The usual metric space definition of convergence in $\mathbb{R}$ as a metric space is
(1) $\lim \limits_{n \to \infty} x_{n}=x$ iff $\forall \epsilon>0$ $\exists N_{\epsilon}$ s.t. $n \geq N_{\epsilon} \implies |x-x_{n}| < \epsilon$.
Basically, this means that starting from some integer $N_{\epsilon}$ all elements of the sequence are in an open ball of radius $\epsilon$ around $x$.
I want to show that this is equivalent to the topological definition of convergence for $\mathbb{R}$. Note that the open sets in $\mathbb{R}$ with the absolute value distance forms a topology since arbitrary unions and finite intersections of open sets are open.
The topological definition of convergence is
(2) $\lim \limits_{n \to \infty} x_{n}=x$ iff for every open neighborhood $B$ of $x$ $\exists N_{B}$ s.t. $n \geq N_{B} \implies x_{n} \in B$.
An open neighbourhood of $x$ is any open set $U \in \tau$ s.t. $x \in U$.
It is easy to show that (1) $\implies$ (2). Consider an open set $A$ and let $x \in A$. Then by definition of open sets in a metric space, every point in $A$ is an interior point, i.e. $\exists$ some open ball $B_{r}(x)$ with $r>0$ that is completely contained in A. Thus, by (1) all points of the sequence $x_{n}$ will eventually lie in $B_{r}$ and thus in $A$. Since $A$ was arbitraty this proves the claim.
Similarly, we can show that (2) $\implies$ (1) since every neighborhood of $x$ of the topology $\tau$ is an open set which contains $x$. In particular, any open ball $B_{\epsilon}(x)$ with $\epsilon>0$ is a neighborhood of $x$. This means that starting from some integer $N_{\epsilon}$ all elements of the sequence are contained in $B_{\epsilon}$ or in other words $\exists N_{\epsilon}$ s.t. $n \geq N_{\epsilon} \implies |x-x_{n}| < \epsilon$. Since $\epsilon$ was arbitrary (1) holds.
I've got a few questions about this.
1) Is my proof correct? I'm a bit worried about the proof for (2) $\implies$ (1).
2)It is easier to think of the extended real line $\overline{\mathbb{R}}$ as a topolocial space rather than a metric space. Now it would make sense if convergence of a seqence in $\mathbb{R} \implies$ convergence in $\overline{\mathbb{R}}$. Is this true? What are the neighborhoods in $\overline{\mathbb{R}}$? Are the neighborhoods simply the open sets, i.e. sets in the topology of the extended real line?
Thanks very much for your help.