Why was the concept of first/second categories in metric spaces introduced?

Question

Let $(X, d) $ be a metric space. $X$ is of first category if it can be expressed as a countable union of nowhere dense subsets. Otherwise, $X$ is called a metric space of second category.

A first category set is intuitively "small " and a second category set is "large". A second category set with no isolated point is uncountable.

Cardinality is purely a set-theoretic characterisation and first/second category is related to the topology on the underlying set.

However I am interested to know whether there are any deeper relations between the cardinality and whether it is first or second category.

Why was the concept of first/second categories in metric spaces introduced?

Note that sets of first category are often called meagre and sets of second category are often called non-meagre. — user829347, Jan 20 '22 at 15:12
One way to express the Baire Category Theorem is that every complete metric space is not first category in itself, so it might have been introduced in relation to that. — user829347, Jan 20 '22 at 15:23
Why close vote? It's very annoying. Atleast give some references or some way to think about that. It may seems nonsense for some of you. But this type of question really troubled me and i am interested to explore more. — Sourav Ghosh, Jan 20 '22 at 15:32
@amWhy The question of why a particular concept was introduced looks at the mathematical motivation for it i.e. for a specific application or to formalise something intuitive. Maybe I'm wrong, but it seems like an appropriate topic for a soft question on MSE to me :) — user829347, Jan 20 '22 at 15:48
@S.G Also note that the definitions of first/second category (meagreness) apply to arbitrary topological spaces, not just metric spaces, as does the definition of a nowhere dense subset, so you might want to look at this in a broader topological context to make sense of where it all came from. — user829347, Jan 20 '22 at 17:32
Yes. Dense, n.w.dense can be characterized entirely by open sets. — Sourav Ghosh, Jan 20 '22 at 17:38
Useful google search: "first category" + Baire + 1899 For terminology, see near beginning of this answer. For an overview of cardinality, measure, category smallness, see this answer. For various ways of "constructively" defining nowhere dense, see this answer (and this for applications to strengthening these notions). — Dave L. Renfro, Jan 20 '22 at 17:41
@Dave L. Renfro Thanks . I will try to read all of mentioned post. — Sourav Ghosh, Jan 20 '22 at 17:45
Regarding the actual problem that led Baire to introduce Baire category and Baire (one) functions, see The genesis of separate versus joint continuity by Zbigniew Piotrowski (1996) and this 4 June 2005 sci.math post. Even more informative (but in French, and might not be freely available) is Notes et documents sur la vie et l'œuvre de René Baire by Pierre Dugac (1976). — Dave L. Renfro, Jan 20 '22 at 18:58
Also, [4] at the end of this answer gives, among other things, a useful introductory survey of Baire category notions. The other parts of that answer mostly deal with Baire functions (see also this answer), which also played an important role in Baire's 1899 Ph.D. dissertation. And now I need to get back to my "day job" . . . — Dave L. Renfro, Jan 20 '22 at 19:10

score 4 · Accepted Answer · answered Jan 21 '22 at 00:35

The Baire category theorem is an abstraction of the following sort of construction, which seems to arise fairly often. You want to construct an object $X$ (of some given sort) that satisfies an infinite list of requirements, say R1, R2, .... Instead of attacking all these requirements at once, you try to handle them one at a time, and you notice (with considerable happiness) that each requirement is amazingly nice, in the sense that, given any object $Y$, you can slightly modify it to satisfy any single R$n$. So you try starting with some arbitrary $Y$, modifying it to get some $Y'$ that satisfies R1, modifying that to get $Y''$ satisfying R2, and --- OOPS: $Y''$ no longer satisfies R1; the second modification ruined what the first had achieved.

But then you notice an improvement on your original happy observation. The R$n$'s are super-nice in the sense that any $Y$ can be modified to a $Y'$ that satisfies R$n$ and continues to satisfy R$n$ when modified sufficiently slightly.

This means that you can modify $Y$ to some $Y'$ that satisfies R1 and this achievement won't be ruined by future modifications if those modifications are slight enough. Now modify $Y'$ very slightly to get $Y''$ satisfying R2 and still satisfying R1, with room to spare. Here "room to spare" means that further, sufficiently slight modifications won't ruin R1 or R2. Continue in this way, producing $Y'''$ and so forth. After $n$ steps, you have $Y^{(n)}$ satisfying R1 through R$n$ and guaranteed to still do so if perturbed extremely slightly. (At each step, modify the previous $Y^{(n)}$ much less than any of your previous $n-1$ modifications.)

OK, so you can satisfy any finitely many of your requirements with a suitable $Y^{(n)}$, but you want to satisfy all infinitely many requirements with a single object $X$. Well, if you've made the modifications in your construction slight enough, the sequence $Y, Y', Y'',\dots,Y^{(n)},\dots$ will be a Cauchy sequence. If the objects you're working with constitute a complete space, then this sequence will converge to an $X$ that is close enough to each $Y^{(n)}$ to guarantee that $X$ satisfies R$n$. So that limit $X$ is the object you want.

The key ingredients in this argument are (1) completeness of the space; (2) availability of a set $G_n$ of objects that satisfy R$n$ and continue to do so when sufficiently slightly modified, i.e., an open set of objects satisfying R$n$; and (3) the possibility of extremely slightly modifying any object to get it into $G_n$. In other words, $G_n$ is dense in the whole space of your objects.

So, isolating the key ingredients, we have the theorem: In a complete metric space, the intersection of any countably many dense open sets is nonempty.

The complement of a dense open set is a nowhere dense closed set, so the theorem is equivalent to: A complete metric space is not the union of countably many nowhere dense (closed) sets. (I put "closed" in parentheses because omitting it doesn't affect the meaning; a set is nowhere dense iff its closure is nowhere dense.) In other words, a complete metric space is of second category.

Why was the concept of first/second categories in metric spaces introduced?

1 Answers1