Sequence Characterization of Open Sets

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

Then the following are equivalent:


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

Proof
$(1) \quad \implies \quad (2)$:

Suppose $G \subseteq X$ is open.

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

$(2) \quad \implies \quad (1)$:

Suppose that $G$ is not open.

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 $G$ is open.

Also see

 * Definition:Open Set (Metric Space)