A useful source of counterexamples in topology is $\mathbb R_\ell$, the set $\mathbb R$ together with the lower limit topology generated by half-open intervals of the form $[a,b)$. For example this space is separable, Lindelöf and first countable, but not second countable. So in particular, this implies that $\mathbb R_\ell$ is not metrizable.

Now Urysohn's metrization theorem says that any regular, second countable space is metrizable. Applying this to $\mathbb Q_\ell\subset \mathbb R_\ell$ gives the (for me rather counterintuitive) fact that there must exist a metric on $\mathbb Q$ inducing the lower limit topology. Which leads to the question:

What is a concrete example of a metric inducing the lower limit topology on $\mathbb Q_\ell$?

It certainly should be possible to go through the steps of the Urysohn metrization theorem to give a (semi-)concrete embedding of $\mathbb Q_\ell$ in $\mathbb R^\omega$ (for instance picking an enumeration of the rationals and constructing functions which separate points from closed sets, then look at the cartesian product of these maps) and then give the metric on $\mathbb Q_\ell$ in terms of this embedding.

But this is really not what I'm looking for, here. I'd be much more interested in a simple metric for $\mathbb Q_\ell$.

Because "simple" is not well-defined, I will definitely also welcome answers which exhibit a metric, but which I would not consider to be simple. On the other hand, if you see a reason for why there may just be no "simple" metrics, please feel free to point this out as well.


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    $\begingroup$ I thought $\mathbb{Q}_{\ell}$ was refering to the $\ell$-adic numbers. $\endgroup$ – M Turgeon Aug 7 '12 at 13:11
  • $\begingroup$ Why is $\mathbb{Q}_l$ second countable? $\endgroup$ – Rob Arthan Aug 7 '12 at 19:54
  • $\begingroup$ @RobArthan: Because the open sets of the form $[p,q)$ for $p,q\in \mathbb Q$ form a basis of the topology. $\endgroup$ – Sam Aug 7 '12 at 20:06
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    $\begingroup$ @MTurgeon: It seems you are not alone in being mislead by the old title. Thanks for pointing it out! My intention was not to mislead anyone, of course, so I have changed it now. $\endgroup$ – Sam Aug 7 '12 at 20:09
  • $\begingroup$ @Sam L. Thank you! I was being stupid. $\endgroup$ – Rob Arthan Aug 7 '12 at 21:04

Let $\nu:\mathbb{Q} \to \mathbb{N}$ be an enumeration of $\mathbb{Q}$. Then $\displaystyle d(x,y):=\sum_{\min(x,y) < r \le \max(x,y)} 2^{-\nu(r)}$ will do.

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    $\begingroup$ @tomasz: Not so. Notice, for instance, that for any $x < 0$, we have $d(x,0) \ge 2^{-\nu(0)}$. $\endgroup$ – Nate Eldredge Aug 7 '12 at 21:26
  • $\begingroup$ @NateEldredge: good point. :) This is a nice example. $\endgroup$ – tomasz Aug 7 '12 at 21:31
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    $\begingroup$ A remark: If we let $f(x) = \sum_{r \le x} 2^{-\nu(r)}$, then $d(x,y) = |f(x) - f(y)|$. In particular, $\mathbb{Q}$ with this metric is isometric (and in particular homeomorphic) to the set $\{f(q) : q \in \mathbb{Q}\} \subset [0,1]$ with the Euclidean metric. $\endgroup$ – Nate Eldredge Aug 7 '12 at 21:47
  • $\begingroup$ This is very nice. Thank you! So given an interval $[p,q)$, the $\epsilon$-ball $B_\epsilon(p)$ will be contained in $[p,q)$ if $\epsilon = \min\{2^{-\nu(p)},2^{-\nu(q)}\}$. Brilliant. $\endgroup$ – Sam Aug 8 '12 at 5:11

EDIT: This one should work out.

