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I am asked to show that the irrational numbers are not a countable union of closed subsets of $\mathbb{R}$ given that if a complete metric space is the countable union of of closed subsets then at least one of them has a nonempty interior.

So far: I assume that the irrationals are the countable union of closed subsets of $\mathbb{R}$. I know that the irrationals are uncountable, so we cannot have this union consist of all singleton subsets. Therefore these closed sets must consist of intervals of irrational numbers.

At this point I cannot see how these intervals must all have a nonempty interior. Since each of these sets is closed, its closure is itself, and and so the closure minus the boundary of the interval should be nonempty given that any interval no matter how small contains infinitely many rationals and irrationals.

Can someone please point me in the right direction and/or point out the flaws in my reasoning?

Thanks

Asaf Karagila
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AnotherPerson
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    There are uncountable closed subsets of the irrationals that look nothing at all like intervals of irrationals. For example, the irrationals contain closed subsets homeomorphic to the middle-thirds Cantor set. – Brian M. Scott Oct 22 '15 at 05:51
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    Do you know that the irrationals are completely metrizable? They’re not complete in the usual metric, but they’re complete in an equivalent metric. That means that the result follows immediately from the Baire category theorem. – Brian M. Scott Oct 22 '15 at 05:53
  • @BrianM.Scott we have not learned about metrizability yet. – AnotherPerson Oct 22 '15 at 05:57
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    Then how on earth are you supposed to use a fact about complete metric spaces? The statement of the problem assumes either a knowledge of what a complete metric space is (at minimum) or the knowledge that the irrationals can be viewed as a complete metric space. – Brian M. Scott Oct 22 '15 at 06:00
  • Actually, one way to show that the irrationals are complete which can be used in this context is the fact that no sequence of irrationals can converge to a rational (unlike in the case of the rationals). – Alekos Robotis Oct 22 '15 at 06:09
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    @Antonios-AlexandrosRobotis: That’s obviously not true: the irrationals are dense in the reals, so every real number is the limit of a sequence of irrationals. In fact the irrationals are completely metrizable because they’re a $G_\delta$ set in the complete metric space of reals. – Brian M. Scott Oct 22 '15 at 06:15
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    For a simple example, take $q\in \mathbb{Q}$ and look at $q+\frac{\sqrt{2}}{n}$ for $n\in\mathbb{N}$. – Alexander Gruber May 02 '16 at 03:51
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    See this post and other questions linked there. – Martin Sleziak Dec 06 '16 at 02:35

1 Answers1

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One statement of the Baire category theorem is that a complete non empty metric space cannot be written as a countable collection of nowhere dense sets. (A set $N$ is nowhere dense if $(\overline{N})^\circ = \emptyset$.)

(Reminds me of a joke: While discussing the double negative, an english professor announced that there is no double positive in the English language to which a smart alec in the back row shouted 'Yea, right'.)

So, suppose $\mathbb{Q}^c = \cup_k C_k$, where the $C_k$ are closed and nowhere dense. Since we can write $\mathbb{Q} = \cup_k \{q_k\}$, where the $\{q_k\}$ are obviously nowhere dense, then we could write $\mathbb{R} = \mathbb{Q}^c \cup \mathbb{Q} = \cup_k (C_k \cup \{q_k\})$, which would contradict the completeness of $\mathbb{R}$.

Hence, if $\mathbb{Q}^c = \cup_k C_k$, at least one $C_k$ must not be nowhere dense, hence $C_k^\circ \neq \emptyset$ and hence contains some non empty interval $(a,b)$. But since this must contain a rational, we have a contradiction.

copper.hat
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