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 $T$.
$\blacksquare$