[proofplan] The two parts share a common technique: the weak* topology on the dual unit ball is metrizable iff a countable family of bounded linear functionals separates points of that ball, which happens iff the predual is separable. We prove (i) $(\Rightarrow)$ by using a countable dense set in $X$ to define a separating countable family of evaluations on $B_{X^*}$, then upgrading the resulting weaker metrizable topology to $w^*$ via the compact-Hausdorff trick (Banach-Alaoglu compactness plus the fact that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism). For (ii) $(\Rightarrow)$, the evaluation map $K \hookrightarrow (B_{C(K)^*}, w^*)$ is an embedding, so metrizability of the target transfers to $K$. For (ii) $(\Leftarrow)$, we exhibit a countable algebra in $C(K)$ separating points of metric $K$ and apply Stone-Weierstrass. Finally (i) $(\Leftarrow)$ follows from (ii) and the isometric embedding $X \hookrightarrow C(B_{X^*}, w^*)$. [/proofplan]

[step:Prove (i) $(\Rightarrow)$: separability of $X$ implies $(B_{X^*}, w^*)$ is metrizable]Suppose $X$ is separable. Fix a countable dense sequence $(x_n)_{n \in \mathbb{N}}$ in $X$. For each $n$, let $\hat{x}_n : X^* \to \mathbb{R}$ (or $\mathbb{C}$) be the canonical evaluation $\hat{x}_n(f) = f(x_n)$. Each $\hat{x}_n$ is $w^*$-continuous on $X^*$ by definition of the weak* topology (it is one of the generating functionals). Define \begin{align*} \sigma := \sigma\bigl(B_{X^*}, \, \{\hat{x}_n|_{B_{X^*}} : n \in \mathbb{N}\}\bigr), \end{align*} the initial topology on $B_{X^*}$ generated by the countable family of restrictions $\hat{x}_n|_{B_{X^*}}$. Since this is a sub-family of the full generating set for $w^*$, we have $\sigma \subseteq w^*|_{B_{X^*}}$. [claim:The countable family $\{\hat{x}_n|_{B_{X^*}}\}$ separates points of $B_{X^*}$] For $f, g \in B_{X^*}$ with $f \neq g$, there exists $n$ with $\hat{x}_n(f) \neq \hat{x}_n(g)$. [/claim] [proof] If $f \neq g$, there exists $x \in X$ with $f(x) \neq g(x)$. By density of $(x_n)$ in $X$, there is a subsequence $x_{n_k} \to x$ in norm. Both $f$ and $g$ are continuous (being elements of $X^*$), so \begin{align*} f(x_{n_k}) \to f(x), \quad g(x_{n_k}) \to g(x). \end{align*} Since $f(x) \neq g(x)$, for sufficiently large $k$, $f(x_{n_k}) \neq g(x_{n_k})$. Pick such an $n_k$. [/proof] By the [Metrizability of Weak Topologies](/theorems/2646), since the countable family $\{\hat{x}_n|_{B_{X^*}}\}$ separates points of $B_{X^*}$, the topology $\sigma$ on $B_{X^*}$ is metrizable. Now we upgrade $\sigma$ to $w^*$. By the **Banach-Alaoglu Theorem**, $(B_{X^*}, w^*)$ is compact (and Hausdorff: weak* is Hausdorff because $X^*$ separates itself via evaluations, but more directly, since the generating family $\{\hat{x} : x \in X\}$ separates points of $X^*$). The identity map \begin{align*} \operatorname{id} : (B_{X^*}, w^*) \to (B_{X^*}, \sigma) \end{align*} is continuous (since $\sigma \subseteq w^*$) and bijective. The domain is compact and the codomain is Hausdorff (a metric topology is Hausdorff). A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Hence $w^*|_{B_{X^*}} = \sigma$, and so $(B_{X^*}, w^*)$ is metrizable.[/step]

