A mathematician wants to prove that 'most' continuous functions on [0,1] are nowhere differentiable, without constructing an explicit example. How does the Baire Category Theorem make this possible?
ABy showing the set of nowhere differentiable functions is open and dense in C[0,1]
BBy showing the set of functions differentiable at even one point is meager in C[0,1], so its complement is comeager
CBy showing C[0,1] is compact, so generic properties hold on a dense subset
DBy constructing an explicit Weierstrass function and proving it is typical
The Baire strategy is: identify the 'exceptions' (functions differentiable somewhere), show this set is meager (a countable union of nowhere dense sets), then invoke Baire to conclude the exceptions don't cover the complete space C[0,1]. What remains — the comeager set — constitutes 'most' functions in the Baire sense. No explicit construction is needed; the theorem provides the existence conclusion for free.
Question 2 Multiple Choice
Why does the Baire Category Theorem fail for ℚ (the rationals with the usual metric)?
Aℚ is uncountable, so it cannot be written as a countable union
Bℚ is not a metric space under the usual absolute value
Cℚ is incomplete — it equals the countable union of singletons {q}, each of which is nowhere dense
Dℚ has no open sets in the subspace topology from ℝ
Each singleton {q} is nowhere dense in ℚ (its closure is itself, which contains no open interval). So ℚ = ∪{q} is a countable union of nowhere dense sets — the exact scenario the theorem says is impossible for complete spaces. The theorem requires completeness, and ℚ is not complete (Cauchy sequences of rationals can converge to irrationals). This shows completeness is not just a technical assumption but the essential hypothesis.
Question 3 True / False
If a complete metric space X is written as a countable union X = A₁ ∪ A₂ ∪ …, then at least one Aₙ must contain an open ball.
TTrue
FFalse
Answer: True
This is essentially what the Baire Category Theorem says. If every Aₙ were nowhere dense (its closure contains no open ball), then their countable union could not be all of X — a complete metric space cannot be meager. So if the union really is all of X, at least one set must fail to be nowhere dense, meaning its closure must contain an open ball.
Question 4 True / False
The Baire Category Theorem implies that meager sets in a complete metric space should be empty.
TTrue
FFalse
Answer: False
Meager sets can be nonempty — even quite large. The rationals ℚ are a meager subset of the complete metric space ℝ (countable union of nowhere dense singletons), yet ℚ is dense in ℝ. What Baire says is that a meager set cannot be *all* of a complete metric space. 'Meager' means topologically negligible, not necessarily small in other senses.
Question 5 Short Answer
What role does completeness play in the proof of the Baire Category Theorem, and why does the theorem fail without it?
Think about your answer, then reveal below.
Model answer: The proof builds a nested sequence of shrinking closed balls, each chosen to avoid the next nowhere dense set in the supposed countable cover. Completeness guarantees that the intersection of this Cauchy sequence of balls is non-empty — producing a point that lies outside every set in the cover, contradicting the assumption that they cover the whole space. Without completeness, the Cauchy sequence might not converge within the space, and the proof fails. ℚ shows the failure: it is covered by countably many nowhere dense singletons, but no convergent point within ℚ can be found because the limit may be irrational.
Completeness is the bridge from 'there exists a Cauchy sequence avoiding all the sets' to 'there exists a point in the space avoiding all the sets.' This is why the theorem holds for complete metric spaces and locally compact Hausdorff spaces, but not for spaces like ℚ where Cauchy sequences can escape the space.