T3 Lindelöf Space is T4 Space/Proof 1
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Theorem
Let $T = \struct {S, \tau}$ be a $T_3$ Lindelöf topological space.
Then:
- $T$ is a $T_4$ space.
Proof
Let $A$ and $B$ be disjoint closed subsets of $T$.
Let $\UU = \set {U \in \tau : U^- \cap B = \O}$.
From Characterization of T3 Space:
- $\forall a \in A : \exists U_a \in \tau: a \in U_a : U_a^- \cap B = \O$
By definition of open cover:
- $\UU$ is an open cover of $A$
From Closed Subspace of Lindelöf Space is Lindelöf Space:
where $\tau_A$ is the subspace topology on $A$.
By definition of Lindelöf space:
- there exists a countable subcover $\family {U_n}_{n \mathop \in \N}$ of $\UU$ for $A$
Let $\VV = \set {V \in \tau : V^- \cap A = \O}$.
Similar to $A$ above, there exists a countable subcover $\family {V_n}_{n \mathop \in \N}$ of $\VV$ for $B$.
From Countable Open Covers Condition for Separated Sets:
- $A$ and $B$ can be separated in $T$
By definition of separated:
- $\exists U, V \in \tau : A \subseteq U, B \subseteq V, U \cap V = \O$.
Since $A$ and $B$ were arbitrary:
- $\forall A, B \in \map \complement \tau, A \cap B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
Hence, by definition, $T$ is a $T_4$ space.
$\blacksquare$
Sources
- 1955: John L. Kelley: General Topology: Chapter $4$: Embedding and Metrization: Lemma $1$
- 1970: Stephen Willard: General Topology: Chapter $4$: Separation and Countability: $\S16$: Countability Properties: Theorem $16.8$