[guided]The strategy has three pieces: 1. Use separability to produce a countable family of $w^*$-continuous functionals on $B_{X^*}$. 2. Show this family separates points of $B_{X^*}$ (so the metrizability theorem applies). 3. Upgrade the resulting weaker metrizable topology to $w^*$ itself using compactness. Step (1) — Countable separating family. Each $x \in X$ defines a $w^*$-continuous functional $\hat{x}: f \mapsto f(x)$ on $X^*$ (this is **the** definition of $w^*$). With a countable dense $(x_n) \subset X$, we get a countable subfamily $\{\hat{x}_n\}$. Step (2) — Separation. We must verify that the countable family $\{\hat{x}_n|_{B_{X^*}}\}$ separates points of $B_{X^*}$. Take $f \neq g$ in $B_{X^*}$. They differ at some $x \in X$. The point is that they must also differ at some $x_n$ near $x$: by continuity (both $f$ and $g$ are bounded linear functionals, in particular continuous), $f(x_{n_k}) \to f(x)$ and $g(x_{n_k}) \to g(x)$ along the dense approximating subsequence $x_{n_k} \to x$. Since the limits differ, eventually the values differ. Now we apply the [Metrizability of Weak Topologies](/theorems/2646), which says: the initial topology on a set $S$ generated by a countable family of point-separating maps to a metric space is metrizable. Hypotheses: countable family ($\{\hat{x}_n\}$ — countable), each map continuous to a metric space ($\hat{x}_n : B_{X^*} \to \mathbb{R}$ or $\mathbb{C}$ — yes), separates points (Claim). Conclusion: $\sigma$ on $B_{X^*}$ is metrizable. Step (3) — Upgrade $\sigma$ to $w^*$. We have $\sigma \subseteq w^*|_{B_{X^*}}$ (since $\sigma$ is generated by a sub-family). To show equality, we use a standard topology trick: > A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. The proof is short: such a map is closed (image of any closed = compact in domain = compact in codomain = closed in Hausdorff), hence open on its inverse, hence a homeomorphism. Applied with $\operatorname{id} : (B_{X^*}, w^*) \to (B_{X^*}, \sigma)$: - Continuous: $\sigma \subseteq w^*|_{B_{X^*}}$ means every $\sigma$-open set is $w^*$-open. So $\operatorname{id}^{-1}(\sigma\text{-open}) = \sigma\text{-open} \subseteq w^*\text{-open}$, i.e.\ $\operatorname{id}$ is $w^*$-to-$\sigma$ continuous. - Bijection: identity. - Domain compact: by the Banach-Alaoglu theorem, the closed unit ball $B_{X^*}$ is $w^*$-compact. - Codomain Hausdorff: $\sigma$ is metrizable, hence Hausdorff. Conclusion: $\operatorname{id}$ is a homeomorphism, so $w^*|_{B_{X^*}} = \sigma$, which is metrizable. We have proved $(B_{X^*}, w^*)$ is metrizable.[/guided]

[step:Prove (ii) $(\Rightarrow)$: separability of $C(K)$ implies $K$ is metrizable]Suppose $C(K)$ is separable. For each $k \in K$, define the evaluation functional \begin{align*} \delta_k : C(K) &\to \mathbb{R} \text{ (or } \mathbb{C}\text{)}, \\ \varphi &\mapsto \varphi(k). \end{align*} Then $\delta_k$ is linear (clear) and continuous: $|\delta_k(\varphi)| = |\varphi(k)| \leq \|\varphi\|_\infty$, so $\|\delta_k\|_{C(K)^*} \leq 1$, i.e.