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I've found in my book that: $$\liminf_{n\to\infty} \ x_{n} = \sup\{\inf\{x_{k}:k\geq n \}:n \in \mathbb{N}\}$$ $$\limsup_{n\to\infty} \ x_{n} = \inf\{\sup\{x_{k}:k\geq n \}:n \in \mathbb{N}\}$$

But I don't understand why. According to my book the definition of $\limsup$ and $\liminf$ is the following:

$$\limsup_{n\to\infty} \ s_{n} = \lim_{N\to\infty} \sup \{s_{n}:n > N \} $$

$$\liminf_{n\to\infty} \ s_{n} = \lim_{N\to\infty} \inf \{s_{n}:n > N \} $$

Arthur
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Jelly Belly
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1 Answers1

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Let $X_n=\{x_{k}:k\geq n \}$. Then $X_{n+1} \subseteq X_n$ and therefore: $$ \inf X_n \le \inf X_{n+1} \le \sup X_{n+1} \le \sup X_n $$ This implies:

  • $a_n=\sup X_n$ is a decreasing sequence and so $\lim_n a_n = \inf_n a_n$.

  • $b_n=\inf X_n$ is an increasing sequence and so $\lim_n b_n = \sup_n b_n$.

lhf
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  • Maybe it's trivial but how do you know that $b_{n}$ is an increasing sequence? – Jelly Belly Nov 26 '15 at 12:19
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    @JellyBelly Because of ${x_n \mid k\geq n+1} \subseteq {x_n \mid k\geq n}$, the $\inf$ of the first set cannot be smaller than the $\inf$ of the latter set, which is to say $b_{n+1} \not<b_n$ for all $n$. – Arthur Nov 26 '15 at 12:28