What is the precise relationship between $\omega+1$ and any convergent sequence?

Question

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.

The convergent sequence can certainly be constant, so there is no bijection in general... — Mariano Suárez-Álvarez, Aug 01 '16 at 05:12
Think. $\omega$ is a limit ordinal. Is the order topology on $\omega +1$ really discrete? — Forever Mozart, Aug 01 '16 at 05:22
@ForeverMozart Well for each $n \in \mathbb{N}$, we can take an open set $(n-1, n+1)$, then the $\mathbb{N}$ part is discrete. For $\omega$, we can similarly take $(\omega-1, \omega+1) = (\omega - 1, \omega]$ which is open....? — Olórin, Aug 01 '16 at 06:20

score 3 · Accepted Answer · edited Apr 13 '17 at 12:20

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.

What is the precise relationship between $\omega+1$ and any convergent sequence?

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