\ $\delta_k \in B_{C(K)^*}$. (Taking the constant function $\varphi = 1_K$ gives $\delta_k(1_K) = 1$, so equality $\|\delta_k\| = 1$.) Define the evaluation map \begin{align*} \Phi : K &\to (B_{C(K)^*}, w^*), \\ k &\mapsto \delta_k. \end{align*} *Continuity of $\Phi$:* by the [Universal Property of Weak Topologies](/theorems/2645), a map into the weak topology is continuous iff its composition with each generating evaluation is continuous. The generating evaluations of $w^*$ on $C(K)^*$ are $\hat{\varphi} : f \mapsto f(\varphi)$ for $\varphi \in C(K)$. The composition is $\hat{\varphi} \circ \Phi : K \to \mathbb{R}$, $k \mapsto \delta_k(\varphi) = \varphi(k)$, which is the function $\varphi \in C(K)$ — continuous by definition. So $\Phi$ is continuous. *Injectivity of $\Phi$:* if $k_1 \neq k_2$ in $K$, then by Urysohn's lemma (applied to compact Hausdorff $K$, which is normal), there is $\varphi \in C(K)$ with $\varphi(k_1) = 0 \neq 1 = \varphi(k_2)$. So $\delta_{k_1}(\varphi) = 0 \neq 1 = \delta_{k_2}(\varphi)$, giving $\delta_{k_1} \neq \delta_{k_2}$. *$\Phi$ is a topological embedding:* $K$ is compact (hypothesis), $\Phi$ is continuous, so $\Phi(K)$ is compact in $(B_{C(K)^*}, w^*)$. Since $w^*$ is Hausdorff (as $C(K)$ separates points of $C(K)^*$), $(B_{C(K)^*}, w^*)$ is Hausdorff. A continuous injection from a compact space to a Hausdorff space is a closed embedding (it is a homeomorphism onto its image: the same compact-Hausdorff trick from Step 1, applied to $\Phi : K \to \Phi(K)$). Hence $K \cong \Phi(K)$ as topological spaces. By Step 1 applied to $X = C(K)$ (separable by assumption), $(B_{C(K)^*}, w^*)$ is metrizable. Subspaces of metric spaces are metrizable (a subspace inherits the restricted metric). So $\Phi(K)$ is metrizable, and hence $K$ is metrizable.[/step]

[guided]We embed $K$ into a known metrizable space — the dual unit ball with weak* topology — and inherit metrizability. The natural embedding is point evaluation. For a fixed $k \in K$, the evaluation functional $\delta_k(\varphi) := \varphi(k)$ is linear and bounded with norm at most $1$ (since $|\varphi(k)| \leq \sup_K |\varphi| = \|\varphi\|_\infty$), so $\delta_k \in B_{C(K)^*}$. Define $\Phi : K \to (B_{C(K)^*}, w^*)$ by $\Phi(k) = \delta_k$. We claim $\Phi$ is a topological embedding. *Continuity.* By the [Universal Property of Weak Topologies](/theorems/2645), continuity of $\Phi$ to the weak* topology reduces to continuity of $\hat{\varphi} \circ \Phi$ for each $\varphi \in C(K)$ (the generating functionals of $w^*$ are $\hat{\varphi} : f \mapsto f(\varphi)$, $\varphi$ ranging over $C(K)$). Compute $\hat{\varphi} \circ \Phi(k) = \delta_k(\varphi) = \varphi(k)$. So $\hat{\varphi} \circ \Phi = \varphi$ as a function $K \to \mathbb{R}$ (or $\mathbb{C}$), which is continuous by hypothesis $\varphi \in C(K)$. Hence $\Phi$ is continuous. *Injectivity.* If $\Phi(k_1) = \Phi(k_2)$, then $\varphi(k_1) = \varphi(k_2)$ for every $\varphi \in C(K)$. We must conclude $k_1 = k_2$. The contrapositive: if $k_1 \neq k_2$ in compact Hausdorff $K$, can we find a continuous function distinguishing them? **Yes** — compact Hausdorff implies normal (the standard hierarchy), and Urysohn's lemma in normal spaces gives a continuous $\varphi : K \to [0,1]$ with $\varphi(k_1) = 0$, $\varphi(k_2) = 1$. So $\delta_{k_1} \neq \delta_{k_2}$. *Embedding.