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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.

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By a result of Sierpinski's, any such space is homeomorphic to the rationals with the subspace topology, and the rationals with that topology are not completely metrizable. Or you could prove directly that a Polish space without isolated points contains a copy of the Cantor set. –  user83827 Oct 17 '11 at 19:25
    
I briefly discussed the result by Sierpiński @ccc is referring to in my answer here and gave references for the (non-trivial) proof. –  t.b. Oct 17 '11 at 19:36
    
I could've sworn that the topic of embedding the Cantor set into a Polish space without isolated points has also come up recently here, but I can't seem to find it. –  user83827 Oct 17 '11 at 19:39
    
A ha! I found what I was thinking of here: math.stackexchange.com/questions/68396 The first version of the question (and the answer by Brian M. Scott) only assumed completeness as I recall, even though compactness somehow entered the mix eventually. In any case, the argument is the same. –  user83827 Oct 17 '11 at 20:04
    
But, is not $\mathbb{Q}$, with the metric of absolute value, a counterexample? –  leo Oct 18 '11 at 8:45

2 Answers 2

up vote 7 down vote accepted

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:

  1. 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.

  2. 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.

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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 $\mathscr{H}(x) = \{H_\sigma \in \mathscr{H}:x \in H\}$; 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 $\mathbb{Q}$.

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Beautiful exposition, thank you! –  t.b. Oct 18 '11 at 23:14
    
An alternate approach is to use a model-theoretic back and forth argument (analogous to the back and forth proof that any countable dense linear order without endpoints is isomorphic to the rationals). But this provided writeup is very easy to follow as well. –  user83827 Oct 19 '11 at 0:23

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