I know that sets of rational and irrational numbers are quite different. In measure, almost no real number is rational and of course, $\mathrm{card}(\mathbb Q) < \mathrm{card}(\mathbb R \setminus \mathbb Q) $ tells us that there are indeed much "more" irrationals than rational nubers.
On the other side, we can observe the following
- between every two different rationals, there are infinitely many irrations
- between every two different irrationals, there are infinitely many rationals
- both sets are dense in $\mathbb R$, i.e. every real number can be written as a limit of a sequence of both rationals or irrationals
- both are disconnected, neither open nor closed
- …
So I wonder, is there any way to distinguish $\mathbb Q$ and $\mathbb R \setminus \mathbb Q$ from a topological perspective as subspaces of $\mathbb R$? Is there any way to explain why the one set is so much bigger by looking at the topology? In what regard are they different?
Best Answer
The irrational numbers are a Baire space (and also the Baire space), and they are completely metrizable. This means that there is a complete metric space which is homeomorphic to the irrational numbers with the subspace topology.
The rational numbers, on the other hand, are not a Baire space and they are not completely metrizable which means they are not homeomorphic to any complete metric space.
From a "local" (Borel) perspective the irrationals are a $G_\delta$ set which is not $F_\sigma$, and the rationals (consequentially) are an $F_\sigma$ set which is not $G_\delta$. This means that the irrationals are the intersection of countably many open sets, but not the union of countably many closed sets.
I should add that being $G_\delta$ is sometimes denoted as $\bf\Pi^0_2$, and being $F_\sigma$ can be denoted as $\bf\Sigma^0_2$.
The two properties are almost the same. It can be shown that $G_\delta$ subsets of a complete metric space (like $\mathbb R$) are exactly those subsets which are completely metrizable. On the other hand, having no isolated points and being completely metrizable means that you're a Baire space (and therefore the rationals are not such space).