Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
5  
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 32 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.

share|improve this answer
    
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
    
@Ant : Thank you --- I'm glad you like it. ${}\qquad{}$ –  Michael Hardy Jul 29 at 18:12

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, beggars never die).

Some of the people eventually get a new job and never show up at the church again. Others are too proud and try not 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 inferior, 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 $\lim \inf A_n \subseteq \lim \sup A_n$.

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

share|improve this answer
    
Publish this!!!!!! –  Jack Bauer Jul 30 at 16:53

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."
share|improve this answer

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.

share|improve this answer
    
Do you expect an undergraduate or beginning graduate student to understand "complete lattice" ? –  Jack Bauer Jul 30 at 16:53
    
@JackBauer: While I don't understand why you are asking about what undergraduate or graduate students can or cannot understand, I do expect a beginning graduate student to understand what a complete lattice is and how it applies to the discussion at hand after learning the definition from, say, Wikipedia. –  Willie Wong Aug 13 at 15:33
    
including the term "partially ordered" ? –  Jack Bauer Aug 15 at 9:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.