[proofplan]
Three things to verify: (1) $d$ is a metric, with the only non-trivial axiom being positivity ($d(x,y) > 0$ for $x \ne y$), which follows from the separating-points hypothesis; (2) the metric topology is finer than $\sigma(X, \mathcal{F})$, shown via the [Universal Property of Weak Topologies](/theorems/2645) — each $f_n$ is uniformly continuous from $(X, d)$ to $(Y_n, d_n)$ because the $n$-th term controls $d_n$; (3) the metric topology is coarser than $\sigma(X, \mathcal{F})$, shown by writing $d$ as a uniformly convergent series of $\sigma(X, \mathcal{F})$-continuous functions, then taking preimages of open balls. Both inclusions together give equality of topologies.
[/proofplan]
[step:Verify that $d$ is a metric on $X$]
Define $\tilde{d}_n(x, y) := \min(d_n(f_n(x), f_n(y)), 1)$ for each $n \in \mathbb{N}$. Each $\tilde{d}_n: X \times X \to [0, 1]$ is a pseudometric (a metric on $X$ would require $f_n$ to be injective, which is not assumed; here we only use the pseudometric properties):
- **Non-negativity and zero on the diagonal.** $\tilde{d}_n(x, x) = \min(d_n(f_n(x), f_n(x)), 1) = \min(0, 1) = 0$.
- **Symmetry.** $\tilde{d}_n(x, y) = \tilde{d}_n(y, x)$ from symmetry of $d_n$.
- **Triangle inequality.** For $x, y, z \in X$, since $d_n$ is a metric on $Y_n$,
\begin{align*}
d_n(f_n(x), f_n(z)) \le d_n(f_n(x), f_n(y)) + d_n(f_n(y), f_n(z)).
\end{align*}
Taking minima with $1$: if both $\tilde{d}_n(x, y) < 1$ and $\tilde{d}_n(y, z) < 1$, the sum is at most $\tilde{d}_n(x, y) + \tilde{d}_n(y, z) < 2$, and $\min(\cdot, 1)$ preserves the inequality. If either term equals $1$, the right-hand side is at least $1 \ge \tilde{d}_n(x, z)$. So $\tilde{d}_n(x, z) \le \tilde{d}_n(x, y) + \tilde{d}_n(y, z)$.
Now $d(x, y) = \sum_n 2^{-n} \tilde{d}_n(x, y)$. The series converges absolutely: $0 \le 2^{-n} \tilde{d}_n(x, y) \le 2^{-n}$, and $\sum 2^{-n} = 1$. So $d: X \times X \to [0, 1]$ is well-defined.
The metric axioms transfer term-by-term: $d \ge 0$, $d(x, x) = 0$, symmetry, and triangle inequality (sum of pseudometric inequalities). It remains to verify positivity for $x \ne y$. Since $\mathcal{F}$ separates points, there exists $n_0 \in \mathbb{N}$ with $f_{n_0}(x) \ne f_{n_0}(y)$. Then $d_{n_0}(f_{n_0}(x), f_{n_0}(y)) > 0$, so $\tilde{d}_{n_0}(x, y) = \min(d_{n_0}(f_{n_0}(x), f_{n_0}(y)), 1) > 0$, hence $d(x, y) \ge 2^{-n_0} \tilde{d}_{n_0}(x, y) > 0$.
So $d$ is a metric.
[/step]
[step:Show each $f_n: (X, d) \to (Y_n, d_n)$ is continuous]
Fix $n \in \mathbb{N}$. We show $f_n$ is uniformly continuous, hence continuous.
Let $\varepsilon > 0$. Set $\delta := 2^{-n} \min(\varepsilon, 1)$. For $x, y \in X$ with $d(x, y) < \delta$,
\begin{align*}
2^{-n} \tilde{d}_n(x, y) \le d(x, y) < \delta = 2^{-n} \min(\varepsilon, 1),
\end{align*}
hence $\tilde{d}_n(x, y) < \min(\varepsilon, 1) \le 1$. The strict inequality $\tilde{d}_n(x, y) < 1$ means the truncation is not active, so $\tilde{d}_n(x, y) = d_n(f_n(x), f_n(y))$, and $d_n(f_n(x), f_n(y)) < \varepsilon$. So $f_n: (X, d) \to (Y_n, d_n)$ is uniformly continuous, hence continuous.
