Equivalence of Definitions of Adherent Point

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Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $H \subseteq S$.


The following definitions of the concept of adherent point of $H$ are equivalent:

Definition from Open Neighborhood

A point $x \in S$ is an adherent point of $H$ if and only if every open neighborhood $U$ of $x$ satisfies:

$H \cap U \ne \O$

Definition from Closure

A point $x \in S$ is an adherent point of $H$ if and only if $x$ is an element of the closure of $H$.

Definition from Neighborhood

A point $x \in S$ is an adherent point of $H$ if and only if every neighborhood $N$ of $x$ satisfies:

$H \cap N \ne \O$


Proof

Definition from Open Neighborhood is equivalent to Definition from Closure

This is shown in Condition for Point being in Closure.

$\Box$


Definition from Open Neighborhood is equivalent to Definition from Neighborhood

Necessary Condition

Let every open neighborhood $U$ of $x$ satisfy:

$H \cap U \ne \O$

Let $N$ be an arbitrary neighborhood of $x$.

By definition of a neighborhood:

$\exists V \in \tau: x \in V \subseteq N$

From Set is Open iff Neighborhood of all its Points, $V$ is an open neighborhood of $x$.

Thus:

$H \cap V \ne \O$

By the contrapositive statement of Subsets of Disjoint Sets are Disjoint:

$H \cap N \ne \O$

Because $N$ was arbitrary, it follows that every neighborhood $N$ of $x$ satisfies:

$H \cap N \ne \O$

$\Box$


Sufficient Condition

Let every neighborhood $N$ of $x$ satisfy:

$H \cap N \ne \O$

By definition, every open neighborhood $U$ of $x$ is a neighborhood of $x$.

So every open neighborhood $U$ of $x$ satisfies:

$H \cap U \ne \O$

$\blacksquare$