# Equivalence of Definitions of Closed Set in Metric Space

## Contents

## Theorem

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

Let $M = \left({A, d}\right)$ 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:

- $\exists \delta \in \R_{>0}: B_\delta \left({b; d}\right) \subseteq A \setminus H$

where $B_\delta \left({b; d}\right)$ denotes the open $\delta$-ball of $b$.

From Intersection with Complement is Empty iff Subset:

- $B_\delta \left({b; d}\right) \cap H = \varnothing$

So by definition of limit point:

- $b \notin H'$

Hence:

- $A \setminus H \cap H' = \varnothing$

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:

- $\exists \delta \in \R_{>0}: B_\delta \left({b; d}\right) \cap H = \varnothing$

and so:

- $B_\delta \left({b; d}\right) \subseteq A \setminus H$

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

- 1962: Bert Mendelson:
*Introduction to Topology*... (previous) ... (next): $\S 2.6$: Open Sets and Closed Sets: Theorem $6.7$