Equivalence of Definitions of Closed Set
Theorem
The following definitions of the concept of Closed Set in the context of topology are equivalent:
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Definition 1
$H$ is closed (in $T$) if and only if its complement $S \setminus H$ is open in $T$.
That is, $H$ is closed if and only if $\paren {S \setminus H} \in \tau$.
That is, if and only if $S \setminus H$ is an element of the topology of $T$.
Definition 2
$H$ is closed (in $T$) if and only if every limit point of $H$ is also a point of $H$.
That is, by the definition of the derived set:
- $H$ is closed (in $T$) if and only if $H' \subseteq H$
where $H'$ denotes the derived set of $H$.
Proof
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Definition 1 implies Definition 2
Let $H$ be a closed set of $T$ by definition 1.
Let $H^\complement$ denote the relative complement of $H$ in $S$.
By definition of closed set in $T$:
- $H^\complement$ is open in $T$
From Set is Open iff Neighborhood of all its Points:
- $\forall x \in S: x \notin H \implies H^\complement$ is a neighborhood of $x$.
By definition of limit point:
- $\forall x \in S: x \notin H \implies x$ is not a limit point of $H$
Thus $H$ is a closed set of $T$ by definition 2.
$\Box$
Definition 2 implies Definition 1
Let $H$ be a closed set of $T$ by definition 2.
Then by definition: $\forall x \in S: x \notin H \implies x$ is not a limit point of $H$
By definition of limit point: $\forall x \in S: x \notin H \implies H^\complement$ is a neighborhood of $x$
By Set is Open iff Neighborhood of all its Points: $H^\complement$ is open in $T$
Thus $H$ is a closed set of $T$ by definition 1.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction: Limit Points
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- 1953: Walter Rudin: Principles of Mathematical Analysis ... (previous) ... (next): $2.23$