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

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

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 a (synthetic) basis for $T$ is:
 * $\displaystyle \BB_T = \set{\prod_{i \mathop \in I} U_i : U_i \in \tau_i, U_i = X_i \text{ for all but finitely many } i \in I }$

From Basis for Topological Subspace a (synthetic) basis for $T_Y$ is:
 * $\BB_Y = \set{U \cap Y: U \in \BB_T}$

That is:

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

From Natural Basis of Tychonoff Topology a (synthetic) basis for $S$ is:
 * $\displaystyle \BB_S = \set{\prod_{i \mathop \in I} V_i : V_i \in \upsilon_i, V_i = Y_i \text{ for all but finitely many } i \in I }$

That is:

Thus $\BB_S = \BB_Y$ and therefore $S = T_Y$.