* A continuous injection from compact to Hausdorff is automatically a closed embedding. The argument is the same compact-Hausdorff trick from Step 1: the map factors as $K \to \Phi(K)$ continuous bijection from compact to Hausdorff, which is a homeomorphism. So $K \cong \Phi(K)$ as topological spaces. Now metrizability transfers: $C(K)$ separable $\Rightarrow$ (by Step 1) $(B_{C(K)^*}, w^*)$ is metrizable. Subspaces of metric spaces are metric (inherit the metric). So $\Phi(K)$ is metrizable, and via the homeomorphism $K \cong \Phi(K)$, so is $K$.[/guided]

[step:Prove (ii) $(\Leftarrow)$: $K$ metrizable implies $C(K)$ separable]Suppose $K$ is metrizable. Fix a metric $d$ on $K$ generating its topology. A compact metric space is separable: cover $K$ by finitely many balls of radius $1/n$ for each $n$, and take the union of centres over $n$ — countable, dense. Let $(k_m)_{m \in \mathbb{N}}$ be a countable dense sequence in $K$. For each $m \in \mathbb{N}$ and $q \in \mathbb{Q}_{>0}$, define \begin{align*} g_{m,q} : K &\to \mathbb{R}, \\ k &\mapsto \max\bigl(q - d(k, k_m), 0\bigr). \end{align*} Each $g_{m,q}$ is continuous on $K$ (as the composition of the continuous distance function $k \mapsto d(k, k_m)$ with continuous operations). The collection $\{g_{m,q} : m \in \mathbb{N}, q \in \mathbb{Q}_{>0}\}$ is countable. Let $\mathcal{A} \subseteq C(K; \mathbb{C})$ be the smallest unital $*$-subalgebra over $\mathbb{C}$ generated by the set $\{g_{m,q}\} \cup \{1_K\}$, i.e.\ the closure under $\mathbb{C}$-linear combinations, products, and complex conjugation, of the unital algebra over $\mathbb{Q}[i]$ generated by $\{g_{m,q}\} \cup \{1_K\}$. (We restrict the field of coefficients to $\mathbb{Q}[i]$ to keep $\mathcal{A}$ countable; the rational-coefficient algebra is dense in the corresponding $\mathbb{C}$-algebra under uniform convergence.) The set $\mathcal{A}$ is countable. We verify the hypotheses of the **Stone-Weierstrass theorem** for the (real or complex) compact Hausdorff space $K$. *Contains constants:* $1_K \in \mathcal{A}$ by construction. *Closed under conjugation:* the generators $g_{m,q}$ and $1_K$ are real-valued, hence equal to their conjugates; algebraic operations preserve closure under conjugation, and we explicitly closed under conjugation. *Separates points of $K$:* let $k_1 \neq k_2$ in $K$, and set $r := d(k_1, k_2) > 0$. By density of $(k_m)$, there is $m$ with $d(k_m, k_1) < r/3$. Choose $q \in \mathbb{Q}_{>0}$ with $r/3 < q < 2r/3$. Then \begin{align*} d(k_1, k_m) < r/3 < q, \quad d(k_2, k_m) \geq d(k_1, k_2) - d(k_1, k_m) > r - r/3 = 2r/3 > q. \end{align*} Hence $g_{m,q}(k_1) = q - d(k_1, k_m) > 0$ and $g_{m,q}(k_2) = 0$. So $g_{m,q}$ separates $k_1$ from $k_2$. By **Stone-Weierstrass** (complex version, for compact Hausdorff $K$): a unital $*$-subalgebra of $C(K; \mathbb{C})$ that separates points is dense in $C(K; \mathbb{C})$ in the uniform norm. Hence $\mathcal{A}$ is dense in $C(K)$. Since $\mathcal{A}$ is countable, $C(K)$ is separable.[/step]

