Convergent sequence in Hausdorff space is homeomorphic to $\{1/n\}\cup \{0\}$? So I was reading this question
How to show any convergent sequence is strongly discrete in Hausdorff space?
and I was wondering if it's true that any convergent sequence $\{a_n\}_{n\in\mathbb{N}}$ (with the addition of the limit point $a$) in a Hausdorff space $X$ where $a_i \neq a_j$ if $i \neq j$ is actually homeomorphic to the space $\{1/n\}\cup\{0\}$? This appears to be the case to me, since we could apply the fact that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism to the map $1/n \mapsto a_n, 0 \mapsto a.$ However, I am not well versed in topology so I would like to confirm whether this is indeed true.
 A: This is quite true. If $a_n$ is a sequence of distinct elements in a Hausdorff space $X$ and $a_n \to a\in X$ where $a\neq a_n$ for all $n$, then
$S=\{a_n \mid n \in \Bbb N\} \cup \{a\}$ is homeomorphic to $K=\{\frac{1}{n}: n \in \Bbb N\} \cup \{0\}$ as a subspace of $\Bbb R$ (which is again homeomorphic to the one-point compactification of the countable discrete space $\alpha D(\omega)$).
Argument (for $S$ and $K$): define $f: K \to S$ by $f(\frac1n)=a_n, f(0)=a$. Then $f$ is continuous at any $x=\frac1n$ as any such $x$ has an open interval $(c_x,d_x)\ni x$ such that $(c_x, d_x) \cap S = \{x\}$ (just take $c_x$ the midpoint between $\frac{1}{n+1}$ and $\frac1n$, and $d_x$ that between $\frac1n$ and $\frac{1}{n-1}$ (or $d_x=2$ if $n=1$), so each $\frac1n$ is an isolated points of $K$ and $f$ is always continuous at isolated points. And if $O$ is an open set of $X$ containing $a$, it contains all $a_n$ for $n \ge N$ for some $N$ (by convergence) and so $f^{-1}[O]$ is also an open neighbourhood of $0$ in $K$ (it contains the tail $\{\frac1n, n \ge N\}$). This show continuity of $f$ everywhere.
Now the fact that $f$ is a bijection from compact $K$ to a Hausdorff $S$ shows it is a homeomorphism.
