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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?

Asaf Karagila
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Dario
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2 Answers2

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  • 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).

Asaf Karagila
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    Note, however that there exist metrisable Baire spaces that are not completely metrisable. See for example http://math.stackexchange.com/questions/243909/what-are-some-motivating-examples-of-exotic-metrizable-spaces/243962#243962 – kahen Jan 23 '13 at 19:18
  • That's an interesting difference... – rschwieb Jan 23 '13 at 19:19
  • @kahen: You are correct. I tried to improve the answer so it won't sound as if all Baire spaces are completely metrizable. – Asaf Karagila Jan 23 '13 at 20:09
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    Thanks very much, this was the kind of characterization I was hoping for. I especially like the completely metrizable-property, because it doesn't mention countability/uncountability in the first place. – Dario Jan 23 '13 at 22:07
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Both spaces have classical characterizations: the rationals $\mathbb{Q}$ is the (up to homeomorphism) unique countable metric space without isolated points. The irrationals are the (up to homeomorphism) unique separable metric space that is completely metrizable, zero-dimensional (it has a base of clopen sets) and nowhere locally compact (no point has a compact neighbourhood). All of these properties $\mathbb{Q}$ also has, except being completely metrizable (because it is not a Baire space).

Henno Brandsma
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