Open Set

Def. Open Set
A set $O \subseteq \mathbb{R}$ is open if $\forall a \in O, \exists V_{\epsilon}(a) \subseteq O.$

Thm. (Intersection and Union of Open Sets)
1. The union of arbitrary collections of open sets is open.
2. The intersection of finite collection of open sets is open.

Proof for 1.
Consider $O = \cup_{\lambda \in \Lambda} O_{\lambda}$ where $O_{\lambda}$ is open for any $\lambda \in \Lambda$.
Let $a \in O$. Then $a \in O_{\lambda '}$ for some $\lambda ' \in \Lambda$.
$\Rightarrow V_{\epsilon}(a) \subseteq O_{\lambda '} \subseteq O$. (QED)

Proof for 2.
Consider $O = \cap_{n \in [N]}O_n$ where $O_n$ is open for any $n \in [N]$. And let $a \in O$, then for each $n \in [N]$,
$\exist \epsilon_n$ such that $V_{\epsilon_n} (a) \subseteq O_n$.
Take $\epsilon = \min (\epsilon_n)_{n \in [N]}$. Then, $\forall n \in [N], V_{\epsilon}(a) \subseteq V_{\epsilon_n}(a) \subseteq O$. (QED)

Def. (Limit Point)
$x \in A$ is a limit point of $A$ if

$\forall \epsilon >0, \exist a \in A (a \neq x) \textup{ s.t } a \in V_{\epsilon}(x).$

Thm.
$x \in A$ is a limit point of $A$ if and only if $\lim_{n \to \infty} a_n = x$ for some $(a_n) \subseteq A$.

Proof (if part).
Take $\epsilon = 1/n$.
Then, we can choose $a_n (\neq x)$ such that $a_n \in V_{1/n}(x)$.
Then for arbitrary $\epsilon > 0$, choose $N \in \mathbb{N}$ so that $1/N < \epsilon$.

Proof (Only if part).
Straightforward due to epsilon-n statement.