Product Space of Subspaces is Subspace of Product Space

Theorem
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.

Let $\displaystyle T = \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.

Let $\family {\struct {Y_i, \nu_i} }_{i \mathop \in I}$ be a family of topological spaces such that:
 * $\forall i \in I : \struct {Y_i, \nu_i}$ is a topological subspace of $\struct {X_i, \tau_i}$

Let $\displaystyle S = \struct {Y, \nu} = \prod_{i \mathop \in I} \struct {Y_i, \nu_i}$ be the product space of $\family {\struct {Y_i, \nu_i} }_{i \mathop \in I}$.

Let $T_Y = \struct {Y, \tau_Y}$ be the topological subspace of $T$.

Then $S = T_Y$.

Proof
From Leigh.Samphier/Sandbox/Cartesian Product of Subsets/Family of Subsets, $Y \subseteq X$.

Thus the topological subspace $T_Y$ is well-defined.

From Natural Basis of Tychonoff Topology,
 * $\\B_T$

is a [Definition:Synthetic Basis|(synthetic) basis]] for $T$.

From Basis for Topological Subspace,
 * $\BB_Y$

is a (synthetic) basis for $T_Y$.

By definition subspace topology,
 * $\forall i \in I: \nu_i = \set{U \cap S_i : U \in \tau_i}$

From Natural Basis of Tychonoff Topology,
 * $\BB_S$

is a (synthetic) basis for $S$.

It remains to show that $\BB_Y = \BB_S$.