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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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    http://en.wikipedia.org/wiki/Baire_category_theorem2012-05-29
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    @srijan: "I need a little more explanation." Of what? Do you have a question about how Baire's theorem applies?2012-05-29
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    @JonasMeyer ya, i want to know how can we apply Baire's theorem here?2012-05-29
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    @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.2012-05-29
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    @JonasMeyer Incomplete metric space may or may not be written as a countable union of nowhere dense sets. I think incomplete metric space Q can be written as a union of one point sets. Am i right?2012-05-29
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    @srijan: Yes, right on both accounts. For another example, the incomplete metric space $(0,1)$ with the standard metric cannot be written as a countable union of nowhere dense subsets. For one thing, an incomplete metric space might be homeomorphic to a complete metric space. For another, Baire's theorem also applies to locally compact Hausdorff spaces, which in some cases are not even metrizable.2012-05-29
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    @JonasMeyer Why the incomplete metric space (0,1) can't be written as a countable union of nowhere dense subsets? I think Nowhere dense means that you can't find even a tiny interval where the points are dense. Is that correct? I would be happy if you can provide me a better reasoning.2012-05-29
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    @srijan: If $X$ is a topological space, then a subset $A$ is called nowhere dense if the closure $\overline A$ has empty interior, or equivalently if the open set $X\setminus\overline{A}$ is dense in $X$. What you said may be equivalent for $\mathbb R$, but it doesn't hurt here to be precise and general. One way to see that $(0,1)$ can't be written as a countable union of nowhere dense sets is to note that the latter condition depends only on the topology, and $(0,1)$ is homeomorphic to the complete metric space $\mathbb R$.2012-05-29
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    @JonasMeyer What it has to do with homeomorphism between (0,1)and $\mathbb{R}$? I am sorry if i may sound stupid but i think completeness is not a topological property so it doesn't matter whether both $\mathbb{R}$ and open interval (0,1) are complete to establish homeomorphism between them.2012-05-29
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    @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.2012-05-29
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    @JonasMeyer oh i got your point. Many many thanks to you for such a long discussion.2012-05-29

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