Sequence converges iff function is continuous on extension of natural numbers (topology)

Question

Let $(X,d)$ be a metric space, and $\{x_n\}$ be a sequence given by $x:\mathbb{N}\rightarrow X$. Show $\{x_n\}$ converges to $c\in X$ if and only if

$$f:\mathbb{N}\cup\{+\infty\}\rightarrow X \ , f(t):=\begin{cases} x(t) & t\neq +\infty \\ c & t=+\infty \end{cases} $$

is continuous.

We define $\mathbb{N}\cup \{+\infty\}$ as the closure of $\mathbb{N}$ in $\mathbb{R}\cup \{\pm\infty\}$, where the topology on $\mathbb{R}\cup \{\pm\infty\}$ is generated from basis elements of the form $[-\infty,a)$ and $(b,\infty]$, for $a,b\in \mathbb{R}$.

I am not really sure how to approach such a problem. I may want to make use of the fact that if $(X,d_1)$ and $(Y,d_2)$ are metric spaces, then a mapping $f:X\rightarrow Y$ is continuous if and only if, whenever $\lim_{n\to\infty}x_n=c$ for a sequence in $X$, then $f\circ x$ converges in $Y$ with $\lim_{n\to\infty}f(x_n)=f(x)$

I apologize for the lack of work, and would appreciate any help/hints. I do believe this is very trivial, and I am probably over thinking the problem.

Answer 1

Assume x(t) converges to c.
Let U be an open subset of X.
If c not in U, then f$^{-1}$(U) is a subset of N,
thus an open subset of N', the extended integers.
If c in U, then x(t) in U for all t > n for some n.
(n,oo] is open within N'.
Thus as f$^{-1}$(U) is (n,oo] and a finite
number of integers, it is open.
Whereupon f is continuous.

Assume f is continuous. Let V be an open nhood of c.
As f(oo) in V, exists open U nhood oo with f(U) subset V.
Thus some n with (n,oo] subset U. f((n,oo]) subset V.
Whereupon x(t) converges to c.

As this proposition holds for all topological spaces,
it is myoptic to prove it just for metric spaces.

Answer 2

Suppose that $x_n \to c$ in $X$. Note that $x$ as a function defined on $\mathbb{N} \cup \{+\infty\}$ is already continuous at all $t \in \mathbb{N}$, because those are isolated points. To check that $x$ is continuous at $+\infty$, we take a neighbourhood $O$ of $c = f(+\infty)$ and then convergence to $c$ gives us $N \in \mathbb{N}$ such that for all $n \ge N: x_n \in O$. But then taking the neighbourhood $(N,+\infty] \cap (N \cup \{+\infty\})$ of $+\infty$ we have that for all $t \in U$, $x(t) = x_n$, (for some $n \ge N$), lies in $O$, or equivalently: $x[U] \subseteq O$. This shows continuity of $x$ at $+\infty$, and thus of $x$.

The reverse implication is entirely similar. From a neighbourhood of $+\infty$ we produce the $N$ needed for convergence.