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Does there exist a metric $d$ on $\mathbb{Q}$ such that $(\mathbb{Q},d)$ is compact?

EDIT

NOT considering $\mathbb{Q}$ as a subspace of $\mathbb{R}$. Can there exist a metric on $\mathbb{Q}$ which makes it compact? We may decide later what does it do to $\mathbb{R}$.

Subhajit Paul
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2 Answers2

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Consider $\{{1\over n},n\in\mathbb{N},n >0\}\cup\{0\}$ it is compact and there exists a bijection between this set and $\mathbb{Q}$.

Tsemo Aristide
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  • Yes, but how to guarantee that that function is a homeomorphism? – Subhajit Paul Apr 29 '20 at 18:44
  • The topology of $\mathbb{Q}$ as a subtopological space of $\mathbb{R}$ is not complete, so it cannot be endowed with the topology of a compact metric space. – Tsemo Aristide Apr 29 '20 at 18:50
  • But completeness is not a topological property whereas compactness is. – Subhajit Paul Apr 29 '20 at 18:55
  • https://math.stackexchange.com/questions/627667/every-compact-metric-space-is-complete – Tsemo Aristide Apr 29 '20 at 18:58
  • So my question is same to ask that does there exist a metric on $\mathbb{Q}$ which makes it complete and totally bounded? – Subhajit Paul Apr 29 '20 at 19:00
  • Completeness was a red herring. $\mathbb Q$ is not compact (in the usual topology), so there is no homeomorphism from it to a compact space. – Robert Israel Apr 29 '20 at 19:21
  • @RobertIsrael please see the edited version of my question. I am sorry if my question was misunderstood. – Subhajit Paul Apr 29 '20 at 19:31
  • So take the metric corresponding to Tsemo's answer. That is, if $f$ is a bijection from $\mathbb Q$ onto ${1/n: n \in \mathbb N, n > 0} \cup {0}$, you take the induced metric $d(r,s) = |f(r) - f(s)|$. – Robert Israel Apr 29 '20 at 19:43
  • @RobertIsrael I am sorry if I am missing something silly. Without the continuity of $f$ how does the compactness of the set flow to $\mathbb{Q}$? – Subhajit Paul Apr 29 '20 at 20:01
  • $f$ is not continuous with respect to the standard topology on $\mathbb Q$, but you don't care about that topology. It is continuous with respect to the topology given by the metric $d$, in fact by construction it is an isometry. – Robert Israel Apr 30 '20 at 15:42
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Here is a slightly more explicit take on Tsemo's answer:

Define the injective map $\phi:\mathbb{Q} \to \mathbb{C}$ by $\phi(0) = 0$ and if $q = { a \over b}$ where $a,b$ are coprime and $a \neq 0, b>0$ then $\phi({a \over b}) = {1 \over b} (-1 + e^{i {7 \over 8} \pi {1 \over a}})$. Let $d(q_1,q_2) = |\phi(q_1)-\phi(q_2)|$.

The following crude picture might help illustrate: enter image description here

Two important characteristics of $\phi$ are that $\lim_{|a| \to \infty} \phi({a \over b}) = 0$ and $\lim_{b \to \infty} \phi({a \over b}) = 0$.

It is straightforward to verify that $d$ is indeed a metric.

It is sufficient to show that $(\mathbb{Q},d)$ is sequentially compact.

Suppose $q_n \in \mathbb{Q}$. Let $q_n = {a_n \over b_n}$ with $b_n \ge 1$, and $a_n,b_n$ coprime.

If $a_n = 0$ infinitely often then it is clear that there is a convergent subsequence (to zero), so we can suppose $|a_n| \ge 1$.

If $(a_n,b_n)$ is bounded in the usual metric then it is clear that $(a_n,b_n)$ has some subsequence that converges to some $(a,b)$ (in fact we have $(a_n,b_n) = (a,b)$ infinitely often).

Otherwise either $|a_n|$ or $b_n$ is unbounded, and hence any corresponding subsequence converges to zero.

Note in particular that if the $q_n$ are all distinct then $q_n \to 0$.

copper.hat
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