Equivalence of Definitions of Separated Sets/Definition 1 implies Definition 2

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Theorem

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

Let $A, B \subseteq S$ satisfy:

$A^- \cap B = A \cap B^- = \empty$

where $A^-$ denotes the closure of $A$ in $T$, and $\empty$ denotes the empty set.


Then there exist $U,V \in \tau$ with:

$A \subset U$ and $U \cap B = \empty$
$B \subset V$ and $V \cap A = \empty$

Proof

From Topological Closure is Closed, $B^-$ is closed in $T$.

Let $U = S \setminus B^-$ be the relative complement of $B^-$.

By the definition of a closed set, $U$ is open in $T$.

From Empty Intersection iff Subset of Relative Complement, $A \subseteq S \setminus B^- = U$

From Relative Complement of Relative Complement, $S \setminus U = B^-$.

By the definition of the closure of a subset, $B \subseteq B^- = S \setminus U$.

From Empty Intersection iff Subset of Relative Complement, $U \cap B = \empty$.


Similarly, let $V = S \setminus A^-$ then $V \in \tau$ with:

$B \subset V$ and $V \cap A = \empty$

$\blacksquare$