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I've written the following proof to show that $\lim \sup(a_n) \geq \lim \inf(a_n)$, however I'm unsure if it's right/holds. Could I get some help poking holes in it, and potentially fixing these holes?

Define $b_n$ to be the supremum of $a_n$ and define $c_n$ to be the infimum of $a_n$.

By definition, $b_n \geq c_n$ (Unsure if this is rigorous enough), which implies $b_n - c_n \geq 0$. Therefore, by a previous proof (This proof I've done shows that if elements of a sequence are within a certain range, the limit must also fall there) $\lim\ b_n - \lim\ c_n \geq 0$ must be true. This implies $\lim \sup \geq \lim \inf$.

I've also considered making a third sequence that is the difference between the supremum and the infimum, and showing that the elements of this third sequence must fall between $[0, \infty)$, however I don't see how that proves the sup is greater than or equal to the inf.

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  • $\begingroup$ Already your first line is flawed. It does not make sense. $\endgroup$ Apr 2, 2017 at 9:10

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I think you are confused about what $\limsup a_n$ means. It is not $\lim_{n\to\infty}(\sup a_n)$. It is $\lim_{m\to\infty}\big(\sup\{a_n\mid n\geq m\}\big)$. Try rewriting your proof based on that definition, and you should get something that works.

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  • $\begingroup$ Should be $\lim_{m\to \infty} \sup\{a_n\mid n\le m\}$. $\endgroup$
    – Lonidard
    Apr 3, 2017 at 13:30
  • $\begingroup$ @Lonidard - No, that would be simply $\sup\{a_n\mid n\in\mathbb N\}$, which depends on $a_1$, unlike $\lim\sup$. $\endgroup$
    – mr_e_man
    Nov 18, 2019 at 4:49

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