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?)
Best Answer
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.