# Why metric space $\mathbb{R}$ with the standard metric cannot be written as a countable union of nowhere dense sets?

I am trying to figure out why the metric space $\mathbb{R}$ with the standard metric cannot be written as a countable union of nowhere dense sets.

Then, another natural question is: Can we write $\mathbb{Q}$ as a countable union of nowhere dense sets?

Can anybody help me with this?

Thanks for giving me time.

Edited: I need a little more explanation. It seems that it is a consequence of Baire category theorem, but I haven't studied this before.

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@srijan: "I need a little more explanation." Of what? Do you have a question about how Baire's theorem applies? –  Jonas Meyer May 29 '12 at 4:16
@JonasMeyer ya, i want to know how can we apply Baire's theorem here? –  srijan May 29 '12 at 4:19
@srijan: $\mathbb R$ with the standard metric is a complete metric space, and Baire's theorem says in part that a (nonempty) complete metric space cannot be written as a countable union of nowhere dense sets. –  Jonas Meyer May 29 '12 at 4:41
@srijan: You are correct that completeness is not a topological property, in the sense that it is not preserved by homeomorphism. That is more or less the point in the example here, because $(0,1)$ and $\mathbb R$ with the standard distance are homeomorphic, but one is complete and the other isn't. But the main reason I brought up homeomorphism in my last comment was to point out a way to see that $(0,1)$ is not a countable union of nowhere dense sets. If this property holds for a topological space, it holds for every homeomorphic space, because being nowhere dense depends only on topology. –  Jonas Meyer May 29 '12 at 5:31

Of course, a measure is not necessary to show that $\mathbb R$ cannot be written as a countable union of nowhere dense sets. Any complete metric space is a Baire space. –  Alex Becker May 29 '12 at 3:47