Countable Open Covers Condition for Separated Sets

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Theorem

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

Let $A, B \subseteq S$

For all $X \subseteq S$, let $X^-$ denote the closure of $X$ in $T$.

Let:

$\UU = \set {U_n : n \in \N}$ be a countable open cover of $A : \forall n \in \N : {U_n}^- \cap B = \O$

Let:

$\VV = \set {V_n : n \in \N}$ be a countable open cover of $B : \forall n \in \N : {V_n}^- \cap A = \O$

Then:

$A$ and $B$ can be separated in $T$

Proof

By definition of cover:

$\ds A \subseteq \bigcup_{n \mathop \in \N} U_n$

We have:

$\ds A \cap \paren{\bigcup_{n \mathop \in \N} {V_n}^-} = \O$

From Subset of Set Difference iff Disjoint Set:

$(1): \quad \ds A \subseteq \paren {\bigcup_{n \mathop \in \N} U_n} \setminus \paren {\bigcup_{n \mathop \in \N} {V_n}^-}$

Similarly:

$(2): \quad \ds B \subseteq \paren {\bigcup_{n \mathop \in \N} V_n} \setminus \paren {\bigcup_{n \mathop \in \N} {U_n}^-}$

For each $n \in \N$, let:

${U_n}' = U_n \setminus \paren {\ds \bigcup_{p \mathop \le n} {V_p}^-}$

For each $n \in \N$, let:

${V_n}' = V_n \setminus \paren {\ds \bigcup_{p \mathop \le n} {U_p}^-}$

Lemma 1

$\forall n, m \in \N : {U_n}' \cap {V_m}' = \O$

$\Box$

Let:

$U = \ds \bigcup_{n \mathop \in \N} {U_n}'$

and

$V = \ds \bigcup_{n \mathop \in \N} {V_n}'$

From Lemma 1:

$U \cap V = \O$

Lemma 2

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

$\Box$

We have:

\(\ds A\)

\(\subseteq\)

\(\ds \paren {\bigcup_{n \mathop \in \N} U_n} \setminus \paren {\bigcup_{n \mathop \in \N} {V_n}^-}\)

From $(1)$ above

\(\ds \)

\(=\)

\(\ds \bigcup_{n \mathop \in \N} \paren {U_n \setminus \paren {\bigcup_{p \mathop \in \N} {V_p}^-} }\)

Set Difference is Right Distributive over Union

\(\ds \)

\(\subseteq\)

\(\ds \bigcup_{n \mathop \in \N} \paren {U_n \setminus \paren {\bigcup_{p \mathop \le n} {V_p}^-} }\)

Set Difference with Subset is Superset of Set Difference

\(\ds \)

\(=\)

\(\ds \bigcup_{n \mathop \in \N} {U_n}'\)

Definition of ${U_n}'$

\(\ds \)

\(=\)

\(\ds U\)

Definition of $U$

Similarly, from $(2)$ above:

\(\ds B\)

\(\subseteq\)

\(\ds V\)

It has been shown:

there exists $U, V \in \tau$ such that $A \subseteq U, B \subseteq V$ and $U \cap V = \O$.

Hence, by definition, $A$ and $B$ can be separated in $T$.

$\blacksquare$