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I was wondering if someone would be so kind to provide a very simple explanation of lim sup and lim inf of s sequence of sets. For a sequence of subsets $A_n$ of a set $X$, the $\limsup A_n= \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N} A_n \right)$ and $\liminf A_n = \bigcup_{N=1}^\infty \left(\bigcap_{n \ge N} A_n\right)$. But i am having a hard time imagining what that really means unions of intersections and intersections of unions i think maybe causing the trouble. I read the version on wikipedia but that did n't resolve this either. Any help would be much appreciated.

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A proof of what? –  Qiaochu Yuan Feb 10 '12 at 21:51
These are the definitions. Are you asking for help understanding them? –  Henno Brandsma Feb 10 '12 at 22:00
Thanks it should have been an explanation, although the book tries to provide a proof, which i could n't follow. Hence the question. –  Hardy Feb 10 '12 at 23:32

4 Answers 4

up vote 17 down vote accepted

A member of $$ \bigcup_{N=1}^\infty \bigcap_{n\ge N} A_n $$ is a member of at least one of the sets $$ \bigcap_{n\ge N} A_n, $$ meaning it's a member of either $A_1\cap A_2 \cap A_3 \cap \cdots$ or $A_2\cap A_3 \cap A_4 \cap \cdots$ or $A_3\cap A_4 \cap A_5 \cap \cdots$ or $A_4\cap A_5 \cap A_6 \cap \cdots$ or $\ldots$ etc. That means it's a member of all except finitely many of the $A$.

A member of $$ \bigcap_{N=1}^\infty \bigcup_{n\ge N} A_n $$ is a member of all of the sets $$ \bigcup_{n\ge N} A_n, $$ so it's a member of $A_1\cup A_2 \cup A_3 \cup \cdots$ and of $A_2\cup A_3 \cup A_4 \cup \cdots$ and of $A_3\cup A_4 \cup A_5 \cup \cdots$ and of $A_4\cup A_5 \cup A_6 \cup \cdots$ and of $\ldots$ etc. That means no matter how far down the sequence you go, it's a member of at least one of the sets that come later. That means it's a member of infinitely many of them, but there might also be infinitely many that it does not belong to.

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I think in the of sup explanation the should be U signs between the subsets, but the rest of the answer was well explained. –  Hardy Feb 10 '12 at 23:39
fixed ${{{{{}}}}}$ –  Michael Hardy Feb 11 '12 at 0:51
thanks for this wonderful explanation! –  under-root Mar 27 '14 at 19:09
great answer! Thank you! –  Ant Jul 24 '14 at 8:58

In terms of sets, we have the following interpretations:

  • $\displaystyle x\in\bigcup_{i\in I} A_i$ means that $x$ is in at least one of the $A_i$ sets.
  • $\displaystyle x\in\bigcap_{i\in I} A_i$ means that $x$ is in all of the $A_i$ sets.

So this means that

  1. $\bigcap_{N\ge1}\bigcup_{n\ge N} A_n$ are all elements somewhere in $A_N,A_{N+1},A_{N+2},\dots$, no matter how large N is, which is logically equivalent to "being in infinitely many of the $A_i$ sets."
  2. $\bigcup_{N\ge1}\bigcap_{n\ge N} A_n$ are all elements in every single one of $A_{N},A_{N+1},A_{N+2},\dots$ for some $N$. This is logically equivalent to every element being "in all but finitely many the $A_i$ sets."
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I just came up with this mnemonic story:

There is a company with employes and one day a whole bunch $\mathcal A$ of them get fired. They become beggars and have to live on the street. One day, the local church decides to start to give out free food for them every week. In the $n$'th week, $A_n$ are the people who show up (this is a sequence of $\mathcal A$-people: $\forall n\ (A_n\subseteq \mathcal A)$. Yes, beggards never die).

Some of the people eventually get a new job and never show up at the church again. Others are to proud and try no to be seen around all the time, but they need to eat so they always come back eventually. Lastly there are the people who have low self esteem, they feel inferiour, and at one point they don't care anymore and start to get their food from the church each week.

lim sup $A_n$ are all the people who don't get another job.

lim inf $A_n$ are the people who become weekly regulars.

Clearly lim inf $A_n$ $\subseteq$ lim sub $A_n$.

A series converges is all the people who can't get another job eventually swallow their pride and befomre regulars too: lim inf $A_n$ = lim sub $A_n$. We call this the limit of $A_n$.

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Are you familiar with the real analysis definition of $$\limsup_{n\to\infty} x_n = \inf_{m\geq 0} \sup_{n\geq m} x_n~?$$

The same definition can be applied to any sequence of elements in a complete lattice. Now apply it to the power set $2^X$ of some base set $X$ with set inclusion as the partial order.

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