# Equivalence of Definitions of Limit Point

## Theorem

The following definitions of the concept of limit point are equivalent:

That is, let $T = \struct {S, \tau}$ be a topological space.

Let $H \subseteq S$.

Let $H^{\complement}$ denote the relative complement of $H$ in $S$.

Then the following conditions are equivalent for any point $x \in S$:

$(1): \quad$ Every open neighborhood $U$ of $x$ satisfies $H \cap \paren {U \setminus \set x} \ne \O$.
$(2): \quad x$ belongs to the closure of $H$ but is not an isolated point of $H$.
$(3): \quad x$ is an adherent point of $H$ but is not an isolated point of $H$.
$(4): \quad H^{\complement} \cup \set x$ is not a neighborhood of $x$.

## Proof

### $({1}) \iff ({2})$

The closure of $H$ is defined as the union of the set of all isolated points of $H$ and the set of all limit points of $H$.

The rest then follows directly from Equivalence of Definitions of Isolated Point.

$\Box$

### $({2}) \iff ({3})$

Follows directly from Equivalence of Definitions of Adherent Point.

$\Box$

### $({1}) \iff ({4})$

The following equivalence holds:

 $\ds$  $\ds$ There exists an open neighborhood $U$ of $x$ such that $H \cap \paren {U \setminus \set x} = \O$ $\ds$ $\leadstoandfrom$ $\ds$ There exists an open neighborhood $U$ of $x$ such that $U \subseteq H^{\complement} \cup \set x$ $\quad$ Modus Ponendo Tollens $\ds$ $\leadstoandfrom$ $\ds$ $H^{\complement} \cup \set x$ is a neighborhood of $x$ $\quad$ Definition of Neighborhood of Point

The result follows from the Rule of Transposition.

$\blacksquare$

### $({1}) \iff ({4})$, Proof Variant

The following equivalence holds:

There exists an open neighborhood $U$ of $x$ such that $H \cap \paren {U \setminus \set x} = \O$

 $\ds \O$ $=$ $\ds H \cap \paren {U \setminus \set x}$ $\ds \leadstoandfrom \ \$ $\ds \O$ $=$ $\ds \paren {U \cap H} \setminus \set x$ $\quad$ Intersection with Set Difference is Set Difference with Intersection $\ds \leadstoandfrom \ \$ $\ds \O$ $=$ $\ds \paren {H \cap U} \setminus \set x$ $\quad$ Intersection is Commutative $\ds \leadstoandfrom \ \$ $\ds \O$ $=$ $\ds U \cap \paren {H \setminus \set x}$ $\quad$ Intersection with Set Difference is Set Difference with Intersection $\ds \leadstoandfrom \ \$ $\ds \O$ $=$ $\ds U \cap \map \complement { \map \complement {H \setminus \set x} }$ $\quad$ Complement of Complement $\ds \leadstoandfrom \ \$ $\ds U$ $\subseteq$ $\ds \map \complement {H \setminus \set x}$ $\quad$ Intersection with Complement is Empty iff Subset $\ds \leadstoandfrom \ \$ $\ds U$ $\subseteq$ $\ds \map \complement {H \cap \map \complement {\set x} }$ $\quad$ Set Difference as Intersection with Complement $\ds \leadstoandfrom \ \$ $\ds U$ $\subseteq$ $\ds \map \complement H \cup \map \complement {\map \complement {\set x} }$ $\quad$ De Morgan's Laws (Set Theory)/Set Complement $\ds \leadstoandfrom \ \$ $\ds U$ $\subseteq$ $\ds \map \complement H \cup \set x$ $\quad$ Complement of Complement

By Definition of Neighborhood of Point, $\map \complement H \cup \set x$ is a neighborhood of $x$

The result follows from the Rule of Transposition.

$\Box$