Point is Isolated iff not Accumulation Point
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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 $H$
- $x$ 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 $H$.
$\blacksquare$