Let $(X, d)$ be a metric space. Let $(x_n)_{n \in \mathbb{N}}$ be a sequence in $\mathbb{N}$, and $x \in X$. Show that:
$x_n \rightarrow x$ w.r.t. $d$ $\iff$ for every $U \subseteq X$ open in $(X, d)$ with $x \in U$, a final part of $(x_n)_{n \in \mathbb{N}}$ lives in $U$
I think I have an idea of how to prove $\implies$, but I'm a bit iffy on proving the other direction. This is what I was thinking:
$U$ is an open set; this means every point $u \in U$ has some neighborhood $N_r (u)$ s.t. $N_r (u) \subseteq U$. Now, we know that a final part of the sequence $S$ lives inside $U$. Does this mean we can take some neighborhood around the final part $S$? If a final part of $S$ living inside $U$, then doesn't that mean the convergence point needs to live inside $U$ as well, since the final part needs to become "final-er" and "final-er"? How should I draw the connection between final parts and open sets? I'd appreciate any help in seeing the connection between final parts of sequences and open sets. Thank you.
$\implies$ Let $U$ be an open set such that $x\in U$. Since $U$ is open, there is a $r>0$ such that $N_r(x)\subset U$. And you know that $x_n\in N_r(x)$ if $n$ is large enough. Therefore $x_n\in U$ if $n$ is large enough.
$\Longleftarrow$ Let $r>0$. Then $N_r(x)$ is an open set and therefore there is a natural $p$ such that $n\geqslant p\implies x_n\in N_r(x)$.