Theorem: Factor Spaces are T4 if Product Space is T4

Let $\SS = \family {\struct{S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces for $\alpha$ in some indexing set $I$.
Let $\ds T = \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Let $T$ be a $T_4$ space.
Then each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_4$ space.

Proof:
Let $T$ be a $T_4$ space.
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 is 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 Composite of Continuous Mappings is Continuous, 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:
{{begin-eqn}}
{{eqn | l = \map {\pr_\alpha^\gets} A \cap \map {\pr_\alpha^\gets} B
      | r = \map {\pr_\alpha^\gets} {A \cap B}
      | c = Preimage of Intersection under Mapping
}}
{{eqn | r = \map {\pr_\alpha^\gets} \O
}}
{{eqn | r = \O
}}
{{end-eqn}} 
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:
{{begin-eqn}}
{{eqn | l = x_\alpha
      | o = \in
      | r = \map {\paren{i \circ h}^\gets} {\map {\pr_\alpha^\gets} A}
}}
{{eqn | ll = \leadstoandfrom
      | l = \map {\paren {i \circ h} } {x_\alpha}
      | o = \in
      | r = \map {\pr_\alpha^\gets} A
      | c = {{Defof|Inverse Image Mapping}}
}}
{{eqn | ll = \leadstoandfrom
      | l = \map {\pr_\alpha} {\map {i \circ h} {x_\alpha} }
      | o = \in
      | r = A
      | c = {{Defof|Inverse Image Mapping}}
}}
{{eqn | ll = \leadstoandfrom
      | l = x_\alpha
      | o = \in
      | r = A
}}
{{end-eqn}}
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.
{{qed}}
{{AoC}}
Category:T4 Spaces
Category:Separation Properties Preserved under Topological Product
Category:Product Spaces
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