T3 Lindelöf Space is T4 Space/Proof 1

Proof
Let $A$ and $B$ be disjoint closed subsets of $T$.

Lemma 1

 * $\forall F \subseteq S : S \setminus F \in \tau : y \in S \setminus F : \exists U \in \tau : U^- \cap F = \O$

where $U^-$ denotes the closure of $U$ in $T$

By definition of regular:
 * $\forall a \in A : \exists U_a, V_a : a \in U_a, B \subseteq V_a : U_a \cap V_a = \O$

We have:
 * $\forall a \in A : a \in U_a \subseteq \ds \bigcup_{a \in A} U_a$

By definition of subset:
 * $A \subseteq \ds \bigcup_{a \in A} U_a$

By definition of open cover:
 * $\family{U_a}_{a \in A}$ is a open cover of $A$

From User:Leigh.Samphier/Topology/Closed Subspace of Lindelöf Space is Lindelöf:
 * $\struct{A, \tau_A}$ is an Lindelöf subspace

where $\tau_A$ is the subspace topology on $A$.

By definition of Lindelöf space:
 * there exists a countable open subcover $\family{U_{a_n}}_{n \in \N}$ of $A$

Lemma 1

 * $\ds A \subseteq \paren{\bigcup_{n \in \N} U_{a_n} } \setminus \paren{\bigcup_{n \in \N} V_{a_n}^-}$

Lemma 2

 * $\ds B \subseteq \paren{\bigcup_{n \in \N} V_{a_n}} \setminus \paren{\bigcup_{n \in \N} U_{a_n}^-}$

For each $n \in \N$, let:
 * $U'_n = U_{a_n} \setminus \paren{\ds \bigcup_{p \le n} V_{a_p}^-}$

where $V_{a_p}^-$ denote the closure of $V_{a_p}$ in $T$.

For each $n \in \N$, let:
 * $V'_n = V_{a_n} \setminus \paren{\ds \bigcup_{p \le n} U_{a_p}^-}$

where $U_{a_p}^-$ denote the closure of $U_{a_p}$ in $T$.

Lemma 3

 * $\forall n, m \in \N : U'_n \cap V'_m = \O$

Let:
 * $U = \ds \bigcup_{n \in \N} U'_n$

and
 * $V = \ds \bigcup_{n \in \N} V'_n$

From Lemma 1:
 * $U \cap V = \O$

Lemma 4

 * $U$ and $V$ are open in $T$

We have:

and