[guided]The strategy has three movements: build a countable family of continuous functions, verify it generates an algebra satisfying Stone-Weierstrass, and conclude separability. *A compact metric space is separable.* This is a standard fact. For each $n \in \mathbb{N}$, by compactness, the open cover $\{B(k, 1/n) : k \in K\}$ has a finite subcover with centres $\{k_1^{(n)}, \dots, k_{N_n}^{(n)}\}$. The union $\bigcup_n \{k_j^{(n)} : 1 \leq j \leq N_n\}$ is countable and dense (any $k \in K$ is within $1/n$ of some $k_j^{(n)}$ for every $n$). *The functions $g_{m,q}$.* The "tent" functions $g_{m,q}(k) = \max(q - d(k, k_m), 0)$ are continuous (distance is $1$-Lipschitz, $\max$ with $0$ preserves continuity), supported in the closed ball $\overline{B}(k_m, q)$, and equal $q$ at $k_m$, decreasing linearly to $0$ at the boundary. For each centre $k_m$ and each rational radius $q$, we get one such function. *The algebra $\mathcal{A}$.* We want a countable set of functions whose closure (in $C(K)$) is all of $C(K)$. The natural candidate is the algebra generated by the $g_{m,q}$ together with $1_K$ — this is automatically a unital algebra. But to keep it countable, we restrict the coefficient field to $\mathbb{Q}[i]$ (or $\mathbb{Q}$ in the real case): the resulting algebra is countable since it consists of finite combinations and products of countably many generators with countably many coefficients. Why does the $\mathbb{Q}[i]$-algebra still suffice? Because $\mathbb{Q}[i]$ is dense in $\mathbb{C}$, and the closure (in uniform norm) of a $\mathbb{Q}[i]$-algebra inside a $\mathbb{C}$-algebra is the same as the closure of the $\mathbb{C}$-algebra. So density follows once we apply Stone-Weierstrass to the $\mathbb{C}$-algebra. *Verifying Stone-Weierstrass hypotheses.* 1. **Unital algebra**: $\mathcal{A}$ contains $1_K$ and is closed under products and sums by construction. 2. **Closed under complex conjugation**: the generators are real-valued, and we explicitly closed under conjugation. 3. **Separates points**: this is the substantive check. Given distinct points $k_1, k_2 \in K$ with separation $r := d(k_1, k_2)$, we want one of the $g_{m,q}$ to take different values at $k_1, k_2$. The construction: pick $k_m$ very close to $k_1$ (within $r/3$ — possible by density of $(k_m)$), and pick a rational radius $q$ between $r/3$ and $2r/3$. Then the tent $g_{m,q}$ contains $k_1$ in its support (with positive value) but excludes $k_2$ (which is more than $q$ from $k_m$, by triangle inequality: $d(k_2, k_m) \geq d(k_1, k_2) - d(k_1, k_m) > r - r/3 = 2r/3 > q$). Stone-Weierstrass concludes $\mathcal{A}$ is dense in $C(K; \mathbb{C})$. Since $\mathcal{A}$ is countable, $C(K)$ is separable.[/guided]

[step:Prove (i) $(\Leftarrow)$: $(B_{X^*}, w^*)$ metrizable implies $X$ separable]Suppose $(B_{X^*}, w^*)$ is metrizable. Set $K := (B_{X^*}, w^*)$. By the **Banach-Alaoglu Theorem**, $K$ is compact, and $w^*$ is Hausdorff (since the family $\{\hat{x} : x \in X\}$ separates points of $X^*$). So $K$ is a compact Hausdorff space, metrizable by hypothesis. By Step 3 (the implication (ii) $(\Leftarrow)$), $C(K)$ is separable. Define \begin{align*} J : X &\to C(K), \\ x &\mapsto \hat{x}|_{B_{X^*}}, \end{align*} where $\hat{x}(f) = f(x)$ for $f \in K = B_{X^*}$. The map $\hat{x}|_{B_{X^*}} : K \to \mathbb{R}$ (or $\mathbb{C}$) is $w^*$-continuous (the evaluation $\hat{x}$ is in the generating family for $w^*$), so $J(x) \in C(K)$. *$J$ is linear:* by linearity of evaluation in $x$. *$J$ is isometric:* \begin{align*} \|J(x)\|_\infty = \sup_{f \in B_{X^*}} |\hat{x}(f)| = \sup_{f \in B_{X^*}} |f(x)| = \|x\|_X, \end{align*} where the last equality is the [Hahn-Banach (Normed Space Version)](/theorems/2629) statement that the norm of a vector in a normed space equals the supremum of the dual norms over the unit ball of the dual. So $J$ is an isometric embedding of $X$ into the separable space $C(K)$. Since subspaces of separable metric spaces are separable (a countable dense set in $C(K)$ projects to a dense set in any subspace via approximation), $J(X) \subseteq C(K)$ is separable. Through the isometric isomorphism $J : X \to J(X)$, $X$ is separable.[/step]

[guided]This implication runs the previous arguments in reverse. Assuming $(B_{X^*}, w^*)$ metrizable, we want to construct a countable dense subset of $X$. *The space $K = (B_{X^*}, w^*)$ is compact Hausdorff metrizable.* Compactness is the **Banach-Alaoglu Theorem**. Hausdorff: the generating family $\{\hat{x} : x \in X\}$ separates points of $X^*$ — if $f \neq g$ then $f - g \neq 0$ in $X^*$, so by definition of the dual norm there is $x \in X$ with $(f-g)(x) \neq 0$, i.e.\ $\hat{x}(f) \neq \hat{x}(g)$. So weak* is Hausdorff. Metrizable: by hypothesis. *$C(K)$ is separable.* Apply Step 3 (the (ii) $(\Leftarrow)$ implication) to $K$. *Embed $X$ into $C(K)$.* The natural map is $J(x) := \hat{x}|_{B_{X^*}}$, restriction of the canonical bidual embedding to the dual unit ball. Each $\hat{x}|_{B_{X^*}}$ is continuous on $(B_{X^*}, w^*)$ since $\hat{x}$ is in the generating family for $w^*$. Linearity is clear. *Isometry.* This is where we use the [Hahn-Banach (Normed Space Version)](/theorems/2629), which states $\|x\|_X = \sup_{f \in B_{X^*}} |f(x)|$ for any $x \in X$. The supremum on the right is precisely $\|J(x)\|_\infty$, so $\|J(x)\|_\infty = \|x\|_X$. *Subspaces of separable metric spaces are separable.* If $C(K)$ has a countable dense set $D$ (in the sup-norm), then for each $\varphi \in J(X)$ and each $n$, pick some $\psi_n \in D$ with $\|\varphi - \psi_n\|_\infty < 1/n$ — this gives a countable subset of $D$ approximating $\varphi$, but to get a countable dense set **inside** $J(X)$ we use a slight refinement: enumerate $D = \{\psi_k\}$ and for each pair $(k, n)$ with $\inf_{\varphi \in J(X)} \|\varphi - \psi_k\|_\infty < 1/n$, pick $\varphi_{k,n} \in J(X)$ with $\|\varphi_{k,n} - \psi_k\|_\infty < 1/n$. Then $\{\varphi_{k,n}\}$ is countable and dense in $J(X)$. (Alternatively: any subspace of a second-countable space is second-countable, and second-countable metric implies separable.) Pulling back via the isometry $J^{-1} : J(X) \to X$, we get a countable dense set in $X$.[/guided]

[step:Conclude] Steps 1 and 4 together prove (i): $X$ separable iff $(B_{X^*}, w^*)$ metrizable. Steps 2 and 3 together prove (ii): $C(K)$ separable iff $K$ metrizable. [/step]