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

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

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, and why this one tests standard counterexample.

See pricing

50 are analysed free — try those first.

The trap it tests

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

Drill statements like this

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

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

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