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I can't get the meaning of "Infinitely Often" as well as "Eventually" statements in the following :

1) if $z > \lim \sup x_n$ , then : $x_n < z$ eventually (that is, for all sufficiently large $n$ )

2) if $z < \lim \sup x_n$ , then : $x_n > z$ infinitely often (that is, for infinitely many $n$)

Any help to get meaning of these both points? The intuition behind?

Especially Lime
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user30614
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  • Try to unpack the definition of $\limsup$ and see exactly what each inequality amounts to. – Arrow Feb 08 '18 at 10:39
  • Maybe helpful: https://math.stackexchange.com/questions/1475536/lim-inf-and-lim-sup-of-events-and-random-variables, https://math.stackexchange.com/questions/1886619/infinitely-often-interpretation-of-lim-sup-of-sequence-of-sets, https://math.stackexchange.com/questions/107931/lim-sup-and-lim-inf-of-sequence-of-sets. – Hans Lundmark Feb 08 '18 at 10:42
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    Why is this question closed? It's the only question on SE about the relations between i.o. and eventually that has a decent answer... I vote to reopen. – Maverick Meerkat Jan 05 '21 at 15:02

2 Answers2

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In terms of convergence of a sequence, it doesn't matter what happens at the start, just the long-term behaviour. You can change the first million terms of a sequence without any effect on its limiting behaviour. So any statement about terms of the sequence that is true except possibly in the first million terms might as well be true if you're only interested in limiting behaviour. And the same would work for any value of a million, i.e. for any statement that's true for all $n\geq m$ , where $m$ is any constant you like. This is what it means to be eventually true.

"Infinitely often" is the opposite: if a statement such as $x_n>z$ is true infinitely often then there's no $m$ after which its negation, $x_n\leq z$, is always true (otherwise $x_n>z$ would be true at most $m$ times, i.e. finitely often). So $\neg(P\text{ eventually})$ is the same as $((\neg P)\text{ infinitely often})$.

lim sup is the answer to a question about which bounds hold eventually. If $z$ is eventually an upper bound, then $x_n\leq z$ for all $n\geq m$ and some $m$. We can turn that around: for a fixed $m$ what is the smallest $z$ which is an upper bound for all $n\geq m$? Call this $z_m$, then the lim sup is the limit of this value as $m\to\infty$. So if $z$ is bigger than the limit of $z_m$, there certainly must be an $m$ for which $z_m<z$, and this gives you (1).

Especially Lime
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Consider the following subsets of $\mathbb N$: $$A = \{2n\mid n\in\mathbb N\},\quad B = \{ n + 100\mid n\in\mathbb N\}.$$ Set $A$ contains infinitely many positive integers, while set $B$ not only contains infinitely many positive integers, it contains all but finitely many positive integers, or you can say that $n\in B$ for sufficiently large $n$ (you can't say the same for $A$).

Can you see the difference and which kind of subset ($A$ or $B$) corresponds to your statements?

Ennar
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  • Thank you. I can now see the difference with respect to $A$ and $B$. $A$ corresponds to statement $2$ and $B$ to $1$. However, it's the inequalities that trouble me. Statement $2$ is okay for me and I read it like this : if $z$ is smaller than the smallest $\sup$, then as there is infinitely many $x$ greater than the smallest $\sup$, then $z<x_n$ for infinitely many $n$. However, for statement $1$, I read : if $z$ is larger than the smallest $\sup$ What does guarantee that $z$ is larger than all the other $\sup$s to say that $z$ is larger than $x_n$ for all large $n$? Any help? – user30614 Feb 09 '18 at 09:20
  • Okay so I guess the answer to my question is that $x_n$ is not necessairly increasing and that $z>\lim \sup x_n$ implies that we can eventually reach an $n$ beyond which all the larger $\sup$s are left in the first sequences. is this correct? – user30614 Feb 09 '18 at 09:44