Sequence Characterization of Open Sets

Theorem
Let $\left({X, d}\right)$ be a metric space.

Let $G \subseteq X$.

Then the following are equivalent:


 * $(1): \quad G \subseteq X$ is an open set of $\left({X, d}\right)$
 * $(2): \quad \forall x \in G: \forall \left\langle{x_n}\right\rangle \in X: x_n \to x: \exists n_0 \in \N: \forall n \ge n_0: \left\langle{x_n}\right\rangle \in G$

$(1)$ implies $(2)$
Let $G \subseteq X$ be open in $\left({X, d}\right)$.

Let $x \in G$.

Let $\left\langle{x_n}\right\rangle$ be a sequence in $X$ such that $x_n \to x$.

By definition of open set, there exists $\epsilon > 0$ such that:
 * $B_\epsilon \left({x}\right) \subseteq G$

where $B_\epsilon \left({x}\right)$ is the open $\epsilon$-ball of $x$.

Since $x_n \to x$, there exists $n_0 \in \N$ such that:
 * $\forall n \ge n_0: d \left({x_n, x}\right) < \epsilon$

Thus:
 * $\forall n \ge n_0: x_n \in B_\epsilon \left({x}\right) \subseteq G$

$(2)$ implies $(1)$
Let $G \subseteq X$.

Let:
 * $\forall x \in G: \forall \left\langle{x_n}\right\rangle \in X: x_n \to x: \exists n_0 \in \N: \forall n \ge n_0: \left\langle{x_n}\right\rangle \in G$

Aiming for a contradiction, suppose $G$ is not open in $\left({X, d}\right)$.

Then:
 * $\exists x \in G: \forall \epsilon \in \R_{>0}: B_\epsilon \left({x}\right) \nsubseteq G$

Therefore, for $n = 1, 2, \ldots$ we can inductively find a sequence:
 * $x_n \in B_{1/n} \left({x}\right) \cap \left({X \setminus G}\right) \implies x_n \notin G$

and:
 * $\forall n \in \N: d \left({x_n, x}\right) < \dfrac 1 n \implies x_n \to x \in G$ and $\forall n \in \N: x_n \notin G$

This contradicts hypothesis $(2)$.

Thus, by Proof by Contradiction, $G$ is open in $\left({X, d}\right)$.

Also see

 * Definition:Open Set (Metric Space)