Countable complete set of limit points Let $(X,d)$ be a metric space with $X$ - countable and such that for any $x\in X,r>0$ there exists $y\in B(x,r)$, $y\neq x$. Can $X$ be complete? I failed to prove that it cannot as well as to construct an example of such a set.
 A: Your hypotheses imply that $X$ doesn't contain an isolated point. This means that points are not open in $X$, hence $X = \bigcup_{x \in X} \{x\}$ is a countable union of nowhere dense sets. By the Baire category theorem this cannot happen if $X$ is completely metrizable, so a countable metric space without isolated points does not admit a complete metric.
In fact, the following stronger results are true:


*

*By a theorem of Sierpiński, every countable metric space without isolated points is homeomorpic to $\mathbb{Q}$ with its usual topology. See my answer here for more on that and for links to the literature. In fact, a subset of a completely metrizable space is completely metrizable if and only if it is a countable intersection of open sets by a classical theorem of P.S. Alexandrov, see this thread for more on that. Since $\mathbb{Q}$ is not a countable intersection of open sets in $\mathbb{R}$ (again by Baire), $\mathbb{Q}$ is not completely metrizable.

*Every complete metric space without isolated points contains a copy of the Cantor set. In particular, it must be uncountable. This is explained by Brian M. Scott in this thread.
A: And because I’d already written up most of it before I checked back to see whether the question had been answered, here’s a (mostly) self-contained proof that if $X$ is a countable metric space without isolated points, then $X$ is homeomorphic to $\mathbb{Q}$. It’s essentially the same argument that one uses to find a copy of the Cantor set in a complete metric space without isolated points, except that one gets only a dense subset of the Cantor set.
Let $D_X = \{d(x,y):x,y \in X\}$; $D_X$ is countable, so $\mathscr{B} = \{B(x,r):r \in \mathbb{R}\setminus D_X\}$ is a clopen base for $\langle X,d \rangle$. Let $\{x_n:n\in\omega \}$ be an enumeration of $X$.
Recursively construct clopen sets $H_\sigma$ for each $\sigma \in \{0,1\}^{<\omega}$ as follows. $H_\varnothing = X$. Given $H_\sigma$, where $\sigma \in \{0,1\}^n$, let $m = \min\{k\in\omega:x_k \in H_\sigma\}$, choose $B(x_m,r) \in \mathscr{B}$ such that $B(x_m,r) \subseteq H_\sigma$, $H_\sigma \setminus B(x_m,r) \ne \varnothing$, and $r < 2^{-n}$, set $H_{\sigma^\frown 0} = B(x_m,r)$, $H_{\sigma^\frown 1} = H_\sigma \setminus B(x_m,r)$, and $x_\sigma = x_m$. Let $\mathscr{H} = \{H_\sigma:\sigma \in \{0,1\}^{<\omega}\}$, and for $n\in\omega$ let $\mathscr{H}_n = \{H_\sigma:\sigma \in \{0,1\}^n\}$; each $\mathscr{H}_n$ is a clopen partition of $X$.
