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

Set of subsequential limits can be uncountable

Let $(a_{n})_{n} \geq_{1}$ be a bounded sequence of real numbers such that $l i m_{n} \to \infty a_{n}$ does not exist. Let S = { $l \in R$ : there exists a subsequence of $(a_{n})$ converging to l}. Which of the following statements are necessarily true?

A.S is the empty set
B.S has exactly one element
C.S has at least two elements✓
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.

See pricing

50 are analysed free — try those first.

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

Set of subsequential limits can be uncountable

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