4.7 Product Spaces
Definition 4.7.1 (Product Topology). Let $\bracsn{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces, then the product topology $\topo$ is the initial topology generated by the projections $\seqi{\pi}$, and:
The family
\[\cb(\ce) = \bracs{\bigcap_{k = 1}^n \pi_{i_k}^{-1}(U_{k}) \bigg | U_{k} \in \topo_{i_k}, \seqf{i_k} \subset I, n \in \nat^+}\]is a base for $\topo$.
For any topological space space $Y$ and $\seqi{f}$ where $f_{i} \in C(Y; X_{i})$ for all $i \in I$, there exists a unique $f \in C(Y; X)$ such that the following diagram commutes
\[\xymatrix{ Y \ar@{->}[d]_{f} \ar@{->}[rd]^{f_i} & \\ X \ar@{->}[r]_{\pi_i} & X_i }\]for all $i \in I$.
Proof. (1): By (3) of Definition 4.1.7.
(U): Let $f \in \prod_{i \in I}f_{i}$, then $f$ is the unique function such that the diagrams commute. For each $\bigcap_{k = 1}^{n} \pi_{i_k}^{-1}(U_{k}) \in \topo$,
which is open in $Y$ by (O2).$\square$
Proposition 4.7.2. Let $\bracsn{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces and $\B$ be a filter base on $\prod_{i \in I}X_{i}$, then $\fB$ converges to $x \in \prod_{i \in I}X_{i}$ if and only if $\pi_{i}(\fB)$ converges to $\pi_{i}(x)$ for all $i \in I$.
Proof. $(\Rightarrow)$: Let $i \in I$ and $U \in \cn(\pi_{i}(x))$, then $\pi_{i}^{-1}(U) \in \cn(x)$. Since $\fB$ converges to $x$, there exists $B \in \fB$ with $B \subset \pi_{i}^{-1}(U)$. In which case, $\pi_{i}(B) \subset U$ and $\pi_{i}(\fB)$ converges to $\pi_{i}(x)$.
$(\Leftarrow)$: Let $U \in \cn(x)$, then there exists $\seqf{i_k}$ and open sets $\seqf{U_k}$ such that $x \in \bigcap_{k = 1}^{n} \pi_{i_k}^{-1}(U_{k}) \subset U$. For each $1 \le k \le n$, since $\pi_{i_k}(\fB)$ converges to $\pi_{i_k}(x)$ for each $1 \le k \le n$, there exists $B_{k} \in \fB$ such that $B_{k} \subset \pi_{i_k}^{-1}(U_{k})$. As $\fB$ is a filter base, there exists $B \in \fB$ with
Therefore $\fB$ converges to $x$.$\square$
Proposition 4.7.3. Let $X$ be a topological space, $\seqi{Y}$ be topological spaces, $\seqi{f}$ where $f_{i} \in C(X; Y_{i})$ for each $i \in I$, then
There exists a unique $f \in C(X; \prod_{i \in I}Y_{i})$ such that the following diagram commutes
\[\xymatrix{ & \prod_{i \in I} Y_i \ar@{->}[rd]^{\pi_i} & \\ X \ar@{->}[rr]_{f_i} \ar@{->}[ru]^{f} & & Y_i }\]for all $i \in I$.
If $X$ is T0 and equipped with the initial topology induced by $\seqi{f}$, then $f$ is an embedding.
Proof. (1): By (U) of Definition 4.7.1.
(2): Let $x, y \in X$ with $x \ne y$. Assume without loss of generality that there exists $U \in \cn(x)$ such that $y \not\in U$. Let $J \subset I$ finite, $\seqj{U}$, and $\seqj{V}$ such that
then $f(x) \in \bigcap_{j \in J}\pi_{j}^{-1}(U_{j})$ but $f(y) \not\in \bigcap_{j \in J}\pi_{j}^{-1}(U_{j})$. Thus $f$ is injective.
Let $U \subset X$ be open. Assume without loss of generality that there exists $J \subset I$ finite and $\bracs{U_j}_{j \in J}$ such that $U = \bigcap_{j \in J}f_{j}^{-1}(U_{j})$. In which case, $\bigcap_{j \in J}\pi_{j}^{-1}(U_{j})$ is open in $\prod_{i \in I}Y_{i}$ and
is a relatively open set.$\square$
Proposition 4.7.4. Let $\seqi{X}$, $\seqi{Y}$ be topological/uniform spaces and $\seqi{f}$ such that $f_{i} \in C(X_{i}; Y_{i})$ is an embedding for all $i \in I$, then
There exists a unique $f \in C(\prod_{i \in I}X_{i}; \prod_{i \in I}Y_{i})$/$f \in UC(\prod_{i \in I}X_{i}; \prod_{i \in I}Y_{i})$ such that the following diagram commutes
\[\xymatrix{ \prod_{i \in I}Y_i \ar@{->}[r]^{\pi_i} & Y_i \\ \prod_{i \in I} X_i \ar@{->}[r]_{\pi_i} \ar@{->}[u]^{f} & X_i \ar@{^{(}->}[u]_{f_i} }\]for all $i \in I$.
$f$ is an embedding.
Proof. (1): By (U) of Definition 4.7.1/Definition 5.2.5.
(2): Consider the following diagram
Since each $X_{i} \to Y_{i}$ is an embedding, the composition
is continuous/uniformly continuous. By (U) of Definition 4.7.1/Definition 5.2.5, $f$ is an embedding.$\square$