[proofplan] We exhibit an explicit countable cover of $\mathcal{C}_1$ by compact sets. For each integer $r \geq 1$, let $B(r) := \{[x] \in \mathcal{C}_1 : \|x^*\|_1 \leq r\}$ denote the closed length ball of radius $r$, where $x^*$ is the tree-reduced representative of $[x]$. The [Tree-Reduced Representative Theorem](/theorems/???) guarantees that every class $[x] \in \mathcal{C}_1$ admits a tree-reduced representative of finite length, so $[x] \in B(\lceil \|x^*\|_1 \rceil)$. Combining this with the [Compactness of Length Balls in $(\mathcal{C}_1, \chi_{\mathrm{pr}})$](/theorems/???) — which asserts that each $B(r)$ is $\chi_{\mathrm{pr}}$-compact — yields the desired countable cover by compacta. [/proofplan] [step:Define the length balls $B(r)$ and identify the candidate countable cover] For each integer $r \geq 1$, define the map \begin{align*} B: \mathbb{N} &\to \mathcal{P}(\mathcal{C}_1) \\ r &\mapsto \{[x] \in \mathcal{C}_1 : \|x^*\|_1 \leq r\}, \end{align*} where $x^*$ denotes the tree-reduced representative of $[x]$ and $\|\cdot\|_1$ denotes the one-variation (length) norm on continuous bounded-variation paths. Our candidate countable cover of $\mathcal{C}_1$ is the family $\{B(r)\}_{r \in \mathbb{N}}$. [/step] [step:Show $\mathcal{C}_1 = \bigcup_{r=1}^\infty B(r)$ via the tree-reduced representative theorem] We prove the set equality by showing both inclusions. The inclusion $\bigcup_{r=1}^\infty B(r) \subseteq \mathcal{C}_1$ is immediate, since every $B(r)$ is by definition a subset of $\mathcal{C}_1$. For the reverse inclusion, fix an arbitrary $[x] \in \mathcal{C}_1$. By the [Tree-Reduced Representative Theorem](/theorems/???), the class $[x]$ contains a tree-reduced representative $x^*$ of finite length, i.e. $\|x^*\|_1 < \infty$. Set \begin{align*} r_0 := \lceil \|x^*\|_1 \rceil \in \mathbb{N}. \end{align*} Then $\|x^*\|_1 \leq r_0$, so $[x] \in B(r_0) \subseteq \bigcup_{r=1}^\infty B(r)$. This establishes $\mathcal{C}_1 \subseteq \bigcup_{r=1}^\infty B(r)$, and hence equality. [guided] We must verify the set equality \begin{align*} \mathcal{C}_1 = \bigcup_{r=1}^\infty B(r). \end{align*} The two inclusions are very different in nature; we examine each in turn. **Forward inclusion: $\bigcup_{r=1}^\infty B(r) \subseteq \mathcal{C}_1$.** This direction is immediate from the definition. Each $B(r)$ was defined as $B(r) := \{[x] \in \mathcal{C}_1 : \|x^*\|_1 \le r\}$, a subset of $\mathcal{C}_1$. A union of subsets of $\mathcal{C}_1$ is a subset of $\mathcal{C}_1$. **Reverse inclusion: $\mathcal{C}_1 \subseteq \bigcup_{r=1}^\infty B(r)$.** This is where the substance lies. We must show every $[x] \in \mathcal{C}_1$ belongs to $B(r)$ for **some** $r$. Why is this not obvious? An element of $\mathcal{C}_1$ is an *equivalence class* of bounded-variation paths under tree-equivalence (the relation that identifies paths differing by tree-like backtracking pieces). Different representatives of the same class can have wildly different one-variations — append a long backtracking excursion (a "tree-like piece") and the one-variation increases without bound, yet the equivalence class is unchanged. So inspecting an arbitrary representative does not yield a finite length bound. The [Tree-Reduced Representative Theorem](/theorems/???) resolves this. The theorem requires only that $[x]$ be a tree-equivalence class of continuous paths of finite one-variation — which is the very definition of $[x] \in \mathcal{C}_1$. Its conclusion is that $[x]$ contains a canonical *tree-reduced* representative $x^*$ with $\|x^*\|_1 < \infty$. This is the minimal-length representative; intuitively, all "wasted" backtracking has been excised. We verify the hypothesis: by definition $[x] \in \mathcal{C}_1$ means $[x]$ is a tree-equivalence class of paths of finite one-variation, so the theorem applies. Fix arbitrary $[x] \in \mathcal{C}_1$. Apply the theorem to obtain $x^*$ with $\|x^*\|_1 < \infty$. Define \begin{align*} r_0 := \lceil \|x^*\|_1 \rceil \in \mathbb{N}, \end{align*} the smallest natural number bounding $\|x^*\|_1$ from above. Since $\|x^*\|_1 < \infty$, the ceiling $r_0$ is finite, and $r_0 \ge 1$ because $\mathbb{N}$ starts at $1$ (we may take $r_0 = 1$ if $\|x^*\|_1 = 0$). Then $\|x^*\|_1 \le r_0$, so by definition $[x] \in B(r_0) \subseteq \bigcup_{r=1}^\infty B(r)$. Since $[x] \in \mathcal{C}_1$ was arbitrary, $\mathcal{C}_1 \subseteq \bigcup_{r=1}^\infty B(r)$. **Combining.