Countable Open Covers Condition for Separated Sets/Lemma 1

Theorem

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

Let $\family {U_n}_{n \mathop \in \N}$ be a family of subsets of $S$.

Let $\family {V_n}_{n \mathop \in \N}$ be a family of subsets of $S$.

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:

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

Then:

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

Proof

Let $n, m \in \N$.

Without loss of generality, let $m \le n$.

We have:

\(\ds {U_m}'\)

\(=\)

\(\ds U_m \setminus \paren {\bigcup_{p \mathop \le m} {V_p}^-}\)

Definition of ${U_m}'$

\(\ds \)

\(\subseteq\)

\(\ds U_m\)

Set Difference is Subset

\(\ds \)

\(\subseteq\)

\(\ds {U_m}^-\)

Set is Subset of its Topological Closure

As $m \le n$, then:

${U_m}^- \in \set { {U_p}^- : p \le n}$

We have:

\(\ds U_m^-\)

\(\subseteq\)

\(\ds \bigcup_{p \mathop \le n} {U_p}^-\)

Set is Subset of Union

Also:

\(\ds {V_n}'\)

\(=\)

\(\ds V_n \setminus \paren {\bigcup_{p \mathop \le n} {U_p}^-}\)

Definition of ${V_m}'$

\(\ds \)

\(\subseteq\)

\(\ds V_n \setminus {U_m}^-\)

Set Difference with Subset is Superset of Set Difference

Finally:

\(\ds {U_m}' \cap {V_n}'\)

\(=\)

\(\ds \paren {U_m \setminus \paren {\bigcup_{p \mathop \le m} {V_p}^-} } \cap \paren {V_n \setminus \paren {\bigcup_{p \mathop \le n} {U_p}^-} }\)

\(\ds \)

\(\subseteq\)

\(\ds {U_m}^- \cap \paren {V_n \setminus {U_m}^-}\)

Set Intersection Preserves Subsets

\(\ds \)

\(=\)

\(\ds \O\)

Set Difference Intersection with Second Set is Empty Set

From Empty Set is Subset of All Sets:

$\O \subseteq {U_m}' \cap {V_n}'$

By definition of set equality:

${U_m}' \cap {V_n}' = \O$

Similarly:

${U_n}' \cap {V_m}' = \O$

The result follows.

$\blacksquare$