Open Set Disjoint from Set is Disjoint from Closure

Theorem

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

Let $A \subseteq S$.

Let $B$ be an open set of $T$ disjoint from $A$.

Then:

$A^- \cap B = \O$

where $A^-$ denotes the closure of $A$.

Since $B \in \tau$, it follows by definition that $S \setminus B$ is closed.

Thus:

$\ds A \cap B$

$=$

$\ds \O$

Definition of Disjoint Sets

$\ds \leadsto \ \ $

$\ds A$

$\subseteq$

$\ds S \setminus B$

$\ds \leadsto \ \ $

$\ds A^-$

$\subseteq$

$\ds \paren {S \setminus B}^-$

$\ds \leadsto \ \ $

$\ds A^-$

$\subseteq$

$\ds S \setminus B$

$\ds \leadsto \ \ $

$\ds A^- \cap B$

$=$

$\ds \O$

$\blacksquare$