[proofplan] We construct a candidate metric $D$ on $\prod_{k=1}^\infty X_k$ by summing weighted bounded metrics $\bar{d}_k = \frac{d_k}{1 + d_k}$ with geometrically decaying coefficients: $D(x,y) = \sum_{k=1}^\infty 2^{-k} \bar{d}_k(x_k, y_k)$. We verify that $D$ is a metric (the triangle inequality reduces to the triangle inequality for each $\bar{d}_k$), then show that the $D$-metric [topology](/page/Topology) coincides with the product topology by proving mutual containment of basis elements: every product-topology subbasis set contains a $D$-ball, and every $D$-ball contains a product-topology basis set. [/proofplan]

[step:Construct bounded metrics from each factor metric]For each $k \in \mathbb{N}$, define the bounded metric \begin{align*} \bar{d}_k: X_k \times X_k &\to [0, 1) \\ (a, b) &\mapsto \frac{d_k(a, b)}{1 + d_k(a, b)}. \end{align*} The function $\bar{d}_k$ is a metric on $X_k$. Symmetry and positive-definiteness are inherited from $d_k$. For the triangle inequality, the function $\varphi: [0, \infty) \to [0, 1)$ defined by $\varphi(t) = \frac{t}{1+t}$ is concave and increasing with $\varphi(0) = 0$, so for all $a, b, c \in X_k$, \begin{align*} \bar{d}_k(a, c) = \varphi(d_k(a, c)) \le \varphi(d_k(a, b) + d_k(b, c)) \le \varphi(d_k(a, b)) + \varphi(d_k(b, c)) = \bar{d}_k(a, b) + \bar{d}_k(b, c), \end{align*} where the first inequality uses that $\varphi$ is increasing and the second uses the subadditivity of concave functions vanishing at the origin: for $s, t \ge 0$, $\varphi(s + t) \le \varphi(s) + \varphi(t)$. Moreover, $\bar{d}_k$ induces the same [topology](/page/Topology) on $X_k$ as $d_k$, since $\varphi$ is a homeomorphism of $[0, \infty)$ onto $[0, 1)$ with inverse $\varphi^{-1}(u) = \frac{u}{1-u}$, so a sequence converges in one metric if and only if it converges in the other.[/step]

[guided]The idea behind replacing $d_k$ with $\bar{d}_k$ is to ensure uniform boundedness: we need the series defining $D$ to converge, so we must prevent any single factor from dominating. The function $\varphi(t) = t/(1+t)$ maps $[0, \infty)$ bijectively onto $[0, 1)$, producing a metric bounded by $1$. Why does $\bar{d}_k$ satisfy the triangle inequality? The key property is that $\varphi$ is **concave** on $[0, \infty)$: we compute $\varphi'(t) = (1+t)^{-2} > 0$ and $\varphi''(t) = -2(1+t)^{-3} < 0$. A concave function $\varphi: [0, \infty) \to \mathbb{R}$ with $\varphi(0) = 0$ is subadditive: for $s, t \ge 0$, \begin{align*} \varphi(s + t) &= \varphi\!\left(\frac{s}{s+t}(s+t) + \frac{t}{s+t}(s+t)\right) \\ &\le \frac{s}{s+t}\varphi(s+t) + \frac{t}{s+t}\varphi(s+t) \end{align*} is not directly helpful. Instead, we use concavity as follows: since $\varphi(0) = 0$ and $\varphi$ is concave, for any $s \ge 0$ and $s + t \ge s$, \begin{align*} \varphi(s + t) = \varphi\!\left(\frac{s}{s+t} \cdot (s+t) + \frac{t}{s+t} \cdot 0 + \frac{t}{s+t} \cdot (s+t)\right). \end{align*} More directly: write $s = (s+t) \cdot \frac{s}{s+t}$, so $\varphi(s) \ge \frac{s}{s+t}\varphi(s+t)$ by concavity (the line from $(0,0)$ to $(s+t, \varphi(s+t))$ lies below the concave graph). Similarly $\varphi(t) \ge \frac{t}{s+t}\varphi(s+t)$. Adding: $\varphi(s) + \varphi(t) \ge \varphi(s+t)$. With subadditivity of $\varphi$ and monotonicity established, the triangle inequality for $\bar{d}_k$ follows: \begin{align*} \bar{d}_k(a, c) &= \varphi(d_k(a, c)) \le \varphi(d_k(a, b) + d_k(b, c)) \le \varphi(d_k(a, b)) + \varphi(d_k(b, c)) = \bar{d}_k(a, b) + \bar{d}_k(b, c). \end{align*} Finally, $\bar{d}_k$ and $d_k$ generate the same topology: for any $\varepsilon > 0$, the ball $B_{\bar{d}_k}(a, \varphi(\varepsilon)) \subset B_{d_k}(a, \varepsilon)$ (since $\bar{d}_k(a,b) < \varphi(\varepsilon)$ implies $d_k(a,b) < \varepsilon$ by monotonicity of $\varphi$), and conversely $B_{d_k}(a, \varphi^{-1}(\varepsilon)) \subset B_{\bar{d}_k}(a, \varepsilon)$. So each $d_k$-ball contains a $\bar{d}_k$-ball around the same centre, and vice versa.[/guided]

