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Definition: A set $M\subset X$ is called "of first category" if it is countable union of nowhere dense sets. Otherwise its called "of second category".

I want to see whether the following sets are of first or second category:

1) $\mathbb Q$

2) $\mathbb R$ \ $\mathbb Q$

3) $\mathbb Z$

4) $\mathbb R$ \ $\mathbb Z$

My attempt:

1) $\mathbb Q$ is of first category because we can write it as $\mathbb Q=\bigcup_{q\in \mathbb Q}\{q\}$. This is a countable union and each $\{q\}$ is closed with empty interior.

2) $\mathbb R$ \ $\mathbb Q$ is of second category. Prove: Suppose that is is of first category, hence $\mathbb R$ \ $\mathbb Q=\bigcup_{n\in \mathbb N}A_n$ with nowhere dense sets $A_n$. Then we can write $\mathbb R=\bigcup_{n\in \mathbb N}A_n\bigcup_{q\in \mathbb Q}\{q\}$ which is a contradiction because every complete metric space if of second category.

3) Actually I thought it is of first category (Same reasoning as in 1) ). But $\mathbb Z$ is a closed subspace of $\mathbb R$, hence also complete, so it must be of second category, I guess.

4) Can be answered when I know 3)


Can someone go through it? Thanks

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  • $\begingroup$ $\mathbb{Z}$ is not a closed subspace of $\mathbb{R}$ because it is not closed under scalar multiplication. $\endgroup$ – user159517 Feb 22 '15 at 19:23
  • $\begingroup$ Oh. I messed up closed subspace and closed set. Thanks :) let me complete the task, above. $\endgroup$ – Duke Feb 22 '15 at 19:25
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Questions 1 and 2 heb been proved quite correctly. The set $\mathbb{Z}$ is first category in $\mathbb{R}$, similarly as 1 is. It is however second category in itself (so the integers as a space in its own right is second category) by Baire's theorem again. $\{n\}$ is nowhere dense in the reals (as the reals have no isolated points) but not as a subset of the integers (as it is isolated, and so has non-empty interior!).

$\mathbb{Z}$ is even closed and nowhere dense, so very much first category. Its complement is again second category by the same argument, which just can be restated as "if $X$ is second catgeory in itself, and $A \subseteq X$ is first category, then $X \setminus A$ is second category in $X$", using the same proof.

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  • $\begingroup$ by in itself you mean you are considering a vector space $\mathbb{Z}$ over $\mathbb{Z}$? In my definition of vector space, you got to have a scalar $\textit{field}$ for a vector space. $\endgroup$ – user159517 Feb 22 '15 at 19:29
  • $\begingroup$ I guess "in itself" just means that we consider the induced topology from the greater space in the smaller one. Correct me if I am wrong, but actually we are not considering vector spaces?? $\endgroup$ – Duke Feb 22 '15 at 19:32
  • $\begingroup$ @user159517 No, I mean the difference between $\mathbb{Z}$ as a subspace of $\mathbb{R}$ or as a topological space in its own right. $\endgroup$ – Henno Brandsma Feb 22 '15 at 19:32
  • $\begingroup$ @Duke No, vector spaces have nothing to do with it. Yes, in the induced topology we can consider it as its own space. $\endgroup$ – Henno Brandsma Feb 22 '15 at 19:33

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