# On “simplifying” the definition of a Baire space

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?)

• 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. – Daniel Fischer Nov 4 '19 at 14:56
• Fair enough, I just wanted the intuition to be there; fixed! – Christopher.L Nov 4 '19 at 15:47

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 $$X = (-\infty, 0) \cap \mathbb{Q} \cup (0,+\infty)$$ 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.
• 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). – Daniel Fischer Nov 4 '19 at 16:24