Equivalence of Definitions of Convergent Sequence in Topology

$(1)$ implies $(2)$
Suppose that $\alpha \in S$ has:


 * $\forall U \in \tau: \alpha \in U \implies \paren {\exists N \in \R_{>0}: \forall n \in \N: n > N \implies x_n \in U}$

Fix $U \in \tau$ with $\alpha \in U$.

Then there exists $N \in \N$ such that $x_n \in U$ for $n > N$.

So $x_n \not \in U$ for $n \in \N$ implies that $n \le N$.

That is:


 * $\set {n \in \N : x_n \not \in U} \subseteq \set {n \in \N : n \le N}$

Since $\set {n \in \N : n \le N}$ is finite, it follows that $\set {n \in \N : x_n \not \in U}$ is finite from Subset of Finite Set is Finite.

$(2)$ implies $(1)$
Suppose that $\alpha \in S$ is such that:


 * $\forall U \in \tau: \alpha \in U \implies \set {n \in \N: x_n \notin U}$ is finite.

Fix $U \in \tau$ with $\alpha \in U$.

Then:


 * $\set {n \in \N : x_n \notin U}$ is finite.

From Subset of Naturals is Finite iff Bounded, there exists $N \in \N$ such that:


 * $\set {n \in \N : x_n \notin U} \subseteq \set {n \in \N : n \le N}$

So, if $n > N$, we must have $x_n \in U$.