T3 Lindelöf Space is T4 Space

Theorem

Let $T = \struct {S, \tau}$ be a $T_3$ Lindelöf topological space.

Then:

$T$ is a $T_4$ space.

Proof 1

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:

$\struct {A, \tau_A}$ is a Lindelöf subspace

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.

$\blacksquare$

Proof 2

From $T_3$ Lindelöf Space is Fully $T_4$ Space:

$T$ is a fully $T_4$ space.

From Fully $T_4$ Space is $T_4$ Space:

$T$ is a $T_4$ space.

$\blacksquare$