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?
Irrational numbers with induced topology Polish
functional-analysisgeneral-topologymetric-spaces
Best Answer
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.