How do we define convergence of a sequence in $\mathbb R\cup\{-\infty,+\infty\}$

Question

Can we define convergence of a sequence in the set $\mathbb R\cup\{-\infty,+\infty\}$?How do we define that?Do we require metric space for this convergence?

Answer 1

If $x\in\Bbb R$ the definition of $x_n\to x$ is the same as in calculus. The definitions of $x_n\to\infty$ and $x_n\to-\infty$ are also the same as in calculus.

(Yes, that works even for $x_n\in\Bbb R\cup\{\infty,-\infty\}$. For example, $x_n\to0$ means for every (real) $\epsilon>0$ there exists $N$ such that $|x_n|<\epsilon$ for every $n>N$. If $x_3=\infty$ then $|x_n|\ge\epsilon$.)

Answer 2

$\overline{\Bbb R}$, the extended reals, is a linearly ordered set and it gets the order topology: at the maximum $+\infty$ it has neighbourhoods of the form $(x, +\infty]$ (any $x < +\infty$ allowed but values $x=n$ for $n \in \mathbb{N}$ suffice for a local base) and at the minumum $-\infty$ we have neighbourhoods $[-\infty, x)$ ($ x > -\infty$ but values of the form $x=-n, n \in \mathbb{Z}$ suffice for a local base again) and all other $x$ have basic neighbourhoods of the form $(a,b)$ with $a < x < b$, as usual for the reals.

Knowing local bases, we also know convergence: for convergence within $\Bbb R$ this is the same as the usual notion, but $x_n \to \infty$ iff for each $K \in \Bbb R$, there exists some $N \in \Bbb N$ such that for all $n \ge N$ we have $x_n \ge K$. For $-\infty$ we have $x_n \le K$ instead.

We could give it a metric, as the space is (regular and) second countable (and order isomorphic to $[0,1]$) but we don't need to do so.