به انگلیسی

Pointwise bounds become one uniform bound

Mathematics · Calculus & Analysis · ages 23-24

Name ______________________ Date ____________

Operators T_n satisfy ||T_n|| <= 5 for all n, and ||x|| = 2. What is the largest possible value of sup over n of ||T_n x||?
Answer: ______________
A student says the rationals form a meagre set in the real line. Is that right?
Circle one: True False
Nested closed intervals start with length 1 and each is half as long as the previous. What is the length of the fifth interval?
Answer: ______________
What does the category theorem forbid in a nonempty complete metric space?
- A dense subset that is also countable, like the rationals
- A Cauchy sequence whose terms stay inside a nowhere dense set
- A countable union of nowhere dense sets filling it
What does the open mapping theorem say about bounded linear operators between Banach spaces?
- Every bounded linear operator between Banach spaces is surjective, even if its image is a proper subspace
- Surjective bounded linear operators between Banach spaces are open maps
- Pointwise bounds on a family give one operator-norm bound without any completeness assumption on the domain
Which sequence shows the rationals are not complete?
- The sequence 1/n, which tends to 0 in the line
- Constant sequences such as 3, 3, 3, and so on
- Decimal approximations of sqrt(2)
- The alternating sequence (-1)^n
A surjective bounded linear operator between Banach spaces sends open sets to open sets. A student says this always holds. Is that right?
Circle one: True False
The Baire argument traps a point using nested closed balls with shrinking radii. Which hypothesis makes that trap work?
- Separability of the space
- Finite dimensionality
- Compactness of the unit ball
- Completeness of the domain
A closed ball centred at a with radius r > 0 lies in E_k. For ||h|| <= r, bound T_n(a+h) and T_n a separately by k. What bound does subtraction give for ||T_n h||?
- k/r
- 2k/r
- 2k
A linear operator between Banach spaces whose graph is closed is bounded.
Circle one: True False

Answer key

For grown-ups. Fold this page away before handing over the rest.

Pointwise bounds become one uniform bound W1-mt_11Rd73trAX-s1

10 · The upper bound is 5 times 2 = 10. It is attained when every T_n is 5 times the identity, so 10 is the largest possible value.
True · Each singleton is nowhere dense and the rationals are countable, so Q is a countable union of nowhere dense sets.
0.0625 · Halving four times from length 1 leaves (1/2)^4 = 0.0625.
A countable union of nowhere dense sets filling it · A nonempty complete metric space is not meagre in itself, so countably many nowhere dense sets cannot cover it.
Surjective bounded linear operators between Banach spaces are open maps · A bounded linear surjection between Banach spaces sends open sets to open sets. Surjectivity is a hypothesis, not a conclusion about every bounded operator.
Decimal approximations of sqrt(2) · The decimals are rational and Cauchy, but their only possible limit, sqrt(2), is not rational, so they show the rationals are not complete.
True · Yes. Baire gives control in an image closure. A convergent series of preimage corrections removes the closure, proving the open mapping theorem.
Completeness of the domain · Baire needs a complete space so that the nested ball centres have a limit. Completeness of the domain is the hypothesis that makes the trap work.
2k · Linearity gives T_n h = T_n(a+h) - T_n a. The triangle inequality adds the two bounds k, giving 2k before scaling.
True · The closed graph is Banach. Its projection onto the domain is a bounded bijection, so open mapping makes the inverse bounded. Composing with the target projection recovers the operator.