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I understand this question is in risk of being a duplicate, but I think the abundance of questions on the topic also suggests there are still some confusion about this. I hope this will contribute in simplifying things somewhat, as I am even more confused after reading through other threads (although it's unlikely I have found and read through all of them).

Baire's Theorem is very new to me, so I am still digesting it. Now, the definition of second category as a set that cannot be written as a union of countably many nowhere dense subsets, seems clear to me. Now, there seems to be many different definitions of a space being a Baire space. See questions/discussions e.g. here, and here. But, one simple characterization I see a lot is that a Baire space is one that satisfies the conclusion of Baire's Category Theorem (which of course also has many formulations).

Now, in our textbook Foundations of Modern Analysis, by A.Friedman, for a course in Functional Analysis, we have the following formulation (Theorem 3.4.2 in Friedman):

BCT 1. A complete metric space is a space of second category.

(A simple formulation, but weirdly, I don't think I've seen it anywhere else.) In the lectures we have used the following:

BCT 2. If X is a complete metric space, and $X=\cup_n F_n,\, F_n=\overline{F_n}$, then there exists $k$ s.t. $F_k$ has nonempty interior.

These two formulations (which are clearly equivalent) of BCT (and thus of something being a Baire space) seems somewhat clear and intuitive, and I would like to use and remember the Baire property of a space in terms of something like "not consisting of meagre sets", as this makes sense.

Now, on Wikipedia, I've found this definition of Baire space:

The definition for a Baire space can then be stated as follows: a topological space X is a Baire space if every non-empty open set is of second category in X.

Question 1: It looks to me as simple as: A Baire space is a space of second category, is this right?

Question 2: (Regardless if the answer to 1 is yes or no) Could (and how would I) prove, more or less directly (i.e. without moving to other formulations of Baire spaces), that an open subset of a second category set is of second category?

(There is a proof that open subsets of Baire spaces are Baire with another common definition of a Baire space as: The interior of every union of countably many closed nowhere dense sets is empty. But, then how is this definition equivalent to mine of just being second category?)

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    $\begingroup$ Rather use "meagre" than "thin". In Several Complex Variables, "thin" sets are sets locally contained in the zero set of a holomorphic function (that's not identically $0$). Then thin sets are meagre, but generically meagre sets aren't thin. $\endgroup$ – Daniel Fischer♦ Nov 4 '19 at 14:56
  • $\begingroup$ Fair enough, I just wanted the intuition to be there; fixed! $\endgroup$ – Christopher.L Nov 4 '19 at 15:47
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The answer to question 1 is "no". A Baire space contains no nonempty open subset that is of the first category, but a space that is of the second category may. Consider $X = (-\infty) \cap \mathbb{Q} \cup (0,+\infty)$ in the subspace topology inherited from $\mathbb{R}$. That space is of the second category since $(0,+\infty)$ is a Baire space, but $(-\infty,0) \cap \mathbb{Q}$ is a countable union of nowhere dense sets, yet open in $X$. Informally, one could say that a Baire space "is of second category at each point", but making that formal would — I think — lead to one of the standard definitions again.

It thus follows that the answer to question 2 as posed is "You can't", for in the example above $(-\infty,0) \cap \mathbb{Q}$ is an open subset of a second category set, but it is of the first category.

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  • $\begingroup$ So, BCT as we (and Friedman) have it doesn't really characterize Baire spaces, but rather 2.cat. spaces, which is weaker? If we add that any open subset needs to be 2.cat., we get Baire in the usual sense? As you might suspect I wanted answ to 2 to move from R to any open interval in your answer here, as we discussed there,but without using Baire and only sec.cat. Then I could perhaps use completeness of closed subsets of compl. spaces or such. And yes, I have a problem of letting things go. $\endgroup$ – Christopher.L Nov 4 '19 at 15:57
  • $\begingroup$ It's a bit more complicated. Complete metric spaces are Baire spaces, not merely second category spaces. Completeness also doesn't just say something about the space as a whole, it says something about the local structure too, like Baire does. In a complete metric space, every point has a neighbourhood basis consisting of complete sets, and that implies that every nonempty open set in a complete metric space is of second category. And yes, one equivalent definition of a Baire space is that every nonempty open set is of second category. $\endgroup$ – Daniel Fischer♦ Nov 4 '19 at 16:06
  • $\begingroup$ In a different direction, the BCT as given by Friedman doesn't characterise Baire spaces because there are Baire spaces that aren't even metrisable (e.g. every locally compact space [just in case, for me "locally compact" implies Hausdorff] is a Baire space). $\endgroup$ – Daniel Fischer♦ Nov 4 '19 at 16:10
  • $\begingroup$ Hmm, well none of these things are things that feels they should be clear to us at this point (in the course) so I have a hard time seing how I should move to any open int. from R there (Unless it's easy to prove R (or any complete space) is Baire). But, I'll ponder it some more, and'll accept your answer when I read it in a bit more detail later. $\endgroup$ – Christopher.L Nov 4 '19 at 16:13
  • $\begingroup$ Moving from $\mathbb{R}$ to an open interval at which level? At the top level it's quite easy because every (nonempty) open interval is homeomorphic to $\mathbb{R}$. Since Baire is a topological concept, it follows that nonempty open intervals are Baire spaces too. At a lower level, say to show that $M \cap (a,b)$ is meagre in $(a,b)$ if $M$ is meagre in $\mathbb{R}$, more needs to be said, but it's mostly not deep (it's of course only easy after one has enough experience with that type of argument). $\endgroup$ – Daniel Fischer♦ Nov 4 '19 at 16:24

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