[step:Define the candidate metric $D$ on the product and verify the metric axioms]Define \begin{align*} D: \prod_{k=1}^\infty X_k \times \prod_{k=1}^\infty X_k &\to [0, \infty) \\ (x, y) &\mapsto \sum_{k=1}^\infty 2^{-k}\, \bar{d}_k(x_k, y_k), \end{align*} where $x = (x_1, x_2, \ldots)$ and $y = (y_1, y_2, \ldots)$ are elements of $\prod_{k=1}^\infty X_k$. **Convergence.** Since $0 \le \bar{d}_k(x_k, y_k) < 1$ for all $k$, each term satisfies $0 \le 2^{-k}\bar{d}_k(x_k, y_k) < 2^{-k}$. The series is bounded above by the geometric series $\sum_{k=1}^\infty 2^{-k} = 1$, so $D(x,y)$ converges absolutely and $D(x,y) \in [0, 1]$ for all $x, y$. **Positive-definiteness.** If $x = y$, then $\bar{d}_k(x_k, y_k) = 0$ for all $k$, so $D(x,y) = 0$. Conversely, if $D(x,y) = 0$, then every term $2^{-k}\bar{d}_k(x_k, y_k) = 0$, hence $\bar{d}_k(x_k, y_k) = 0$ for all $k$, which gives $x_k = y_k$ for all $k$, so $x = y$. **Symmetry.** Each $\bar{d}_k$ is symmetric, so $D(x,y) = D(y,x)$. **Triangle inequality.** For all $x, y, z \in \prod_{k=1}^\infty X_k$, \begin{align*} D(x, z) &= \sum_{k=1}^\infty 2^{-k}\,\bar{d}_k(x_k, z_k) \le \sum_{k=1}^\infty 2^{-k}\bigl(\bar{d}_k(x_k, y_k) + \bar{d}_k(y_k, z_k)\bigr) = D(x, y) + D(y, z), \end{align*} where the inequality applies the triangle inequality for $\bar{d}_k$ termwise, and the rearrangement of the series is justified by absolute convergence.[/step]

[guided]The metric $D$ is a weighted $\ell^1$-combination of bounded metrics. The geometric weights $2^{-k}$ serve two purposes: they ensure convergence of the series (since each $\bar{d}_k$ is bounded by $1$), and they make the contributions of later factors progressively less significant — reflecting the product [topology](/page/Topology)'s requirement that basic [open sets](/page/Open%20Set) constrain only finitely many coordinates. The verification of the metric axioms is carried out in the exact version above. The triangle inequality — the most substantive axiom to check — follows by applying the triangle inequality for each $\bar{d}_k$ inside the sum and using linearity of summation (justified by absolute convergence — the partial sums of non-negative terms form a monotone sequence).[/guided]

[step:Show that every product-topology subbasis set is $D$-open]A subbasis for the product [topology](/page/Topology) on $\prod_{k=1}^\infty X_k$ consists of sets of the form $\pi_j^{-1}(V)$ where $j \in \mathbb{N}$ and $V \subset X_j$ is open. We show each such set is open in the $D$-metric topology. Fix $j \in \mathbb{N}$ and $V \subset X_j$ open. Let $x \in \pi_j^{-1}(V)$, so $x_j \in V$. Since $V$ is open in the $d_j$-topology (which coincides with the $\bar{d}_j$-topology), there exists $\varepsilon > 0$ such that $B_{\bar{d}_j}(x_j, \varepsilon) \subset V$. Set $\delta = 2^{-j}\varepsilon$. If $D(x, y) < \delta$, then \begin{align*} 2^{-j}\,\bar{d}_j(x_j, y_j) \le D(x, y) < \delta = 2^{-j}\varepsilon, \end{align*} so $\bar{d}_j(x_j, y_j) < \varepsilon$, giving $y_j \in B_{\bar{d}_j}(x_j, \varepsilon) \subset V$, hence $y \in \pi_j^{-1}(V)$. Therefore $B_D(x, \delta) \subset \pi_j^{-1}(V)$, and $\pi_j^{-1}(V)$ is $D$-open. Since the product topology is generated by these subbasis sets and every subbasis set is $D$-open, every product-[open set](/page/Open%20Set) is $D$-open. That is, $\tau_{\mathrm{prod}} \subset \tau_D$.[/step]

