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I've got two definitions of $\limsup_{n \to \infty} a_n$ and I'm looking for proof of equivalence:

  1. Let $A$ be the set of a limit points of sequence $a_n$. Then $\limsup_{n \to \infty} a_n = \sup A$.
  2. $\limsup_{n \to \infty} a_n = \inf_{n \in \mathbb{N}} \left( \sup_{m \ge n} a_n \right)$.

My attempt was as follows: if $g = \sup A$, then for all $g' \in A$ we have $g' \le g$ and for all $\varepsilon > 0$ there exist $g''$ such that $g'' > g - \varepsilon$. Of course $\inf_{n \in \mathbb{N}} \left( \sup_{m \ge n} a_n \right)$ exists because $\sup_{m \ge n} a_n$ is decreasing, but I can't say that it is a subsequence of $a_n$ and I got stuck. Also I was attempting proof by contradiction, but I have no idea how can I start.

José Carlos Santos
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Marcinek665
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1 Answers1

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Let $ h = \inf_{n} ({\sup_{m \geq n}} a_n)$. Since the sequence $({\sup_{m \geq n}} a_n)$ is bounded and decreasing, it actually follows that $h = \lim_{n \to \infty} ({\sup_{m \geq n}} a_n)$.

Given that $g = \sup A$, where $A$ is the set of limit points of the sequence.

Note that $a_{l} \leq \sup_{m \geq l} a_m$. We know that for $\epsilon > 0$, there is a limit point $a \in A > g-\epsilon$, by the supremum property. Let $a_l \to a$, then $a_l \leq (\sup_{m \geq l} a_m)$. Taking limits on both sides (the RHS is a decreasing bounded sequence so we are fine), we get $a \leq h$. Taking $\epsilon \to 0$, we get $g \leq h$.

The other way:

I claim $h \in A$. To see this, note that by definition of $h$ being the limit, for all $\epsilon > 0$, there exists $N$ large enough so that $h + \epsilon > \sup_{m \geq N} a_m > h - \epsilon$. Now, by definition of supremum, there exists an $M$ large enough so that $h + \epsilon > \sup_{m \geq N} a_m \geq a_M > h-\epsilon$, since the $a_m$.

For $\epsilon = 1,\frac 12,\frac 13...$ take the corresponding sequence of $a_M$s, it can be seen that these $a_M$s converge to $h$, completing the argument. Therefore, $h \leq g$, so $h=g$.

Sarvesh Ravichandran Iyer
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