Product Space of Subspaces is Subspace of Product Space

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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 $\ds T = \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ 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 $\ds S = \struct {Y, \upsilon} = \prod_{i \mathop \in I} \struct {Y_i, \upsilon_i}$ 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 Cartesian Product of Family of Subsets, $Y \subseteq X$.

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


From Natural Basis of Product Topology, a (synthetic) basis for $T$ is:

$\ds \BB_T = \set {\prod_{i \mathop \in I} U_i : U_i \in \tau_i, U_i = X_i \text{ for all but finitely many } i \in I}$

From Basis for Topological Subspace a (synthetic) basis for $T_Y$ is:

$\BB_Y = \set {U \cap Y: U \in \BB_T}$

That is:

\(\ds \BB_Y\) \(=\) \(\ds \set {\paren{ \prod_{i \mathop \in I} U_i} \bigcap \paren{\prod_{i \mathop \in I} Y_i}: U_i \in \tau_i, U_i = X_i \text{ for all but finitely many } i \in I}\)
\(\ds \) \(=\) \(\ds \set {\prod_{i \mathop \in I} \paren {U_i \cap Y_i} : U_i \in \tau_i, U_i = X_i \text{ for all but finitely many } i \in I}\) General Case of Cartesian Product of Intersections


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 Product Topology a (synthetic) basis for $S$ is:

$\ds \BB_S = \set {\prod_{i \mathop \in I} V_i : V_i \in \upsilon_i, V_i = Y_i \text{ for all but finitely many } i \in I}$

That is:

\(\ds \BB_S\) \(=\) \(\ds \set {\prod_{i \mathop \in I} \paren{U_i \cap Y_i} : U_i \in \tau_i, U_i \cap Y_i = Y_i \text{ for all but finitely many } i \in I}\)
\(\ds \) \(=\) \(\ds \set {\prod_{i \mathop \in I} \paren{U_i \cap Y_i} : U_i \in \tau_i, U_i = X_i \text{ for all but finitely many } i \in I}\) as $Y_i \subseteq X_i$ for all $i \in I$

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

$\blacksquare$

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