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Part CCSIR NET June 2024set-of-subsequential-limits-can-be-uncountable

Set of subsequential limits can be uncountable

Let (an)n≥1(a_{n})_{n}\ge_{1} be a bounded sequence of real numbers such that limn→∞anlim_{n}\to\infty a_{n} does not exist. Let S = {l∈Rl \in \mathbb{R} : there exists a subsequence of (an)(a_{n}) converging to l}. Which of the following statements are necessarily true?

  1. A.S is the empty set
  2. B.S has exactly one element
  3. C.S has at least two elements✓ Correct
  4. D.S has to be a finite set

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.

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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: limsup(aₙ + bₙ) = limsup aₙ + limsup bₙ

More on this topic

The chapter behind this: limsup and liminf — three equivalent definitions and the algebra — free to read

From The Real Line › limsup, liminf and subsequential limits

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