> [!definition] > > Let $\bracs{(X_i, \topo_i)}_{i \in I}$ be a family of [[Topological Space|topological spaces]] and $X = \prod_{i \in I}X_i$ be their [[Cartesian Product|Cartesian product]]. The **product topology** $\topo$ on $X$ is the [[Weak Topology|weak topology]] generated by the projection maps $\pi_i: X \to X_i$, $f \mapsto f(i)$, which coincides with the topology generated by the **cylinder sets** > $ > \ce = \bracs{ > \pi_i^{-1}(U_i): U_i \in \topo_i, i \in I > } > $ > which is the smallest topology that allows the projection maps to be [[Continuity|continuous]]. > > Since a generated topology consists of finite intersections of these cylinder sets, a [[Base|base]] for the product topology is > $ > \cb = \bracs{\prod_{i \in I}U_i: U_i \in \topo_i, U_i \ne X_i \text{ for finitely many }i \in I} > $ > [!theorem] > > Let $\bracs{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces, $(X, \topo)$ be their product, and $(Y, \topo_Y)$ be another topological space. A mapping $f: Y \to X$ is [[Continuity|continuous]] if and only if $\pi_i \circ f: Y \to X_i$ is continuous for all $i \in I$ (continuous in each coordinate). > > *Proof*. Suppose that $f$ is continuous, then since $\pi_i$ is continuous as well, $\pi_i \circ f$ is continuous. > > Suppose that each $\pi_i \circ f$ is continuous, then for any [[Open Set|open set]] $U_i \in \topo_i$, we have > $ > (\pi_i \circ f)^{-1}(U_i) = f^{-1} \circ \braks{\pi_i^{-1}(U_i)} \in \topo_Y > $ > As the preimage of all cylinder sets are open, and as the cylinder sets generate the product topology, $f$ is continuous. > [!theorem] > > Let $\bracs{(X_i, \topo_i)}_{i \in I}$ the same topological space for all $i \in I$, and $(X, \topo)$ be the product topology. Let $\seq{f_n} \subset X$ be a [[Sequence|sequence]] in $X$. Then $f_n \to f$ [[Limit|converges]] in the product topology, if and only if $\pi_i(f_n) \to \pi_i(f)$ for all $i \in I$ ($f_n \to f$ [[Pointwise Convergence|pointwise]]). > > *Proof*. Let $J \subset I$ be a finite subset and define > $ > N\paren{\bracs{U_j}_{j \in J}} = \bigcap_{j \in J}\pi_j^{-1}(U_j) > $ > where each $U_j$ is an open [[Neighbourhood|neighbourhood]] of $\pi_j(f)$, then $N$ is also an neighbourhood of $f$. > > Let > $ > \cn = \bracs{N\paren{\bracs{U_j}_{j \in J}}: > U_j\text{ neighbourhood of }\pi_j(f), > J \subset I \text{ finite}} > $ > be the collection of all such sets. Let $U \in \topo$ be an open neighbourhood of $f$, then $U$ is a union of finite intersections of cylinder sets. Since $f \in U$, > $ > \exists \bracs{U_j}_{j \in J}, J \subset I \text{ finite}, U_j \text{ open}: f \in \bigcap_{j \in J}\pi_j^{-1}(U_j) > $ > which means that $\pi_j(f) \in U_j$ for all $j \in J$, and $N(\bracs{U_j}_{j \in J}) \in \cn$ is an open neighbourhood of $f$. Therefore $\cn$ is a [[Neighbourhood Base|neighbourhood base]] of $f$. > > Suppose that $f_n \to f$ pointwise. Let $U$ be an open neighbourhood of $f$, then > $ > \exists N\paren{\bracs{U_j}_{j \in J}} \in \cn: f \in N\paren{\bracs{U_j}_{j \in J}} \subset U > $ > Since $f_n \to f$ pointwise, for each $U_j$ there exists some $N_j \in \nat$ such that $\pi_j(f_n) \in U_j$ for all $n \ge N_j$. Take $N = \max_{j \in J}N_j$, then $f_n \in N\paren{\bracs{U_j}_{j \in J}} \subset U$ for all $n \ge N$, and $f_n \to f$ in the product topology. > > Suppose that $f_n \to f$ in the product topology. Let $i \in I$, and $U_i \in \topo_i$ be a neighbourhood of $\pi_i(f)$, then $\pi^{-1}_i(U_i) \in \topo$ is a neighbourhood of $f$. Since $f_n \to f$, there exists $N \in \nat$ such that $f_n \in \pi_1^{-1}(U_i)$ for all $n \ge N$, which means that $\pi(f_n) \in U_i$ for all $n \ge N$. Therefore $f$ converges pointwise. > [!definition] > > Let $(X_i, d_k)$ be a list of [[Metric Space|metric spaces]]. Then let > $ > d^\prime_k(x_k, y_k) = \frac{d_k(x_k, y_k)}{1 + d_k(x_k, y_k)} > $ > and define > $ > D(x, y) = \sum_{k = 1}^{\infty}\frac{d^\prime_k(x_k, y_k)}{2^k} > $ > Then the product topology on $X = \prod_{k = 1}^{\infty}X_i$ is also a metric topology.