Questions: Baire Category Theorem for Metric Spaces

Question 1

The rational numbers ℚ are dense in ℝ — every open interval contains a rational. Does this make ℚ a 'large' set in the Baire category sense?

• A. Yes; a dense set is non-meager by definition in any metric space
• B. Yes; density implies non-meagerness because dense sets 'fill' the space
• C. No; ℚ is meager — a countable union of singletons, each nowhere dense — even though it is dense in ℝ
• D. No; ℚ is meager because countable sets are always meager in any metric space

Answer: (C)

Density and meagerness are independent properties. ℚ is dense in ℝ (every open interval contains a rational), yet it is meager: enumerate the rationals q₁, q₂, ..., and each singleton {qₙ} is nowhere dense (its closure is itself, containing no open interval). Their countable union is ℚ — meager by definition. This shows that a set can be topologically 'everywhere present' (dense) yet categorically 'small' (meager). Baire category measures a different dimension of size than density.

Question 2

The Baire Category Theorem guarantees that a complete metric space X cannot be written as a countable union of nowhere-dense sets. What does this directly imply?

• A. Each individual nowhere-dense set in X has positive Lebesgue measure
• B. The union of countably many nowhere-dense sets must miss at least one point of X
• C. The intersection of the nowhere-dense sets is empty
• D. The nowhere-dense sets must form a finite collection

Answer: (B)

The theorem says the countable union of nowhere-dense sets cannot equal all of X — it must miss at least a point (in fact, a dense Gδ set of points). This is the direct content of the theorem. Option A is wrong because nowhere-dense sets can have positive measure. Option C is wrong because the intersection of the Fₙ could be non-empty. Option D is wrong because the theorem applies to countably infinite collections, not just finite ones.

Question 3

A meager set in ℝ cannot be dense in ℝ.

• T. True
• F. False

Answer: False

ℚ is the standard counterexample: it is both meager (a countable union of nowhere-dense singletons) and dense in ℝ (every open interval contains a rational). Density and meagerness measure different properties. Density says the set comes arbitrarily close to every point; meagerness says the set is a 'thin' union of sets with no interior. A set can be simultaneously thin (meager) and everywhere present (dense).

Question 4

The Baire Category Theorem has an equivalent formulation: in a complete metric space, every countable intersection of dense open sets is itself dense.

• T. True
• F. False

Answer: True

This is the standard dual formulation. Taking complements: a set is nowhere dense if and only if its complement contains a dense open set. So a countable union of nowhere-dense sets corresponds (via complement) to a countable intersection of dense open sets. The theorem saying 'the union doesn't cover X' is equivalent to saying 'the intersection is non-empty (and in fact dense).' Both formulations capture the same content: complete metric spaces are topologically 'large.'

Question 5

Why does the Baire Category Theorem require completeness, and what breaks down in an incomplete metric space?

Model answer: The proof constructs a Cauchy sequence by choosing nested closed balls, each shrinking in radius and each avoiding the next nowhere-dense set Fₙ. Completeness guarantees this Cauchy sequence converges to a limit point — and that limit point avoids every Fₙ, providing the contradiction. Without completeness, the Cauchy sequence might not converge in the space, and the contradiction fails. The rationals ℚ confirm this is necessary: ℚ is not complete, and ℚ is a countable union of singletons (each nowhere dense), which would violate the Baire conclusion — showing completeness cannot be dropped.

The key step 'the Cauchy sequence converges' is exactly where completeness enters. This is not a technicality — it is the heart of the proof. The theorem is false for incomplete spaces, so the condition is necessary, not just sufficient. Understanding this also shows why the theorem is a result about complete metric spaces specifically, not about arbitrary metric or topological spaces.