Equivalence of Definitions of Closed Set in Metric Space

From ProofWiki
Jump to navigation Jump to search

Theorem

The following definitions of the concept of Closed Set in the context of Metric Spaces are equivalent:

Let $M = \struct {A, d}$ be a metric space.

Let $H \subseteq A$.

Definition 1

$H$ is closed (in $M$) if and only if its complement $A \setminus H$ is open in $M$.

Definition 2

$H$ is closed (in $M$) if and only if every limit point of $H$ is also a point of $H$.


Proof

Let $H'$ denote the set of limit points of $H$.

Definition 1 implies Definition 2

Let $H$ be closed in $M$ by definition 1.

Then by definition $A \setminus H$ is open in $M$.

Let $b \in M: b \in A \setminus H$.

Then by definition of open set:

$\exists \delta \in \R_{>0}: \map {B_\delta} {b; d} \subseteq A \setminus H$

where $\map {B_\delta} {b; d}$ denotes the open $\delta$-ball of $b$.

From Intersection with Complement is Empty iff Subset:

$\map {B_\delta} {b; d} \cap H = \O$

So by definition of limit point:

$b \notin H'$

Hence:

$A \setminus H \cap H' = \O$

By Intersection with Complement is Empty iff Subset:

$H' \subseteq H$

Thus $H$ contains all its limit points.

That is, $H$ be closed in $M$ by definition 2.

$\Box$


Definition 2 implies Definition 1

Let $H$ be closed in $M$ by definition 2.

That is:

$H' \subseteq H$

Then by Set Complement inverts Subsets:

$A \setminus H \subseteq A \setminus H'$

Let $b \in A \setminus H$.

Then by definition of relative complement:

$b \notin H'$

Then by definition of limit point:

$\exists \delta \in \R_{>0}: \map {B_\delta} {b; d} \cap H = \O$

and so:

$\map {B_\delta} {b; d} \subseteq A \setminus H$

As $b$ is arbitrary, it follows that $A \setminus H$ is open in $M$.

Hence by definition, $H$ is closed in $M$ by definition 1.

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Equivalence_of_Definitions_of_Closed_Set_in_Metric_Space&oldid=506982"