** The two inclusions give $\mathcal{C}_1 = \bigcup_{r=1}^\infty B(r)$. Why-not: One might attempt to bound $\|x\|_1$ for an arbitrary representative $x$. This fails because tree-equivalence does not preserve one-variation. Only the *minimal* (tree-reduced) representative provides a length bound that depends only on the equivalence class. The Tree-Reduced Representative Theorem is doing the heavy lifting — without it, a class might fail to belong to any $B(r)$. Connection to technique: This is the standard exhaustion-by-balls strategy for proving $\sigma$-compactness. The novelty here is that the "ball" is defined by an invariant of the equivalence class (the tree-reduced length), not by a metric ball in some ambient space. [/guided] [/step] [step:Invoke compactness of each length ball $B(r)$] By the [Compactness of Length Balls in $(\mathcal{C}_1, \chi_{\mathrm{pr}})$](/theorems/???), for every $r \in \mathbb{N}$ the set $B(r)$ is compact in the topology $\chi_{\mathrm{pr}}$. The hypotheses of that theorem require only that $r$ be a positive real, which is satisfied by every $r \in \mathbb{N}$. [guided] We need each $B(r)$ to be compact in the topology $\chi_{\mathrm{pr}}$. The choice of topology matters: $\chi_{\mathrm{pr}}$ is the **projective signature topology** — the initial topology on $\mathcal{C}_1$ induced by the family of signature-projection maps $\pi_m: [x] \mapsto S(x)^{(m)}$ to the finite-dimensional spaces $V^{\otimes m}$. Compactness in $\chi_{\mathrm{pr}}$ is much weaker than compactness in (say) the one-variation topology — the projective topology is coarse, with fewer open sets, hence more compact sets. We cite the [Compactness of Length Balls in $(\mathcal{C}_1, \chi_{\mathrm{pr}})$](/theorems/???). To verify its hypotheses, we recall its statement: for every positive real $r > 0$, the set $\{[x] \in \mathcal{C}_1 : \|x^*\|_1 \le r\}$ is compact in $\chi_{\mathrm{pr}}$. Our $B(r)$ for $r \in \mathbb{N}$ is exactly this set (with $r$ a positive integer, hence a positive real). The hypothesis $r > 0$ is satisfied because $r \in \mathbb{N}$ and $\mathbb{N}$ starts at $1 > 0$. Therefore $B(r)$ is $\chi_{\mathrm{pr}}$-compact for every $r \in \mathbb{N}$. Why is the cited compactness theorem a non-trivial result, deserving of a full external reference? The space $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ is infinite-dimensional, so closed-and-bounded does not imply compact (the Heine–Borel theorem fails outside finite dimensions). Compactness of $B(r)$ uses two specific facts: - The signature levels $S(x)^{(m)}$ are uniformly bounded on $B(r)$ by factorial decay $\|S(x)^{(m)}\| \le r^m / m!$ (the [Factorial Decay of Signature Levels](/theorems/2493) — the very theorem this proof depends on indirectly via the cited compactness result), so the image of $B(r)$ in each $V^{\otimes m}$ is bounded, hence pre-compact in the finite-dimensional space $V^{\otimes m}$. - The projective topology then turns this level-wise pre-compactness into compactness via Tychonoff's theorem on the product space $\prod_m V^{\otimes m}$. Why-not: in the stronger one-variation topology, $B(r)$ is generally **not** compact — the unit one-variation ball in $\mathcal{C}_1$ is not compact in its own one-variation topology, just as in any infinite-dimensional Banach space. The compactness here is a feature of the coarser topology $\chi_{\mathrm{pr}}$. [/guided] [/step] [step:Conclude $\sigma$-compactness from a countable cover by compacta] By Step 2, $\mathcal{C}_1 = \bigcup_{r=1}^\infty B(r)$ is a countable union of subsets of $\mathcal{C}_1$. By Step 3, each $B(r)$ is $\chi_{\mathrm{pr}}$-compact. By the [Definition of $\sigma$-Compact Space](/page/Sigma-Compact%20Space), a topological space is $\sigma$-compact if and only if it is a countable union of compact subsets. Therefore $(\mathcal{C}_1, \chi_{\mathrm{pr}})$ is $\sigma$-compact, completing the proof. [/step]