Product Space is T3 1/2 iff Factor Spaces are T3 1/2
Theorem
Let $\mathbb S = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of topological spaces for $\alpha$ in some indexing set $I$ with $S_\alpha \ne \O$ for every $\alpha \in I$.
Let $\ds T = \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\mathbb S$.
Then $T$ is a $T_{3 \frac 1 2}$ space if and only if each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_{3 \frac 1 2}$ space.
Proof
Necessary Condition
Suppose $T$ is a $T_{3 \frac 1 2}$ space.
Since $S_\alpha \ne \O$ we also have $S \ne \O$ by the axiom of choice.
Let $\alpha \in I$ be arbitrary.
From Subspace of Product Space is Homeomorphic to Factor Space:
- $\struct {S_\alpha, \tau_\alpha}$ is homeomorphic to a subspace $T_\alpha$ of $T$.
From $T_{3 \frac 1 2}$ Property is Hereditary:
- $T_\alpha$ is $T_{3 \frac 1 2}$.
From $T_{3 \frac 1 2}$ Space is Preserved under Homeomorphism:
- $\struct {S_\alpha, \tau_\alpha}$ is $T_{3 \frac 1 2}$.
Because $\alpha \in I$ was arbitrary then the result follows.
$\Box$
Sufficient Condition
Let $\struct{S_\alpha, \tau_\alpha}$ is a $T_{3 \frac 1 2}$ space for each $\alpha \in I$.
Let $x \in S$.
Let $F$ be a closed subset of $S$ such that $x \notin F$.
By definition of a closed subset:
- $S \setminus F \in \tau$
By definition of the product topology, there exists an open set $B$ of the natural basis containing $x$ which is disjoint from $F$.
By definition of the natural basis, $B$ is of the form:
- $\map {\pr_{\alpha_1}^\gets} {U_1} \cap \dotsb \cap \map {\pr_{\alpha_n}^\gets} {U_n}$
where:
- $\pr_{\alpha_k}$ is the $\alpha_k$-th projection from $S$
- $U_k$ is open in $S_{\alpha_k}$ for all $1 \le k \le n$
By definition of a $T_{3 \frac 1 2}$ space, for each $1 \le k \le n$ there exists a continuous mapping:
- $f_k: S_{\alpha_k} \to \closedint 0 1$
such that:
- $\map {f_k} {x_{\alpha_k} } = 1$
and:
- $\map {f_k} {S_{\alpha_k} \setminus U_k} = 0$
Let $g_k = f_k \circ \pr_{\alpha_k}$ be the composite mapping of $f_k$ with $\pr_{\alpha_k}$ for each $1 \le k \le n$.
From Composite of Continuous Mappings is Continuous each $g_k: S \to \closedint 0 1$ is a continuous mapping.
We define $g: S \to \closedint 0 1$ by setting:
- $\map g y = \min \set {\map {g_k} {y_{\alpha_k} }: k = 1, \dotsc, n}$
From Minimum Rule for Continuous Functions:
- $g$ is continuous.
Now:
\(\ds \map g x\) | \(=\) | \(\ds \min \set{\map {g_k} x : k = 1, \dotsc, n}\) | Definition of $g$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \min \set {\map {f_k} {\map {\pr_{\alpha_k} } x} : k = 1, \dotsc, n}\) | Definition of Composite Mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds \min \set {\map {f_k} {x_{\alpha_k} } : k = 1, \dotsc, n}\) | Definition of Projection | |||||||||||
\(\ds \) | \(=\) | \(\ds \min \set {1 : k = 1, \dotsc, n}\) | Definition of $f_k$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | Definition of Minimum |
Let $y \in F$.
By definition of disjoint sets:
- $\exists j \in \set {1, \dotsc, n} : y \notin \map {\pr_{\alpha_j}^\gets} {U_j}$
By definition of the inverse image mapping of $\pr_{\alpha_j}$:
- $y_{\alpha_j} \notin U_j$
Thus:
- $\map {f_j} {y_{\alpha_j} } = 0$
So:
\(\ds \map g y\) | \(=\) | \(\ds \min \set {\map {g_k} y : k = 1, \dotsc, n}\) | Definition of $g$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \min \set {\map {f_k} {\map {\pr_{\alpha_k} } y} : k = 1, \dotsc, n}\) | Definition of Composite Mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds \min \set {\map {f_k} {y_{\alpha_k} } : k = 1, \dotsc, n}\) | Definition of Projection | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | Definition of Minimum and $\map {f_k} {y_{\alpha_k} } = 0$ |
Therefore $T$ is a $T_{3 \frac 1 2}$ space.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces