Factor Spaces are T4 if Product Space is T4

Theorem
Let $\mathbb S = \family {\struct{S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces for $\alpha$ in some indexing set $I$.

Let $\displaystyle T = \struct {S, \tau} = \displaystyle \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\mathbb S$.

Let $T$ be a $T_4$ space.

Then each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_4$ space.

Proof
Since $S_\alpha \ne \O$ we also have $S \ne \O$ by the axiom of choice.

Let $z \in S$.

From Subspace of Product Space Homeomorphic to Factor Space, every $\struct {S_\alpha, \tau_\alpha}$ is homeomorphic to the subspace $T_\alpha$ of $T$ defined by:
 * $T_\alpha = \set {x \in S: \forall \beta \in I \setminus \set \alpha: x_\beta = z_\beta}$

and the homeomorphism $h : S_\alpha \to T_\alpha$ is given by:
 * $\forall x_\alpha \in S_\alpha : \map h {x_\alpha} = x : \forall \beta \in I : x_\beta = \begin {cases} x_\alpha & \beta = \alpha \\ z_\beta & \beta \ne \alpha \end {cases}$

From Inclusion Mapping is Continuous, the inclusion mapping $i: T_\alpha \to S$ is continuous.

From Continuity of Composite Mapping, the composite mapping $i \circ h : S_\alpha \to S$ is continuous.

Let $\pr_\alpha: S \to S_\alpha$ be the projection of $S$ to $S_\alpha$.

For all $x_\alpha \in S_\alpha$:
 * $\map {\pr_\alpha} {\map {i \circ h} {x_\alpha}} = x_\alpha$

Let $A, B \subseteq S_\alpha$ be an arbitrary pair of disjoint closed sets in $\struct {S_\alpha, \tau_\alpha}$.

Then:

From Projection from Product Topology is Open and Continuous:General Result
 * $\pr_\alpha: S \to S_\alpha$ is continuous.

From Continuity Defined from Closed Sets:
 * $\map {\pr_\alpha^\gets} A, \map {\pr_\alpha^\gets} B$ are disjoint closed sets in $T$

By definition of $T_4$ space:
 * $\exists U, V \in \tau : \map {\pr_\alpha^\gets} A \subseteq U, \map {\pr_\alpha^\gets} B \subseteq V : U \cap V = \O$

By definition of continuity:
 * $\map {\paren {i \circ h}^\gets} U, \map {\paren{i \circ h}^\gets} V$ are open sets in $S_\alpha$.

From Preimage of Intersection under Mapping:
 * $\map {\paren {i \circ h}^\gets} U, \map {\paren{i \circ h}^\gets} V$ are disjoint open sets in $S_\alpha$.

From Preimage of Subset is Subset of Preimage:
 * $\map {\paren {i \circ h}^\gets} {\map {\pr_\alpha^\gets} A} \subseteq \map {\paren{i \circ h}^\gets} U$
 * $\map {\paren {i \circ h}^\gets} {\map {\pr_\alpha^\gets} B} \subseteq \map {\paren{i \circ h}^\gets} V$

Now:

From the definition of set equality:
 * $\map {\paren{i \circ h}^\gets} {\map {\pr_\alpha^\gets} A} = A$

Similarly:
 * $\map {\paren{i \circ h}^\gets} {\map {\pr_\alpha^\gets} B} = B$

It follows that $A$ and $B$ are contained in the disjoint open sets $\map {\paren {i \circ h}^\gets} U, \map {\paren {i \circ h}^\gets} V$, respectively.

Since $A$ and $B$ were an arbitrary pair of disjoint closed sets in $\struct {S_\alpha, \tau_\alpha}$ then $\struct {S_\alpha, \tau_\alpha}$ is a $T_4$ space.