[guided]To show the product topology is coarser than the $D$-metric topology, it suffices to check on a subbasis. The product topology on a [countable](/page/Countable%20Set) product $\prod_{k=1}^\infty X_k$ has a standard subbasis: the sets $\pi_j^{-1}(V)$ where $V$ is open in $X_j$. Why does the estimate work? The key observation is that the $j$-th term of the series defining $D$ controls the $j$-th coordinate: since all terms are non-negative, $2^{-j}\bar{d}_j(x_j, y_j) \le D(x,y)$. So if $D(x,y)$ is small, then $\bar{d}_j(x_j, y_j)$ is small (up to the factor $2^j$). This allows us to convert a $D$-ball at $x$ into a constraint on the $j$-th coordinate alone. Specifically, for $x \in \pi_j^{-1}(V)$ with $B_{\bar{d}_j}(x_j, \varepsilon) \subset V$, the $D$-ball of radius $\delta = 2^{-j}\varepsilon$ around $x$ satisfies $B_D(x, \delta) \subset \pi_j^{-1}(V)$, because $D(x,y) < 2^{-j}\varepsilon$ forces $\bar{d}_j(x_j, y_j) < \varepsilon$.[/guided]

[step:Show that every $D$-ball contains a product-topology basis set]Fix $x \in \prod_{k=1}^\infty X_k$ and $\varepsilon > 0$. We produce a product-[topology](/page/Topology) [open set](/page/Open%20Set) $U$ with $x \in U \subset B_D(x, \varepsilon)$. Choose $N \in \mathbb{N}$ large enough that \begin{align*} \sum_{k=N+1}^\infty 2^{-k} < \frac{\varepsilon}{2}. \end{align*} This is possible since $\sum_{k=N+1}^\infty 2^{-k} = 2^{-N} \to 0$ as $N \to \infty$. For each $k \in \{1, \ldots, N\}$, set $\varepsilon_k = \frac{\varepsilon}{2N}$ and define \begin{align*} U = \bigcap_{k=1}^N \pi_k^{-1}\!\bigl(B_{\bar{d}_k}(x_k, \varepsilon_k)\bigr). \end{align*} This is a finite intersection of subbasis sets, hence a basic open set in the product topology, and $x \in U$. For any $y \in U$, we have $\bar{d}_k(x_k, y_k) < \varepsilon_k$ for $k = 1, \ldots, N$ and $\bar{d}_k(x_k, y_k) < 1$ for all $k$, so \begin{align*} D(x, y) &= \sum_{k=1}^N 2^{-k}\,\bar{d}_k(x_k, y_k) + \sum_{k=N+1}^\infty 2^{-k}\,\bar{d}_k(x_k, y_k) \\ &< \sum_{k=1}^N 2^{-k} \cdot \frac{\varepsilon}{2N} + \sum_{k=N+1}^\infty 2^{-k} \cdot 1 \\ &\le N \cdot \frac{\varepsilon}{2N} + \frac{\varepsilon}{2} = \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Therefore $U \subset B_D(x, \varepsilon)$, which shows that every $D$-ball is product-open, i.e., $\tau_D \subset \tau_{\mathrm{prod}}$.[/step]

[guided]This is the more delicate direction: we must show that the $D$-metric topology is not finer than the product topology. The difficulty is that a $D$-ball constrains *all* coordinates simultaneously, while a product-topology basic open set constrains only *finitely many*. The strategy is a **tail truncation argument**. We split the series $D(x,y) = \sum_{k=1}^\infty 2^{-k}\bar{d}_k(x_k, y_k)$ into a finite head (indices $1, \ldots, N$) and a tail (indices $N+1, N+2, \ldots$). We choose $N$ large enough that the tail is unconditionally small — less than $\varepsilon/2$ — regardless of the values $y_k$ for $k > N$. This is possible because $\bar{d}_k \le 1$ and $\sum_{k=N+1}^\infty 2^{-k} = 2^{-N}$, which can be made arbitrarily small. For the head, we constrain each of the finitely many coordinates $k = 1, \ldots, N$ to lie in a $\bar{d}_k$-ball of radius $\varepsilon_k = \varepsilon/(2N)$. The total contribution from the head is then \begin{align*} \sum_{k=1}^N 2^{-k} \cdot \frac{\varepsilon}{2N} \le N \cdot \frac{\varepsilon}{2N} = \frac{\varepsilon}{2}, \end{align*} using the crude bound $2^{-k} \le 1$. Adding head and tail gives $D(x,y) < \varepsilon$. The set $U = \bigcap_{k=1}^N \pi_k^{-1}(B_{\bar{d}_k}(x_k, \varepsilon_k))$ is a **finite** intersection of subbasis sets, hence a basic open set in the product topology. This is where the product topology's defining feature — constraining only finitely many coordinates — aligns with the tail truncation.[/guided]

[step:Conclude that $D$ metrizes the product topology] The inclusions $\tau_{\mathrm{prod}} \subset \tau_D$ (from the subbasis step) and $\tau_D \subset \tau_{\mathrm{prod}}$ (from the ball containment step) give $\tau_D = \tau_{\mathrm{prod}}$. Therefore $D$ is a metric on $\prod_{k=1}^\infty X_k$ that induces the product [topology](/page/Topology), proving that the [countable](/page/Countable%20Set) product is metrizable. [/step]