Point is Isolated iff not Accumulation Point

Theorem

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

Let $H \subseteq S$.

Let $x \in H$.

Then:

$x$ is an isolated point in $H$

if and only if:

$x$ is not an accumulation point of $H$

Proof

Sufficient Condition

Let $x \in H$ be an isolated point in $H$.

Then by definition of isolated point:

$\exists U \in \tau: H \cap U = \set x$

That is, by definition of uniqueness:

$\lnot \forall U \in \tau: \paren {x \in U \implies \exists y \in S: \paren {y \in H \cap U \land x \ne y} }$

Hence by Characterization of Derivative by Open Sets:

$x \notin A'$

where $A'$ denotes the derivative of $A$.

Thus by definition of derivative:

$x$ is not an accumulation point of $H$.

$\Box$

Necessary Condition

Let $x \in H$ not be an accumulation point of $H$.

Thus by definition of derivative:

$x \notin A'$

Hence:

\(\ds \lnot \forall U \in \tau: \, \)

\(\ds \)

\(\)

\(\ds \paren {x \in U \implies \exists y \in S: \paren {y \in H \cap U \land x \ne y} }\)

Characterization of Derivative by Open Sets

\(\ds \exists U \in \tau: \, \)

\(\ds \)

\(\)

\(\ds \lnot \paren {x \in U \implies \exists y \in S: \paren {y \in H \cap U \land x \ne y} }\)

Denial of Universality

\(\ds \exists U \in \tau: \, \)

\(\ds \)

\(\)

\(\ds \paren {x \in U \land \lnot \exists y \in S: \paren {y \in H \cap U \land x \ne y} }\)

[[ Conjunction with Negative is Equivalent to Negation of Conditional]]

\(\ds \exists U \in \tau: \, \)

\(\ds \)

\(\)

\(\ds \paren {x \in U \land \forall y \in S: \lnot \paren {y \in H \cap U \land x \ne y} }\)

Denial of Existence

\(\ds \exists U \in \tau: \, \)

\(\ds \)

\(\)

\(\ds \paren {x \in U \land \forall y \in S: \paren {y \in H \cap U \implies x = y} }\)

Conditional is Equivalent to Negation of Conjunction with Negative

\(\ds \exists U \in \tau: \, \)

\(\ds \)

\(\)

\(\ds H \cap U = \set x\)

Definition of Unique, and $x \in H$

Thus by definition of isolated point:

$x$ is an isolated point in $H$.

$\blacksquare$