In a complete metric space,first category sets are 'meagre' in the sense that they cannot contain any non-empty open set.

Question

I had just started studying Baire's Category Theorem and I was at first having problem with the seemingly technical definitions of nowhere dense sets,first category and second category sets.But slowly it was revealed to me that these definitions are not coming out of the blue,they have some significance.The idea of nowhere dense is quite simple to interpret,it is nowhere dense in the sense that it is not dense in any non-empty open set.But first category and second category sets are a bit complicated to be appreciated at first glance.Sometimes we say that first category sets are 'meagre' or small in the sense that they are countable(small enough) union of nowhere dense(sparse) sets.That would seem quite convincing but at the next glance you would figure out something else.Suppose we consider the metric space $\mathbb Q$ with usual distance $|.|$,consider an enumeration $(r_n)$ of this set.Then $\mathbb Q=\large\cup_{n\in \mathbb N}$$\{ r_n\}$,where each of the singletons in nowhere dense as $\mathbb Q$ has no isolated points.So,$\mathbb Q$ is of first category,but notice that here it is not small in that sense because it is the entire space.So,the thing small is not understood properly here.Now,it is best understood when we work with complete metric space.We know that complete metric spaces are of second category.Now here if we have a first category set,then it is clear that it cannot contain any open non-empty set(By Baire's Category theorem):

We can proceed directly that if $X$ is a complete metric space and $U\subset A$,$U$ being a non-empty open set,then $U$ is of second category.So any superset of $U$ in particular $A$ must be of second category.So,a first category set in a complete metric space cannot contain a non-empty open set.

So,a first category set is indeed 'meagre' in true sense of the term when we consider a complete metric space.Also,we can say that first category set is 'meagre' because in complete metric space,its complement is of second category.

So,I think the terms 'meagre' or 'small' in terms of category is best understood and in fact justified when we look at complete metric spaces.In fact,Baire developed his theorem for $\mathbb R$ which is a complete metric space.So,it think he thought of these terminologies like 'meagre'.Is it correct?

Answer 1

It's certainly true that notions like meagre vs second category are only really "meaningful" in nice complete (or locally compact Hausdorff spaces, which are also Baire) spaces. There is another analogy one can consider too: the null sets (in $\Bbb R$) in the Lebesgue measure. This is a $\sigma$-ideal of "small" subsets of $\Bbb R$. One can think of the meagre sets as the smallest $\sigma$-ideal of the reals that contains all nowhere dense subsets, so all small sets wrt the topology (they indeed do not contain open sets, their complement is even dense! but a set like the irrationals in the reals (often denoted $\Bbb P$) also contains no open sets and has dense complement but is second category, so not meagre, so meagre-ness is more subtle, it's more $\Bbb Q$-like and less $\Bbb P$-like.

If you're interested in this stuff, there is an accessible short book by Oxtoby called "Measure and Category", in which he explores the similarities and analogies of these notions of meagre-ness vs measure $0$; both are important in analysis, but there are some subtle and nice differences too. A (classic) book to keep in mind, IMO.