What is the precise relationship between $\omega+1$ and any convergent sequence? Let $\omega+1$ be defined as:
$$\omega+1 = \mathbb{N} \cup \{\omega\}$$
Where $\omega \geq n, \forall n \in \mathbb{N}$, where $\geq$ is the usual ordering relationship 

What is the relationship between $\omega+1$
  and the set 
$$\mathbb{X} = \{x_n | n \in \mathbb{N}\} \cup \{x\}$$ Where $x_n \to x$, as $n \to
> \infty$

I can see that $\omega+1$ bijects into $\{x_n | n \in \mathbb{N}\} \cup \{x\}$ via $$f: \omega +1 \to \mathbb{X},  f(i) = x_i, i \in \mathbb{N}, f(\omega) = x$$
Is this a homeomorphism? 
I think it will depend on whether $\omega+1$ with its order topology is discrete. If it is, then $f$ is a homeomorphism. 
Is there some other relationship I might be missing? What are some applications of this relationship? Any link/info will be much appreciated.
 A: Let $X$ be a topological space, $(x_n)$ be a sequence in $X$, and $x\in X$.  Then $(x_n)$ converges to $x$ iff the map $f:\omega+1\to X$ defined by $f(n)=x_n$ and $f(\omega)=x$ is continuous.  (Note that $f$ need not be injective, so you certainly shouldn't expect it to be a homeomorphism to its image in general.)
The proof is straightforward.  The key observation is that a set $U\subseteq \omega+1$ is open iff $\omega\not\in U$ or $U$ is cofinite.  So to say that $f$ is continuous is just to say that if $V\subseteq X$ is open and $\omega\in f^{-1}(V)$, then $f^{-1}(V)$ is cofinite (if $\omega\not\in f^{-1}(V)$, then $f^{-1}(V)$ is automatically open).  But $\omega\in f^{-1}(V)$ iff $f(\omega)=x\in V$, so this is just saying that $f^{-1}(V)$ is cofinite for any neighborhood $V$ of $x$.  That is, every neighborhood of $x$ contains $x_n$ for all but finitely many $n$.  That is, $(x_n)$ converges to $x$.
This fact is occasionally useful.  For instance, it implies that any sequential space is compactly generated, since $\omega+1$ is compact.  This fact can also be generalized to nets; for an overview of how this works and an application see this answer of mine.
