# 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 $\ds T = \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ 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 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:

 $\ds \map {\pr_\alpha^\gets} A \cap \map {\pr_\alpha^\gets} B$ $=$ $\ds \map {\pr_\alpha^\gets} {A \cap B}$ Preimage of Intersection under Mapping $\ds$ $=$ $\ds \map {\pr_\alpha^\gets} \O$ $\ds$ $=$ $\ds \O$
$\pr_\alpha: S \to S_\alpha$ is continuous.
$\map {\pr_\alpha^\gets} A, \map {\pr_\alpha^\gets} B$ 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 $S_\alpha$.
$\map {\paren {i \circ h}^\gets} U, \map {\paren{i \circ h}^\gets} V$ are disjoint open sets in $S_\alpha$.
$\map {\paren {i \circ h}^\gets} {\map {\pr_\alpha^\gets} A} \subseteq \map {\paren{i \circ h}^\gets} U$
$\map {\paren {i \circ h}^\gets} {\map {\pr_\alpha^\gets} B} \subseteq \map {\paren{i \circ h}^\gets} V$

Now:

 $\ds x_\alpha$ $\in$ $\ds \map {\paren{i \circ h}^\gets} {\map {\pr_\alpha^\gets} A}$ $\ds \leadstoandfrom \ \$ $\ds \map {\paren {i \circ h} } {x_\alpha}$ $\in$ $\ds \map {\pr_\alpha^\gets} A$ Definition of Inverse Image Mapping $\ds \leadstoandfrom \ \$ $\ds \map {\pr_\alpha} {\map {i \circ h} {x_\alpha} }$ $\in$ $\ds A$ Definition of Inverse Image Mapping $\ds \leadstoandfrom \ \$ $\ds x_\alpha$ $\in$ $\ds A$

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.

$\blacksquare$

## Axiom of Choice

This theorem depends on the Axiom of Choice.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.