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Part CCSIR NET December 2025h-is-the-classical-hilbert-cube-compact-closure-and-an-infinite-orthonormal-set-can-never-fit-inside-one

H is the classical hilbert cube compact closure and an infinite orthonormal set can never fit inside one

Let ℓ2\ell^{2} denote the vector space of all square summable sequences {ana_{n}}n≥1_{n}\ge1 of real numbers with the inner product ⟨{ana_{n}},{bnb_{n}}⟩ =∑n= \sum_{n}₌1∞anbn_{1}^\infty a_{n}b_{n}. Let H = {{ana_{n}}∈ℓ2\in\ell^{2} : |ana_{n}|≤1/n for all positive integers n}. Which of the following statements are true?

  1. A.H contains an orthonormal basis of ℓ2\ell^{2}.
  2. B.H is a linear subspace of ℓ2\ell^{2}.
  3. C.H is a bounded subset of ℓ2\ell^{2}.✓
  4. D.H is a convex subset of ℓ2\ell^{2}.✓

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

See pricing

50 are analysed free — try those first.

Related counterexample: L¹[0,1] ⊆ L²[0,1]

More on this topic

The chapter behind this: L^p spaces: inequalities and inclusions — free to read

From Lebesgue Measure and Integration › L^p spaces essentials

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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