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The definition of "being of first category) that I am using is provided below. I usually see this mentioned without specifying in which space it is meant.

For example, Engelen says in one of his paper "Characterizations of the countable infinite product of rationals and some related problems" that "$\mathbb{Q}^\omega$ is clearly of first category". But in another his paper, he says $\mathbb{Q}^\omega$ is of first category in the Cantor space $2^\omega$. Both he says without proof, which confuses me.

Question 1: Is $\mathbb{Q}^\omega$ of first category (in itself)? And how to prove that?

Question 2: Is this notion by default used for the space itself?

Thank you!

Definitions

  1. $\mathbb{Q}^\omega$ = the set of all rational sequences with the standard product topology

  2. A subset of a topological space $X$ is said to be of first category in $X$ if it is a countable union of nowhere dense subsets of $X$. ($X$ is called nowhere dense (or rare) in $X$ if its closure has empty interior)

Tereza Tizkova
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  • Your second definition is what it means for a set to be of first category ( in the sense of Baire) a set which is not of first category, is then said to be of second category – Mittens May 01 '22 at 20:15
  • @OliverDíaz So this definition is for sets and it doesnt matter if they are subspaces of anything? While "first category in something" refers to being also subspace of something? I think I still dont understand. – Tereza Tizkova May 01 '22 at 20:34
  • @OliverDíaz So the condition $(int(\overline{E_n} \cap A)) \cap A = \emptyset$ is equivalent to "Closure of $E_n$ in $A$ has empty interior" which is equivalent to $E_n$ is nowhere dense in $A$? – Tereza Tizkova May 02 '22 at 09:27
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    Errata: A set may be of first category without being a (sub)-space of first category. See the posting – Mittens May 02 '22 at 16:55
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    When a topological space $(X,\tau)$ is given, a subset $A$ is a set of first category in $X$ if it can be expressed as a countable union of nowhere dense sets in $X$; A topological space $(X,\tau)$ is of a space of first category if $X$, as a subset of itself, is a set of first category. – Mittens May 02 '22 at 17:01

1 Answers1

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Question 1: Yes, $\mathbb{Q}^{\omega}$ is of first category in itself. Consider the following definition for a finite sequence of rational numbers $(q_0,\dots,q_n)$: $$ [(q_0,\dots,q_n)] = \{ x \in \mathbb{Q}^{\omega}: x_i = q_i, 0\leq i \leq n \} $$ Clearly $\mathbb{Q}^{\omega}$ is the countable union of all the sets of the form $[(q_0,\dots,q_n)] $. Also, you can prove that the sets of the form $[(q_0,\dots,q_n)] $ are closed in $\mathbb{Q}^{\omega}$ and have empty interior.

Question 2: This notion is always relative to the topological space in which you are working since it is defined for subsets of a topological space, so unless you have a larger topological space on which you are working, it is assumed that "of first category" is meant to be "of first category relative to itself".

Elma
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  • Thank you, that is nice answer! I am still not sure how would you prove that the sets of $[(q_0,\dots,q_n)] $ are closed in $\mathbb{Q}^\omega$and have empty interior? – Tereza Tizkova May 02 '22 at 09:43