Sequence Convergence
# Definition 1
Let $(X, \tau)$ be a Topological Space and let $(x_n) \subset X$. We say $x_{n}$ converges to $x \in X$ if $\forall V \in \tau$ so that $x \in V$, there exists $N \in \mathbb{N}$ so that $x_{n} \in V$ $\forall n \geq N$. If we have convergence, we write
$$\lim\limits_{n \to \infty} x_{n} = x$$ or $$x_{n} \to x \text{ (as } n \to \infty)$$
# Definition 2
Let $(M, d)$ be a Metric Space and let $(x_n) \subset M$. We say $x_{n}$ converges to $x \in X$ if $\forall \epsilon > 0$ there exists $N > 0$ s.t. $d(x_{n}, x) < \epsilon$ for all $n \geq N$.
# Properties
Definition 1 is true If and Only If Definition 2 is as well on the Metric Topology of $M$:
Proof $(\Rightarrow)$: Suppose Definition 1 holds and $x_{n} \to x \in M$. Let $\epsilon > 0$. Since $B_\epsilon(x)$ is Open in $M$, there exists $N \in \mathbb{N}$ so that for all $n \geq N$ we have that $x_{n} \in B_{\epsilon}(x)$. By definition of $B_{\epsilon}(x)$, this means $d(x_{n}, x) < \epsilon$ $\checkmark$.
($\Leftarrow$): Suppose Definition 2 holds and $x_{n} \to x \in M$. Let $V \in \tau$ contain $x$. Since A Set is Open in the Metric Topology iff it contains a Ball around each Point, we know there exists $\epsilon > 0$ so that $B_{\epsilon}(x) \subset V$. By Definition 2, there exists $N \in \mathbb{N}$ so that for all $n \geq N$, $d(x_{n}, x) < \epsilon$. In other words, $x_{n} \in B_{\epsilon}(x)$ and thus $x_{n} \in V$. $\checkmark$ $\blacksquare$
TODO This could probably reduced to single line about the relationship between sequence convergence and a Topological Basis.