Here's another option: write your rationals as "mixed fractions," that is as their integer floor plus a fractional part and define $$d\left(a\frac{p}{q},b\frac{r}{s}\right)=\begin{cases}|a-b|,& a\neq b\\ d'\left(\frac{p}{q},\frac{r}{s}\right), &\textrm{otherwise} \end{cases}.$$

Define the distance between pure fractions separately. Assume WLOG that $\frac{p}{q}\leq \frac{r}{s}$. Then set $$d'\left(\frac{p}{q},\frac{r}{s}\right)=\max\left(\left|\frac{p}{q}-\frac{r}{s}\right|,\frac{1}{m}\right), \frac{k}{m} \in \left(\frac{p}{q},\frac{r}{s}\right]$$ We see, for instance, that this gives $\left[\frac{1}{2},\frac{5}{6}\right)$ open, since the distance from anything smaller than $\frac{1}{2}$ to $\frac{1}{2}$ is $\frac{1}{2},$ but each distance from $\frac{1}{2}$ to something less than $\frac{5}{6}$ is no more than $\frac{1}{3}$.

The symmetry and homogeneity axioms are immediate. Let's consider the triangle inequality. The only case in which it doesn't follow from that for the standard absolute metric is when we must show $d'\left(\frac{p}{q},\frac{r}{s}\right)+d'\left(\frac{r}{s},\frac{t}{u}\right)\geq d'\left(\frac{p}{q},\frac{t}{u}\right)$ and $d'\left(\frac{p}{q},\frac{t}{u}\right)=\frac{1}{m}$. But if the inequality failed, we'd have to have the interval $\left(\frac{p}{q},\frac{t}{u}\right]$ contain something with denominator $m$ bigger than any denominator in the intervals on the left-hand side-which is absurd, since depending on the ordering we have one of $ \left(\frac{p}{q},\frac{t}{u}\right]=\left(\frac{p}{q},\frac{r}{s}\right]\cup\left(\frac{r}{s},\frac{t}{u}\right], \left(\frac{p}{q},\frac{t}{u}\right]\subset \left(\frac{p}{q},\frac{r}{s}\right],$ or $\left(\frac{p}{q},\frac{t}{u}\right]\subset\left(\frac{r}{s},\frac{t}{u}\right]$.

So we have a metric. To get some half-open interval $[x,y)=\left[a\frac{p}{q},b\frac{r}{s}\right)$, take the union of all the $[m,m+1) \subset [a,b)$. Then for the least such $m$, construct $[x,m)$ by the infinite union $\bigcup_{i=0}^\infty\left[a\frac{p}{q+i},a\frac{p}{q+i}+\frac{1}{q+i+1}\right)$, getting all these intervals by the argument above about $\left[\frac{1}{2},\frac{2}{3}\right)$. Get $\left[b,b\frac{r}{s}\right)$ as the ball of radius $\frac{r}{s}$ around $b$, union the three pieces together, and we're done.

  • $\begingroup$ Do you mean $a\neq b$ in the first formula? $\endgroup$ – Cheerful Parsnip Aug 7 '12 at 21:10
  • $\begingroup$ Yep, thanks @JimConant. $\endgroup$ – Kevin Carlson Aug 7 '12 at 21:11
  • $\begingroup$ @tomasz: You're right, this won't do it. If it would work for the rationals, it would work for the reals. $\endgroup$ – Kevin Carlson Aug 7 '12 at 21:18
  • $\begingroup$ This seems to work. It's nowhere near as elegant as Alexander Shamov's answer, though, especially considering the many cases (some of which you omitted) with triangle inequality... Still, kudos on your determination. :) $\endgroup$ – tomasz Aug 7 '12 at 22:40
  • $\begingroup$ @tomasz, I agree-Alexander's solution just wasn't intuitive to me, so I thought it wouldn't be completely useless to share one with a few more of the strings still showing. $\endgroup$ – Kevin Carlson Aug 7 '12 at 23:05

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