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Prove/Disprove $$\lim \sup a_n = -\lim \inf -a_n$$

Intuitively it seems like a proof, as $\sup(a_n)=-\inf(-a_n)$ and then taking the limit from both sides, but it is informal and I am not sure that it is correct.

Dirk
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newhere
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  • It might be informal, but it's a good start. If you hope to make it more formal, look through the proof of $\sup(a_n) = -\inf(-a_n)$, and see whether you actually can just insert $\lim$ in front of everything. If you can, then good. If you can't, then look at the exact step where that happens. That's either something that needs reworking, or a hint for a counterexample. – Arthur Dec 09 '19 at 09:54
  • Whats your definition of limsup? – IAmNoOne Dec 09 '19 at 10:04
  • @IAmNoOne $\inf_{n\geq 0}\sup_{m\geq n }a_n$ – newhere Dec 09 '19 at 10:06

2 Answers2

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This question is pretty basic in a certain sense, so basic it is just stated on Wikipedia without comment or proof. If you want to prove it, you will prove it from a very specific set of definitions and properties.

So, if you need help, please explain on what kind of assumptions you already work on. Have you, for example already established basic properties of inf/sup and of lim? If not: proof those with help of their definitions first; In a second step use the fact, that you can pull out the factor $-1$

don-joe
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Using that $\inf(A) = -\sup(-A)$ and $\sup(A) = -\inf(-A)$ for every subset $A$ of the extended real numbers, we get $$\liminf_{n \to \infty}(-a_n) = \sup_{n=1}^\infty \inf_{k=n}^\infty (-a_k) = \sup_{n=1}^\infty (-\sup_{k=n}^\infty a_k) = -\inf_{n=1}^\infty \sup_{k=n}^\infty a_j = -\limsup_{n \to \infty} a_n$$

J. De Ro
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