I read that the irrational numbers with topology induced by the standard topology on $\mathbb{R}$ is a Polish space (i.e. a separable, completely metrizable topological space). Now the metric induced by the standard metric on $\mathbb{R}$ would not make the space complete. So what metric could you use that induces the topology induced by the standard topology on $\mathbb{R}$ and makes the irrational numbers complete?
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see https://en.wikipedia.org/wiki/Baire_space_(set_theory)#Relation_to_the_real_line – Abdelmalek Abdesselam Apr 28 '21 at 14:36
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I don't know the full details so I can't post an answer, but I think there's something you can do by mapping Baire space to the irrationals via continued fractions. You'd have to worry about some details regarding sign and zeroes, but I believe this approach will be fruitful. – paul blart math cop Apr 28 '21 at 14:39
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It’s a general fact that a $G_\delta$ subset of a complete metric space is completely metrisable. And the irrationals are a $G_\delta$ is their complement is a countable set (and thus $F_\sigma$).
A concrete metric that works: let $\{q_n\mid n \in \Bbb N\}$ be an enumeration of the rationals, and define $$d(x,y)= |x-y| + \sum_{n=1}^\infty 2^{-n}\min(1, \left|\frac{1}{|x-q_n|} - \frac{1}{|y-q_n|}\right|)$$
An alternative is to show that the irrationals are homeomorphic to $\Bbb N^{\Bbb N}$ via continued fractions. And a countable product of Polish spaces is Polish.
Henno Brandsma
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1Alright thanks, that confirms my feeling that there's not an obvious simple metric for this :D – joinijo Apr 28 '21 at 14:54
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2@joinijo you have to ensure a sequence stays away from the rationals to make it complete. – Henno Brandsma Apr 28 '21 at 14:55
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1For the alternative approach Martin Sleziak’s two answers here present the continued fraction argument in detail and also Arnold Miller’s alternative construction of a homeomorphism between $\Bbb Z^\omega$ and $\Bbb R\setminus\Bbb Q$. – Brian M. Scott Apr 28 '21 at 19:28