Equivalence of Definitions of Closed Set

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Theorem

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


Let $T = \left({S, \tau}\right)$ 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 $\left({S \setminus H}\right) \in \tau$.

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 $T = \left({S, \tau}\right)$ 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.

$\blacksquare$


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