Completeness and Category
Baire Category Theorem
You should know: completeness of reals
Overview
The Baire category theorem states that in a complete metric space, the intersection of any countable collection of dense open sets is still dense; equivalently, a complete metric space cannot be written as a countable union of nowhere-dense sets. This gives a rigorous, topological notion of 'most' points or 'most' functions having a given property, without appealing to measure — a set is called meager (or of first category) if it is a countable union of nowhere-dense sets, and Baire's theorem says a complete metric space is never itself meager. The theorem is the engine behind several surprising existence proofs, most famously that the set of continuous nowhere-differentiable functions is 'topologically generic' (comeager) in C[0,1] — such pathological functions are not rare exceptions but overwhelmingly the norm from a category standpoint.
Intuition
A 'nowhere dense' set is one so thin it doesn't even fill up any small ball densely — think of a single point, a line segment in the plane, or the Cantor set. A 'meager' set is built from countably many such thin pieces stacked together — still, intuitively, 'small' in a topological sense (analogous to, but distinct from, having measure zero). Baire's theorem says that a complete metric space (like ℝ, or C[0,1]) can never be exhausted by countably many such thin pieces — no matter how you try to cover the whole space with countably many nowhere-dense sets, something is always left over, and in fact a dense set of points is left over. This gives a powerful 'most functions/points are like this' argument: if you can show the 'bad' functions (say, differentiable ones) form a meager set, then the 'good' (generic) functions — nowhere differentiable ones — must be comeager, hence in some sense the overwhelming majority.
Formal Definition
Let (X, d) be a complete metric space (or more generally, a locally compact Hausdorff space). Then:
Worked Examples
Each singleton {x} is nowhere dense in ℝ (a single point contains no interval, and its closure is itself, with empty interior).
If ℝ were countable, it would equal the countable union of these singletons, ℝ = ⋃ₓ{x}, expressing ℝ as a countable union of nowhere-dense sets.
But ℝ is a complete metric space, so by Baire's theorem it cannot be a countable union of nowhere-dense sets — contradiction. Hence ℝ is uncountable.
Answer: ℝ is uncountable — assuming otherwise makes ℝ a countable union of nowhere-dense singletons, contradicting Baire's theorem.
Practice Problems
Is the set of rational numbers ℚ meager in ℝ? Explain.
The Baire category theorem requires the space to be:
Sketch the proof that in a complete metric space X, a countable intersection of dense open sets Uₙ is dense.
Quiz
Summary
- In a complete metric space, a countable intersection of dense open sets is still dense — equivalently, the space is never a countable union of nowhere-dense sets.
- This gives a topological notion of 'smallness' (meager/first-category) distinct from measure-zero, used to prove existence results without exhibiting explicit examples.
- Classic applications include proving ℝ is uncountable and that continuous nowhere-differentiable functions are topologically generic (comeager) in C[0,1].
Mathematics