Existence of a metric space where each open ball is closed and has a limit point Show that there exists a metric space in which every open ball is closed and contains a limit point.
I think that the space $\{\frac{1}{n}\mid n\in\mathbb{N},n>0\}\cup \{0\}$ with the standard Euclidean metric is an answer, but it is not true open ball with center 1 is closed.

Let $Y$ be any set with more than one element, and let $X=Y^{\Bbb N}$. If $x=\langle x_n:n\in\Bbb N\rangle$ and $y=\langle y_n:n\in\Bbb N\rangle$ are distinct points of $X$, let $e(x,y)=\min\{n\in\Bbb N:x_n\ne y_n\}$. Now define
$$d:X\times X\to\Bbb R:\langle x,y\rangle\mapsto\begin{cases}
0,&\text{if }x=y\\
2^{-e(x,y)},&\text{if }x\ne y\;.
\end{cases}$$
It’s a standard exercise to show that $d$ is a metric on $X$. Let $\epsilon>0$ and $x\in X$ be arbitrary. Then the set $\{n\in\Bbb N:2^{-n}<\epsilon\}$ has a smallest element, say $m$, and 
$$B(x,\epsilon)=\{y\in X:d(x,y)\le 2^{-m}\}\;,$$
which is evidently a closed set. 
Finally, let $x=\langle x_n:n\in\Bbb N\rangle$ be arbitrary. For each $n\in\Bbb N$ let $y_n\in Y\setminus\{x_n\}$, and define the point $x^{(n)}=\langle x_k^{(n)}:k\in\Bbb N\rangle\in X$ by
$$x_k^{(n)}=\begin{cases}
y_n,&\text{if }k=n\\
x_k,&\text{if }k\ne n\;.
\end{cases}$$
Then $e\left(x,x^{(n)}\right)=n$, so $d\left(x,x^{(n)}\right)=2^{-n}$, and the sequence $\langle x^{(n)}:n\in\Bbb N\rangle$ converges to $x$ in $X$. Thus, $\langle X,d\rangle$ is a metric space with no isolated points in which every open ball is closed.
As a side note, if $Y$ is finite, $X$ is homeomorphic to the middle-thirds Cantor set, and if $Y$ is countably infinite, $X$ is homeomorphic to $\Bbb R\setminus\Bbb Q$, the space of irrational numbers.