Observe first that for each $\sigma \in \{0,1\}^{<\omega}$, $x_{\sigma^\frown 0} = x_\sigma$: if $x_\sigma = x_m$, $H_{\sigma^\frown 0}$ does not contain any $x_k$ with $k<m$. It follows that for every $n \ge |\sigma|$ there is $\tau \in \{0,1\}^n$ such that $x_\tau = x_\sigma$. Next, observe that for each $x \in X$ there is a $\sigma \in \{0,1\}^{<\omega}$ such that $x = x_\sigma$. If not, let $m \in \omega$ be minimal such that $x_m \ne x_\sigma$ for any $\sigma \in \{0,1\}^{<\omega}$. Choose $n \in \omega$ large enough so that for each $k<m$ there is a $\sigma_k \in \{0,1\}^n$ such that $x_k = x_{\sigma_k}$, and moreover so that $2^{-n}<d(x_k,x_m)$ for each $k<m$. Then for each $k<m$ we have $x_m \notin H_{\sigma_k^\frown 0} \in \mathscr{H}_{n+1}$. Let $H_\tau \in \mathscr{H}_{n+1}$ contain $x_m$. Since $x_k = x_{\sigma_k} \in H_{\sigma_k^\frown 0}$ for each $k<m$, $m = \min\{i \in \omega:x_i \in H_\tau\}$, and hence $x_m = x_\tau$.
For each $x \in X$ let $\mathscr{H}(x) = \{H_\sigma \in \mathscr{H}:x \in H_\sigma\}$; it follows from the preceding observations that $\mathscr{H}(x)$ contains sets of arbitrarily small diameter and is therefore a clopen base at $x$. Let $$\sigma_x = \bigcup\limits_{H_\sigma \in \mathscr{H}(x)}\sigma \in \{0,1\}^\omega\;;$$ it also follows from those observations that $\sigma_x(k) = 0$ for all sufficiently large $k \in \omega$. Let $\Phi = \{\sigma \in \{0,1\}^\omega:\exists n\in\omega\,\forall k>n (\sigma(k)=0)\}$, and let $\varphi:X\to\Phi:x \mapsto \sigma_x$. Clearly $\varphi$ is injective, and since $H_\tau$ is non-empty (in fact infinite) for each $\tau \in \{0,1\}^{<\omega}$, $\varphi$ is in fact a bijection. Moreover, if $\Phi$ is topologized as a subspace of the product $\{0,1\}^\omega$, $\varphi$ is a homeomorphism: for each $\tau \in \{0,1\}^{<\omega}$, $\varphi[H_\tau] = \{\sigma \in \Phi:\sigma\upharpoonright \text{dom }\tau = \tau\}$, and clearly $\{\varphi[H]:H \in \mathscr{H}\}$ is a base for $\Phi$.
There’s a fairly easy homeomorphism between $\Phi$ and the dyadic rationals in $[0,1)$, and hence between $\Phi$ and $\mathbb{Q}\cap [0,\to)$, but it doesn’t seem to be particularly easy to prove directly that this is homeomorphic to $\mathbb{Q}$; it’s easier to do a little extra work in $\{0,1\}^\omega$ first. Let $f:\{0,1\}^\omega\to \{0,1\}^\omega$ be the autohomeomorphism that ‘flips’ each odd-numbered coordinate: for $\sigma \in \{0,1\}^\omega$, $$[f(\sigma)](k) = \begin{cases}
\sigma(k),&k\text{ even}\\
1-\sigma(k),&k\text{ odd}\;.
\end{cases}$$ Clearly $f[\Phi]$ is homeomorphic to $X$. Note also that since $\Phi$ is dense in $\{0,1\}^\omega$, so is $f[\Phi]$.
It’s well-known that $\{0,1\}^\omega$ is homeomorphic to the middle-thirds Cantor set $C$ via the map $$\sigma \mapsto \sum_{k\ge 0}\frac{2\sigma(k)}{3^k}\;.$$ Let $Y$ be the image of $f[\Phi]$; then $Y$ is dense in $C$. Moreover, every point of $Y$ has a ternary expansion whose digits from some point on are alternately $0$ and $2$, so $0,1\notin Y$, and $Y$ contains none of the endpoints of the deleted intervals. It follows that every point of $Y$ is a two-sided limit point of $Y$ and hence (1) that $Y$ with the order that it inherits from $\mathbb{R}$ is a dense linear order without endpoints, and (2) that the topology on $Y$ as a subspace of $C$ (or $\mathbb{R}$) is identical to the order topology on $Y$. It’s well known that up to order-isomorphism there is only one countable dense linear order without endpoints, so $Y$ is order-isomorphic to $\mathbb{Q}$ and therefore also homeomorphic to $\mathbb{Q}$. Thus (finally!) $X$ is homeomorphic to $\mathbb{Q}$.
