_p := S(\mathcal{C}_p)$ is the *signature space*, a subset of the countable product. **Bijectivity.** $S$ is injective on $\mathcal{C}_p$ by the [uniqueness theorem for signatures](/theorems/???): two paths have the same signature iff they are tree-equivalent, which is precisely the equivalence relation defining $\mathcal{C}_p$. By construction $S$ surjects onto $\mathcal{S}_p$, so \begin{align*} S : \mathcal{C}_p \to \mathcal{S}_p \end{align*} is a bijection. **Homeomorphism.** For the topological transfer we need $S$ to be a homeomorphism, not just a bijection. The projective topology $\chi_{\mathrm{pr}}$ is, by definition, the *initial topology* on $\mathcal{C}_p$ generated by the family $\{\pi_i \circ S : i \geq 0\}$ of level-projections through the signature. Equivalently, $\chi_{\mathrm{pr}}$ is the pullback of the subspace product topology on $\mathcal{S}_p$: \begin{align*} U \in \chi_{\mathrm{pr}} \iff S(U) \text{ is open in } \mathcal{S}_p \text{ (subspace topology)}. \end{align*} This is exactly the condition that $S$ is a homeomorphism onto its image. **Why this rephrasing helps.** Topological properties preserved by homeomorphism — and *separability* and *metrisability* are both such properties — are inherited identically by homeomorphic spaces. So \begin{align*} (\mathcal{C}_p, \chi_{\mathrm{pr}}) \text{ is separable/metrisable} \iff \mathcal{S}_p \text{ (subspace topology) is separable/metrisable}. \end{align*} The right-hand side is the question we will actually attack: it concerns a subspace of an explicit countable product of finite-dimensional spaces, where standard descriptive-topology theorems apply directly. **What is left to prove.** It now suffices to show that $\mathcal{S}_p$, viewed with its subspace topology inherited from $\prod_{i \geq 0} V^{\otimes i}$, is separable and metrisable. Steps 3-5 do this by first checking that the ambient product $\prod_{i \geq 0} V^{\otimes i}$ is Polish (Steps 3-4) and then using that subspaces of Polish spaces are separable and metrisable (Step 5). [/guided] [/step] [step:Verify that each level $V^{\otimes k}$ is a Polish space] For each $k \geq 0$, $V^{\otimes k}$ is a finite-dimensional real vector space (as a tensor power of the finite-dimensional $V$). Equip it with any norm; all norms on a finite-dimensional real vector space induce the same topology and that topology is metrisable, separable, and complete. Concretely, $V^{\otimes k} \cong \mathbb{R}^{d^k}$ where $d = \dim V$, and $\mathbb{R}^{d^k}$ with the Euclidean metric is [completely metrisable](/theorems/???) and separable (rational coordinates form a countable dense set). Hence each $V^{\otimes k}$ is a [Polish space](/theorems/???). [guided] Polish spaces — separable and completely metrisable — are the natural setting for the descriptive arguments to follow. The first job is to verify that each tensor level $V^{\otimes k}$ is Polish. The key facts: $V$ is finite-dimensional real, so $V^{\otimes k}$ is finite-dimensional real of dimension $(\dim V)^k$. On a finite-dimensional real vector space, all norms induce the same topology — this is the [equivalence of norms in finite dimensions](/theorems/???) — and that topology is the Euclidean one under any choice of basis. The Euclidean space $\mathbb{R}^N$ is separable (rationals are countable and dense in each coordinate, and the product of the rationals is countable and dense in $\mathbb{R}^N$) and complete in the Euclidean metric. Both properties are intrinsic to the topology, not the choice of metric. So each $V^{\otimes k}$, with its unique vector space topology, is Polish. [/guided] [/step] [step:Apply that a countable product of Polish spaces is Polish] The product space \begin{align*} \Pi := \prod_{i = 0}^\infty V^{\otimes i} \end{align*} is a countable product (indexed by $\mathbb{N}_0$) of Polish spaces. By the [theorem that a countable product of Polish spaces is Polish](/theorems/???), $\Pi$ is Polish in the product topology. In particular $\Pi$ is separable and metrisable. [guided] We have a countable family $(V^{\otimes i})_{i \geq 0}$ of Polish spaces. The standard fact is: **Theorem.** A countable product of Polish spaces is Polish in the product topology. Verifying the hypotheses: the index set $\mathbb{N}_0$ is countable, and each factor $V^{\otimes i}$ has been shown Polish in the previous step. So we may apply the theorem to conclude $\Pi$ is Polish. The construction of a complete compatible metric on $\Pi$ is the standard weighted-supremum construction: if $d_i$ is a complete bounded metric on $V^{\otimes i}$ (replacing any unbounded $d_i$ by $\min(d_i, 1)$), then \begin{align*} d_\Pi\bigl((a_i), (b_i)\bigr) := \sum_{i = 0}^\infty 2^{-i} d_i(a_i, b_i) \end{align*} is a complete metric on $\Pi$ inducing the product topology, and the countable union of countable dense sets in each factor (one element per coordinate, all but finitely many fixed) gives a countable dense subset. For our purposes only the conclusion matters: $\Pi$ is Polish. [/guided] [/step] [step:Inherit separability and metrisability for $\mathcal{S}_p$ as a subspace of $\Pi$] The subset $\mathcal{S}_p \subset \Pi$ inherits the subspace topology. Two general facts: - *Subspaces of metrisable spaces are metrisable*: if $d$ is a metric inducing the topology on $\Pi$, then the restriction $d|_{\mathcal{S}_p \times \mathcal{S}_p}$ is a metric inducing the subspace topology on $\mathcal{S}_p$. - *Subspaces of separable metric spaces are separable*: this is the [hereditary separability of separable metric spaces](/theorems/???). (More generally, any second-countable space — and Polish spaces are second-countable — has all subspaces second-countable, hence separable.) Combining both, $\mathcal{S}_p$ in its subspace product topology is separable and metrisable. By the homeomorphism $S : (\mathcal{C}_p, \chi_{\mathrm{pr}}) \to \mathcal{S}_p$ established above, separability and metrisability transfer back to $(\mathcal{C}_p, \chi_{\mathrm{pr}})$. Together with Hausdorffness from the first step, this establishes all three claimed properties of $(\mathcal{C}_p, \chi_{\mathrm{pr}})$. [guided] We have $\mathcal{S}_p \subset \Pi$ with the subspace topology, and $\Pi$ is Polish. We do **not** need to claim $\mathcal{S}_p$ is itself Polish — only separable and metrisable, which are weaker. Metrisability is hereditary in the strongest sense: any subspace of any metric space inherits the restricted metric, and that metric induces the subspace topology. So $\mathcal{S}_p$ is metrisable. Separability is more delicate in general: a subspace of a separable space need not be separable in arbitrary topological spaces. But for *metric* spaces it is automatic. The cleanest justification is via second-countability: a separable metric space has a countable basis (take metric balls of rational radius around a countable dense set), and second-countability is hereditary (intersect each basis element with the subspace). Second-countable implies separable. Since $\Pi$ is separable metric (in fact Polish), so is $\mathcal{S}_p$. Finally, $S$ is a homeomorphism from $(\mathcal{C}_p, \chi_{\mathrm{pr}})$ to $\mathcal{S}_p$, and homeomorphisms preserve every topological property — Hausdorff, separable, metrisable. We have already proved Hausdorffness directly. The two other properties transfer from $\mathcal{S}_p$. Hence $(\mathcal{C}_p, \chi_{\mathrm{pr}})$ is Hausdorff, separable, and metrisable. [/guided] [/step]