Equivalence of Definitions of T4 Space
Theorem
The following definitions of the concept of $T_4$ space are equivalent:
Let $T = \struct {S, \tau}$ be a topological space.
Definition by Open Sets
$T = \struct {S, \tau}$ is a $T_4$ space if and only if:
- $\forall A, B \in \map \complement \tau, A \cap B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
That is, for any two disjoint closed sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively.
Definition by Closed Neighborhoods
$T = \struct {S, \tau}$ is $T_4$ if and only if each open set $U$ contains a closed neighborhood of each closed set contained in $U$.
Proof
Definition by Open Sets implies Definition by Closed Neighborhoods
Let $T$ satisfy the definition by open sets of a $T_4$ space.
Let $A$ be a closed set in $T$, and let $U_A$ be an open neighborhood of $A$.
By definition of open neighborhood:
- $A \subseteq U_A$
Let $B := \relcomp S {U_A}$.
By Intersection with Complement is Empty iff Subset:
- $A \cap B = \O$
As $B$ is the complement of $U_A$ in $S$, it is by definition closed in $T$.
Thus $A$ and $B$ are disjoint closed sets in $T$.
Also, from Set Complement inverts Subsets:
- $B \subseteq \relcomp S A$
By assumption, there exist disjoint open sets $U$ and $V$ such that $A \subseteq U$ and $B \subseteq V$.
From Set Intersection Preserves Subsets:
- $B \subseteq \map \complement A \cap V$
Note that the latter set, being an intersection of open sets, is itself open.
Then, from Complement of Complement, De Morgan's Laws: Complement of Union and Set Complement inverts Subsets:
- $A \cup \map \complement V \subseteq U_A$
From Subset of Union we also have:
- $A \subseteq A \cup \map \complement V$
Since $\relcomp S A \cap V$ is open, $A \cup \relcomp S V$ is closed.
Hence we have found a closed neighborhood for $A$ in $U_A$, as desired.
Hence it is concluded that $T$ satisfies the definition by closed neighborhoods of a $T_4$ space.
$\Box$
Definition by Closed Neighborhoods implies Definition by Open Sets
Let $T$ satisfy the definition by closed neighborhoods of a $T_4$ space.
Let $A$ and $B$ be disjoint closed sets in $T$.
Then from Empty Intersection iff Subset of Complement, we have:
- $A \subseteq \map \complement B$
and the latter is open in $T$.
Applying the assumption, we find a closed neighborhood $C_A$ of $A$ contained in $\map \complement B$.
From Empty Intersection iff Subset of Complement and Set Complement inverts Subsets we establish:
- $A \cap \map \complement {C_A} = \O$
- $B \subseteq \map \complement {C_A}$
Similarly, we find a closed neighborhood $C_B$ of $B$ contained in $\map \complement {C_A}$.
Then from Intersection with Complement is Empty iff Subset:
- $B \cap \map \complement {C_B} = \O$
But from Set Complement inverts Subsets, we have:
- $\map \complement {C_A} \subseteq \map \complement A$
and so from Subset Relation is Transitive:
- $C_B \subseteq \map \complement A$
Finally, another application of Set Complement inverts Subsets shows:
- $A \subseteq \map \complement {C_B}$
Since $C_B \subseteq \map \complement {C_A}$, Empty Intersection iff Subset of Complement shows that $C_A$ and $C_B$ are disjoint sets.
They are also open sets, being the complement of closed sets.
Above, we established also that:
- $A \subseteq \map \complement {C_B}$
- $B \subseteq \map \complement {C_A}$
Hence it is concluded that $T$ satisfies the definition by open sets as well.
$\blacksquare$
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