[/step]
[step:Conclude the metric topology is finer than $\sigma(X, \mathcal{F})$]
Let $\mathcal{T}_d$ denote the metric topology on $X$ induced by $d$. From Step 2, each $f_n: (X, \mathcal{T}_d) \to (Y_n, d_n)$ is continuous. Apply the [Universal Property of Weak Topologies](/theorems/2645) with $Z = (X, \mathcal{T}_d)$ and $g = \operatorname{id}_X: Z \to X$: the hypothesis is that $f_n \circ g = f_n: (X, \mathcal{T}_d) \to (Y_n, d_n)$ is continuous for every $n$, which holds. The conclusion is that $g = \operatorname{id}_X: (X, \mathcal{T}_d) \to (X, \sigma(X, \mathcal{F}))$ is continuous.
Continuity of the identity from $\mathcal{T}_d$ to $\sigma(X, \mathcal{F})$ means $\sigma(X, \mathcal{F}) \subseteq \mathcal{T}_d$ as collections of subsets of $X$: every $\sigma(X, \mathcal{F})$-open set is $\mathcal{T}_d$-open.
[/step]
[step:Show each $\tilde{d}_n: (X, \sigma(X, \mathcal{F})) \times (X, \sigma(X, \mathcal{F})) \to \mathbb{R}$ is continuous]
Each $f_n: (X, \sigma(X, \mathcal{F})) \to (Y_n, d_n)$ is continuous by the definition of $\sigma(X, \mathcal{F})$ (Step 1 of the proof of the universal property; equivalently, taking $g = \operatorname{id}_X$ on $(X, \sigma(X, \mathcal{F}))$ in the forward direction). The product map
\begin{align*}
f_n \times f_n: (X, \sigma(X, \mathcal{F})) \times (X, \sigma(X, \mathcal{F})) &\to (Y_n, d_n) \times (Y_n, d_n) \\
(x, y) &\mapsto (f_n(x), f_n(y))
\end{align*}
is continuous (continuity of a map into a product is equivalent to continuity of both component projections, both of which are $f_n$ on one factor of the source).
The metric $d_n: (Y_n, d_n) \times (Y_n, d_n) \to \mathbb{R}$ is continuous (a standard property: $|d_n(a, b) - d_n(a', b')| \le d_n(a, a') + d_n(b, b')$ from two applications of the triangle inequality). The truncation $r \mapsto \min(r, 1)$ is continuous on $\mathbb{R}$. Composition of continuous maps gives
\begin{align*}
\tilde{d}_n: (X, \sigma(X, \mathcal{F})) \times (X, \sigma(X, \mathcal{F})) &\to [0, 1] \\
(x, y) &\mapsto \min(d_n(f_n(x), f_n(y)), 1)
\end{align*}
is continuous on the product topology.
[/step]
[step:Show $d: (X, \sigma(X, \mathcal{F})) \times (X, \sigma(X, \mathcal{F})) \to \mathbb{R}$ is continuous as a uniform limit]
Define partial sums $S_N: X \times X \to \mathbb{R}$ by $S_N(x, y) := \sum_{n=1}^N 2^{-n} \tilde{d}_n(x, y)$. Each $S_N$ is continuous on $(X, \sigma(X, \mathcal{F})) \times (X, \sigma(X, \mathcal{F}))$ as a finite sum of continuous functions (Step 4) times constants.
The convergence $S_N \to d$ is uniform on $X \times X$: for all $(x, y)$,
\begin{align*}
|d(x, y) - S_N(x, y)| = \sum_{n=N+1}^\infty 2^{-n} \tilde{d}_n(x, y) \le \sum_{n=N+1}^\infty 2^{-n} = 2^{-N} \to 0.
\end{align*}
A uniform limit of continuous functions on a topological space is continuous (standard fact: if $S_N \to d$ uniformly and each $S_N$ is continuous, then $d$ is continuous; the proof is the same $\varepsilon/3$ argument as for metric spaces, using uniform closeness of $S_N$ and pointwise continuity of $S_N$). Hence $d: (X, \sigma(X, \mathcal{F}))^2 \to \mathbb{R}$ is continuous.
[/step]
[step:Conclude the metric topology is coarser than $\sigma(X, \mathcal{F})$]
Fix $x_0 \in X$ and $r > 0$. The open ball $B_d(x_0, r) := \{x \in X : d(x_0, x) < r\}$ is the preimage of the open interval $(-\infty, r) \subseteq \mathbb{R}$ under the map
\begin{align*}
d(x_0, \cdot): (X, \sigma(X, \mathcal{F})) &\to \mathbb{R} \\
x &\mapsto d(x_0, x).
\end{align*}
This map is continuous: it is the composition of $x \mapsto (x_0, x)$ (continuous, with constant first component) with $d: X \times X \to \mathbb{R}$ (continuous on $\sigma(X, \mathcal{F})^2$ by Step 5). So $B_d(x_0, r) \in \sigma(X, \mathcal{F})$.
Open balls form a basis for the metric topology $\mathcal{T}_d$, so every $\mathcal{T}_d$-open set is a union of $\sigma(X, \mathcal{F})$-open sets, hence $\sigma(X, \mathcal{F})$-open. Therefore $\mathcal{T}_d \subseteq \sigma(X, \mathcal{F})$.
Combined with Step 3, $\mathcal{T}_d = \sigma(X, \mathcal{F})$, so $\sigma(X, \mathcal{F})$ is metrizable, induced by $d$.
[guided]
The proof has a symmetric structure: each direction of the topology equality corresponds to "the identity map is continuous in this direction", and each is established using the universal property of $\sigma(X, \mathcal{F})$ in different ways.
**Direction $\sigma(X, \mathcal{F}) \subseteq \mathcal{T}_d$ (Step 3): the metric makes the $f_n$ continuous.** This is the universal property in its most direct form. We need to check each $f_n: (X, d) \to (Y_n, d_n)$ is continuous; the $\sigma(X, \mathcal{F})$ topology is by definition the **coarsest** making the $f_n$ continuous, so any other topology making them continuous is finer. The $n$-th term $2^{-n} \tilde{d}_n$ in $d$ controls $d_n(f_n(x), f_n(y))$ when both are below $1$, and the prefactor $2^{-n}$ scales the threshold appropriately.
**Direction $\mathcal{T}_d \subseteq \sigma(X, \mathcal{F})$ (Steps 4-6): the metric is itself made of $\sigma$-continuous pieces.** Here we work in the opposite direction. The metric $d$ is a sum of $\sigma(X, \mathcal{F})$-continuous functions $\tilde{d}_n$ (continuous because $f_n$ is $\sigma$-continuous and $d_n$ is continuous on $Y_n^2$). The series converges uniformly (the tail beyond $N$ is bounded by $2^{-N}$), and uniform convergence preserves continuity. So $d$ is $\sigma(X, \mathcal{F})$-continuous on $X^2$, which means open balls $\{d(x_0, \cdot) < r\}$ are $\sigma(X, \mathcal{F})$-open, hence so is every $\mathcal{T}_d$-open set.
**Why the truncation $\min(\cdot, 1)$?** Two reasons. First, it bounds each term by $1$, making the series $\sum 2^{-n} \tilde{d}_n$ converge regardless of whether the $d_n$ are bounded or unbounded on $f_n(X)$. Second, the truncation does not alter the topology generated by each $\tilde{d}_n$ (truncating a metric at any constant produces a uniformly equivalent metric on bounded distances; the topology is unchanged), so we can swap $d_n$ for $\tilde{d}_n$ without losing topological information.
**Why the $2^{-n}$ weights?** They are not topologically essential (any positive sequence with $\sum c_n < \infty$ would do), but they ensure absolute convergence of the series uniformly in $(x, y)$, which is what gives the uniform-limit-of-continuous-is-continuous step.
**Why "separates points" is needed.** Without it, $d(x, y) = 0$ for some $x \ne y$, so $d$ is only a pseudometric, not a metric. The induced "topology" would not be Hausdorff, but $\sigma(X, \mathcal{F})$ might also fail to be Hausdorff in that case (it is Hausdorff iff $\mathcal{F}$ separates points, since each $Y_n$ is Hausdorff being metric). So the hypothesis is exactly what is needed to make both topologies Hausdorff and metric, simultaneously.
**Where countability of $\mathcal{F}$ is used.** The series $\sum 2^{-n} \tilde{d}_n$ has only countably many terms; for an uncountable family $\mathcal{F}$, no such single metric can exist in general (the topology may fail to be first-countable, hence not metrizable).
[/guided]
[/step]
