In our course Baire theorem is defined as follows.
If complete metric space is represented as countable union of closed sets, then at least one of these closed sets contains one type of ball( open or closed ).
I would like to understand what is motivation for this theorem to grasp it better.
Why should author publish it ?
Thank you for taking thought to read this post !
Many important theorems in analysis depend on the Baire theorem (and thus on completeness of certain metric spaces), a lot of examples can be found in this thread.
As a more topological motivation: an alternative equivalent formulation is that if $O_n$, $n \in \mathbb{N}$ is a sequence of open and dense subsets of a space $X$, then $\bigcap_n O_n$ is still dense in $X$. It is quite easy to see that every finite such intersection is still (open and) dense, so the countable case is the first possible extension/generalisation of this fact.
Also, a set such that its closure does not contain an open ball is "small" (all finite sets in most metric spaces are like that), and such sets are closed under finite unions. In measure theory and in analysis generally we also like our families to be closed under countable unions as well, so it is natural to study all countable unions of "small" sets. Such sets are called "meagre" (or US spelling "meager"), and like sets of measure $0$ we intuitively consider them "small" or "thin" too. But we'd like open sets to be "big" topologically, and this is what the Baire theorem says in another alternative form: "non-empty open sets are not meagre".
But mostly, its importance lies in the many applications.
$O_n, n∈Nn \in \mathbb{N}$ is a sequence of open and dense subsets of a space $X$. \
I don't follow from here.
If $O_n$ is dense in X for $n \in \mathbb{N}$, then does it not follow that $ \forall n, O_n = \bar O_n = X $ ? If this so so, then does not $\bigcap_n O_n = O_n$ ?
– flowian Mar 31 '18 at 09:36