[proofplan] We prove both directions of the equivalence by reducing net convergence in the product to convergence in each factor. The forward direction is immediate from the [continuity](/page/Continuity) of projections: composing a convergent net with a continuous map preserves convergence. The reverse direction uses the subbasis characterisation of the product [topology](/page/Topology) — every basic [open set](/page/Open%20Set) in the product is a finite intersection of preimages $\pi_\alpha^{-1}(U_\alpha)$, so coordinatewise convergence forces the net to eventually enter every such basic open neighbourhood. [/proofplan]

[step:Compose the convergent net with each continuous projection to get coordinatewise convergence]Suppose $(x_\lambda)_{\lambda \in \Lambda} \to x$ in $X$. Fix $\beta \in A$. By the [Universal Property of the Product Topology](/theorems/962), the projection $\pi_\beta: X \to X_\beta$ is [continuous](/page/Continuity). Since the continuous image of a convergent net is convergent, $\pi_\beta(x_\lambda) \to \pi_\beta(x)$ in $X_\beta$.[/step]

[guided]Suppose $(x_\lambda)_{\lambda \in \Lambda} \to x$ in $X = \prod_{\alpha \in A} X_\alpha$. We want to show that the $\beta$-th coordinate converges in $X_\beta$ for every $\beta \in A$. Fix $\beta \in A$. By the [Universal Property of the Product Topology](/theorems/962), the product [topology](/page/Topology) is defined precisely so that every projection $\pi_\beta: X \to X_\beta$ is continuous. Now let $V \subset X_\beta$ be any [open set](/page/Open%20Set) containing $\pi_\beta(x)$. Since $\pi_\beta$ is continuous, the preimage $\pi_\beta^{-1}(V)$ is open in $X$ and contains $x$. By definition of net convergence, there exists $\lambda_0 \in \Lambda$ such that $x_\lambda \in \pi_\beta^{-1}(V)$ for all $\lambda \ge \lambda_0$. Applying $\pi_\beta$, we obtain $\pi_\beta(x_\lambda) \in V$ for all $\lambda \ge \lambda_0$. Since $V$ was an arbitrary open neighbourhood of $\pi_\beta(x)$ in $X_\beta$, this establishes $\pi_\beta(x_\lambda) \to \pi_\beta(x)$. This direction uses only the continuity of projections and does not require the product topology to be the *coarsest* such topology — any topology making all projections continuous would suffice.[/guided]

[step:Show coordinatewise convergence forces the net into every subbasis neighbourhood]Suppose $\pi_\alpha(x_\lambda) \to \pi_\alpha(x)$ in $X_\alpha$ for every $\alpha \in A$. We show $(x_\lambda) \to x$ in the product [topology](/page/Topology). A subbasis for the product topology consists of sets of the form $\pi_\beta^{-1}(U_\beta)$ where $\beta \in A$ and $U_\beta \subset X_\beta$ is open. Let $G \subset X$ be an [open set](/page/Open%20Set) containing $x$. By definition of the product topology, $G$ contains a basic open neighbourhood of $x$ of the form \begin{align*} B = \pi_{\beta_1}^{-1}(U_{\beta_1}) \cap \pi_{\beta_2}^{-1}(U_{\beta_2}) \cap \cdots \cap \pi_{\beta_m}^{-1}(U_{\beta_m}) \end{align*} where $\beta_1, \ldots, \beta_m \in A$ and each $U_{\beta_j} \subset X_{\beta_j}$ is open with $\pi_{\beta_j}(x) \in U_{\beta_j}$. For each $j \in \{1, \ldots, m\}$, the hypothesis $\pi_{\beta_j}(x_\lambda) \to \pi_{\beta_j}(x)$ provides an index $\lambda_j \in \Lambda$ such that $\pi_{\beta_j}(x_\lambda) \in U_{\beta_j}$ for all $\lambda \ge \lambda_j$. Since $\Lambda$ is a directed set and $\{1, \ldots, m\}$ is finite, there exists $\lambda_0 \in \Lambda$ with $\lambda_0 \ge \lambda_j$ for all $j = 1, \ldots, m$. For every $\lambda \ge \lambda_0$, we have $\pi_{\beta_j}(x_\lambda) \in U_{\beta_j}$ for all $j$, hence $x_\lambda \in B \subset G$. Since $G$ was an arbitrary open neighbourhood of $x$, we conclude $(x_\lambda) \to x$.[/step]

[guided]Suppose $\pi_\alpha(x_\lambda) \to \pi_\alpha(x)$ in $X_\alpha$ for every $\alpha \in A$. We must show that $(x_\lambda) \to x$ in $X$ with the product topology. To establish net convergence, we must show: for every open set $G \subset X$ containing $x$, there exists $\lambda_0 \in \Lambda$ such that $x_\lambda \in G$ for all $\lambda \ge \lambda_0$. Instead of working with arbitrary open sets directly, we exploit the product topology's structure. The product topology on $X = \prod_{\alpha \in A} X_\alpha$ has the subbasis $\mathcal{S} = \{\pi_\beta^{-1}(U_\beta) : \beta \in A,\, U_\beta \in \tau_\beta\}$. By definition, the basic open sets are finite intersections of subbasis elements. Every open set $G$ containing $x$ contains some basic open neighbourhood of $x$, which has the form \begin{align*} B = \pi_{\beta_1}^{-1}(U_{\beta_1}) \cap \pi_{\beta_2}^{-1}(U_{\beta_2}) \cap \cdots \cap \pi_{\beta_m}^{-1}(U_{\beta_m}) \end{align*} for finitely many indices $\beta_1, \ldots, \beta_m \in A$ and open sets $U_{\beta_j} \subset X_{\beta_j}$ with $\pi_{\beta_j}(x) \in U_{\beta_j}$ for each $j$. This is the step where the *finiteness* of the intersection is essential — and where the product topology differs from the box topology. The box topology would allow arbitrary (possibly infinite) intersections of conditions, and the argument below would fail because a directed set need not have an upper bound for infinitely many elements. For each $j \in \{1, \ldots, m\}$, we use the hypothesis that $\pi_{\beta_j}(x_\lambda) \to \pi_{\beta_j}(x)$ in $X_{\beta_j}$. Since $U_{\beta_j}$ is an open neighbourhood of $\pi_{\beta_j}(x)$, there exists $\lambda_j \in \Lambda$ such that \begin{align*} \pi_{\beta_j}(x_\lambda) \in U_{\beta_j} \quad \text{for all } \lambda \ge \lambda_j. \end{align*} We now have finitely many indices $\lambda_1, \ldots, \lambda_m \in \Lambda$. Since $\Lambda$ is a directed set, for any finite collection of elements there exists an upper bound: choose $\lambda_0 \in \Lambda$ with $\lambda_0 \ge \lambda_j$ for all $j = 1, \ldots, m$. For every $\lambda \ge \lambda_0$, we have $\lambda \ge \lambda_j$ for each $j$, so $\pi_{\beta_j}(x_\lambda) \in U_{\beta_j}$ for all $j = 1, \ldots, m$. By definition of the basic open set $B$, this means $x_\lambda \in B \subset G$. Since $G$ was an arbitrary open neighbourhood of $x$ in $X$, we conclude $(x_\lambda) \to x$ in the product topology.[/guided]

[step:Specialize to sequences in countable products] The "in particular" statement follows immediately: a sequence is a net indexed by $(\mathbb{N}, \le)$, which is a directed set. Therefore the equivalence above applies to sequences without modification. [/step]