From Planetmath

A meager or Baire first category set in a topological space is one which is a countable union of nowhere dense sets.

A Baire second category set is one which contains a countable union of open and dense sets.

From Wikipedia:

A subset of a topological space X is called

  • nowhere dense in X if the interior of its closure is empty
  • of first category or meagre in X if it is a union of countably many nowhere dense subsets
  • of second category or nonmeagre in X if it is not of first category in X

I was wondering

  1. according to Wikipedia's definition, is any subset of a topological space either of first category or of second category?
  2. are the definitions for second category set in Planetmath and Wikipedia consistent with each other?
  3. Wikipedia says these definitions are used for "historical definition" of Baire space. I was wondering if they are archaic i.e. no longer in use?

Thanks and regards!

  • $\begingroup$ What is your doubt in answering question 1? $\endgroup$ Feb 6 '12 at 13:42
  • $\begingroup$ @wildildildlife: no now. I felt a little unsure. $\endgroup$
    – Tim
    Feb 6 '12 at 13:49
  • $\begingroup$ With the correction to Planetmath, replacing "union" with "intersection" the def'ns from the 2 sources are not identical for all spaces. Consider the co-finite topology on $N$: Any $S\subset N$ is open iff $(S=\phi)\lor (N\backslash S $ is finite ). Then $N=\cup \{\{n\} : n\in N\}$ is meager in $N$. But $X$ is dense open in $X$ so according to Planetmath it is second category ,but not according to Wikipedia. I think the answers given so far are of good quality. $\endgroup$ Nov 22 '15 at 23:21

so to recap: the more modern definition

1) $X$ is a Baire space iff every countable union of closed nowhere dense sets has empty interior.

Equivalent by taking complements (note that a set $A$ is nowhere dense iff its complement $X \setminus A$ contains an open dense subset) to my favourite formulation, which seems to be more commonly used among topologists:

1') $X$ is a Baire space iff every countable intersection of open and dense subsets is dense.

(note that in any space a finite intersection of open and dense subsets is open and dense, so the countable intersection is the first "interesting" question, in a way.)

And what they call the historical definition:

2) Every non-empty open subset of $X$ is of second category.

The article call it historical because it uses a notion "category" of a subset (a subset is either first category or second category, and not both, by definition), which has fallen in some disuse. Nowhere dense sets and meagre sets (the countable unions of nowhere dense subsets) are still normal usage. Note that a first category subset is now called meagre, and the notion of "second category" is not used as much (but still occurs), so it's good to know it. But definition 1) and 2) are easily proved to be equivalent, so they give rise to the same spaces being called Baire. So we have a trivial reformulation of the "classical" definition 2 as:

2') Every non-empty open subset of $X$ is non-meagre.

Or, stated more "positively"

2'') Every meagre set has empty interior.

(otherwise the non-empty interior is a subset of a meagre set, and thus meagre etc.)

which brings us back to definition 1) again.

It's just that the Wikipedians do not like the category terminology (because it might confuse people with category theory as a branch of maths) and so choose to reformulate everything using meagre and non-meagre instead.

  • $\begingroup$ Regardless of possible confusion with category theory (which I think is neglible), I find it a bad naming: 'of first category' is totally undescriptive ('nowhere dense' and 'meagre' do not suffer from this). In the same spirit, I also dislike 'Type X' in the theory of Von Neumann Algebras, and '$T_i$' indicating a certain separation axiom. $\endgroup$ Feb 6 '12 at 13:39
  • 1
    $\begingroup$ @wildildildlife Agreed, it's non-descriptive, as is "normal" (for any property, really). The $T_i$ at least has the advantage of being a hierarchy, in the sense that the higher ones imply the lower ones, at least in the definitions I was taught, where $T_4$ = normal + $T_1$ etc. $\endgroup$ Feb 6 '12 at 17:44
  • $\begingroup$ Unfortunately the $T_i$ are no longer required to be a hierarchy in modern usage, so it's necessary to check any given author's definitions. (E.g. $T_3$ might mean that points are completely separated from closed sets,without regard to $T_0,T_1$ or $T_2$.) There is also the rarely used Urysohn space, which could be called $T_{2\frac {1}{2}}$ space. In Urysohn space, any 2 distinct points have nbhds whose closures are disjoint. $\endgroup$ Nov 22 '15 at 23:45

a. It is true you are either a countable union of nowhere dense sets or you are not. Thus, any set is is either of first category or second category.

b. It is stated in the wrong way. It should have said "contains a countable intersection of open dense sets", not union. Note that the complement of an open dense set is closed nowhere dense (and vice versa).

c. The notation is still in use.

  • $\begingroup$ In addition to the correction you made in (b), the Planetmath definition seems to be conflating "$A$ is second category" and "$A$'s complement is first category." $\endgroup$
    – user83827
    Feb 5 '12 at 1:37
  • $\begingroup$ @ccc: I was wondering if "A is second category" and " A's complement is first category" are equivalent? $\endgroup$
    – Tim
    Feb 5 '12 at 2:31
  • $\begingroup$ No, there are in general plenty of second category sets with second category complement. For example, $(0,1)$ as a subset of the reals. $\endgroup$
    – user83827
    Feb 5 '12 at 2:38
  • $\begingroup$ Indeed. The fact that a set of reals can be nonmeagre and have a nonmeagre complement is part of the reason why I think the meagre/nonmeagre/comeagre terminology should be preferred. In addition it's more "natural" (in terms of language): "X is comeagre" sounds much nicer than "X is the complement of a set of the 1st category". $\endgroup$
    – kahen
    Feb 5 '12 at 6:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.