Equivalence of Definitions of Isolated Point

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Theorem

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


Let $T = \left({S, \tau}\right)$ be a topological space.

Let $H \subseteq S$ be a subset of $S$.

Definition 1

$x \in H$ is an isolated point of $H$ if and only if:

$\exists U \in \tau: U \cap H = \left\{{x}\right\}$

That is, if and only if there exists an open set of $T$ containing no points of $H$ other than $x$.

Definition 2

$x \in H$ is an isolated point of $H$ if and only if $x$ is not a limit point of $H$.

That is, if and only if $x$ is not in the derived set of $H$.


Proof

Definition 1 implies Definition 2

Let $x$ be an isolated point of $H$ by definition 1.

Then by definition:

$\exists U \in \tau: U \cap H = \left\{{x}\right\}$

Thus we have an open set in $T$ such that $x \in U$ contains no other point of $H$ than $x$.

Thus, by definition, $x$ is not a limit point of $H$.


Thus $x$ is an isolated point of $H$ by definition 2.

$\Box$


Definition 2 implies Definition 1

Let $x$ be an isolated point of $H$ by definition 2.

Aiming for a contradiction, suppose $x$ is a limit point of $H$.

Then by definition every open set $U \in \tau$ such that $x \in U$ contains some point of $H$ other than $x$.

That is:

$\forall U \in \tau: x \in U \implies \exists y \in S, y \ne x: y \in U \cap H$

That is:

$\not \exists U \in \tau: U \cap H = \left\{{x}\right\}$

because all $U$ with $x$ in them are such that there is at least one point in $U \cap H$ apart from $x$.

Thus by Proof by Contradiction $x$ is not a limit point of $H$.

That is, $x$ is an isolated point of $H$ by definition 1.

$\blacksquare$


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