You are currently browsing the tag archive for the ‘Baire category theorem’ tag.

The notion of what it means for a subset E of a space X to be “small” varies from context to context. For instance, in measure theory, when is a measure space, one useful notion of a “small” set is that of a null set: a set E of measure zero (or at least contained in a set of measure zero). By countable additivity, countable unions of null sets are null. Taking contrapositives, we obtain

Lemma 1.(Pigeonhole principle for measure spaces) Let be an at most countable sequence of measurable subsets of a measure space X. If has positive measure, then at least one of the has positive measure.

Now suppose that X was a Euclidean space with Lebesgue measure m. The Lebesgue differentiation theorem easily implies that having positive measure is equivalent to being “dense” in certain balls:

Proposition 1.Let be a measurable subset of . Then the following are equivalent:

- E has positive measure.
- For any , there exists a ball B such that .

Thus one can think of a null set as a set which is “nowhere dense” in some measure-theoretic sense.

It turns out that there are analogues of these results when the measure space is replaced instead by a complete metric space . Here, the appropriate notion of a “small” set is not a null set, but rather that of a nowhere dense set: a set E which is not dense in any ball, or equivalently a set whose closure has empty interior. (A good example of a nowhere dense set would be a proper subspace, or smooth submanifold, of , or a Cantor set; on the other hand, the rationals are a dense subset of and thus clearly not nowhere dense.) We then have the following important result:

Theorem 1.(Baire category theorem). Let be an at most countable sequence of subsets of a complete metric space X. If contains a ball B, then at least one of the is dense in a sub-ball B’ of B (and in particular is not nowhere dense). To put it in the contrapositive: the countable union of nowhere dense sets cannot contain a ball.

**Exercise 1.** Show that the Baire category theorem is equivalent to the claim that in a complete metric space, the countable intersection of open dense sets remain dense.

**Exercise 2. **Using the Baire category theorem, show that any non-empty complete metric space without isolated points is uncountable. (In particular, this shows that Baire category theorem can fail for incomplete metric spaces such as the rationals .)

To quickly illustrate an application of the Baire category theorem, observe that it implies that one cannot cover a finite-dimensional real or complex vector space by a countable number of proper subspaces. One can of course also establish this fact by using Lebesgue measure on this space. However, the advantage of the Baire category approach is that it also works well in infinite dimensional complete normed vector spaces, i.e. Banach spaces, whereas the measure-theoretic approach runs into significant difficulties in infinite dimensions. This leads to three fundamental equivalences between the *qualitative* theory of continuous linear operators on Banach spaces (e.g. finiteness, surjectivity, etc.) to the *quantitative* theory (i.e. estimates):

- The uniform boundedness principle, that equates the qualitative boundedness (or convergence) of a family of continuous operators with their quantitative boundedness.
- The open mapping theorem, that equates the qualitative solvability of a linear problem Lu = f with the quantitative solvability.
- The closed graph theorem, that equates the qualitative regularity of a (weakly continuous) operator T with the quantitative regularity of that operator.

Strictly speaking, these theorems are not used much directly in practice, because one usually works in the reverse direction (i.e. first proving quantitative bounds, and then deriving qualitative corollaries); but the above three theorems help explain *why* we usually approach qualitative problems in functional analysis via their quantitative counterparts.

## Recent Comments