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Let $a_{n}$ be a sequence of real numbers and let E be the set of limit points. Is it the case that $\limsup a_{n} = \sup E$.

Attempt:

My initial answer is yes, if I start with the simpler case of $a_{n}$ being a bounded sequnece of real numbers, let $b_{n}:=\sup\{a_{k}:k\geq n\}$ then since $b_{n}$ converges, every subsequence will converge to the same limit. Let $(a_{n_{k}})$ be a convergent subsequence. Then we have $a_{n_{k}} \leq b_{n_{k}}$ for all $k \in \mathbb{N}$. Taking limits the we get $\lim_{k \to \infty}a_{n_{k}} \leq \limsup_{n \to \infty}a_{n}$.

I don't think the same argument will work if the sequence is unbounded? (Corrected thanks to Steven Stadnicki)

learningmathematics
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  • I presume your last question means 'if the sequence is unbounded'; for the definition to make sense, though, you have to consider $\infty$ to be a valid limit point (e.g., it is a limit point of the sequence $a_n$ that alternates between 0 and $n$ based on the parity of $n$; i.e., $a_n={0,2,0,4,0,6,\ldots}$). – Steven Stadnicki Jul 22 '21 at 19:32
  • @StevenStadnicki Yes, unbounded, will correct the typo. Yes, agree about the infinity point, but isn't that the general convention anyway, since people say the limsup always exists? – learningmathematics Jul 22 '21 at 19:35

1 Answers1

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The unbounded case is relatively straightforward: since we have $\limsup a_n=\infty$, for each $M$ and each $n_0$ there's some $n\geq n_0$ with $a_n\geq M$. Define a sequence $b_i$ such that $b_i$ is the first number $\gt b_{i-1}$ such that $a_{b_i}\geq i$, and note that this defines a subsequence of $a_n$ whose limit is $\infty$.

Steven Stadnicki
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