Htykuut

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 $limsup A_n= bigcap_{N=1}^infty left( bigcup_{nge N} A_n right)$ and $liminf A_n = bigcup_{N=1}^infty left(bigcap_{n ge N} A_nright)$. But O 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 didn't resolve this either.
Any help would be much appreciated.

A member of
$$
bigcup_{N=1}^infty bigcap_{nge N} A_n
$$
is a member of at least one of the sets
$$
bigcap_{nge N} A_n,
$$
meaning it's a member of either $A_1cap A_2 cap A_3 cap cdots$ or $A_2cap A_3 cap A_4 cap cdots$ or $A_3cap A_4 cap A_5 cap cdots$ or $A_4cap A_5 cap A_6 cap cdots$ or $ldots$ etc. That means it's a member of all except finitely many of the $A$.

A member of
$$
bigcap_{N=1}^infty bigcup_{nge N} A_n
$$
is a member of all of the sets
$$
bigcup_{nge N} A_n,
$$
so it's a member of $A_1cup A_2 cup A_3 cup cdots$ and of $A_2cup A_3 cup A_4 cup cdots$ and of $A_3cup A_4 cup A_5 cup cdots$ and of $A_4cup A_5 cup A_6 cup cdots$ 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.

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: $forall n (A_nsubseteq mathcal A)$. Yes, beggars never die).

1. Some of the people eventually get a new job and never show up at the church again.




2. 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.




3. 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. (Categories 2 and 3). Thus, $lim inf A_n^C$ are all the people who eventually get a new job. (Category 1)

$lim inf A_n$ are the people who become weekly regulars. (Category 3)

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 $A_n$.

(This is a late answer, but hopefully it may add another perspective to the discussion.)

When one thinks about the problem of defining the limit of a sequence of sets,
there are two easy cases: if the sequence is increasing, or if it's decreasing.

For example, when defining improper integrals $iint_D$ in multivariable calculus, one looks at sequences $D_1 subseteq D_2 subseteq D_3 subseteq dotsb$ which exhaust $D$, meaning that

• each $D_i$ is a subset of $D$, and


• each element of $D$ is contained in $D_n$ for all sufficiently large $n$ (in other words, $D=bigcup_{n=1}^infty D_n$).

In this situation, students often ask if they can write $D_n to D$, and the answer is usually “well, we haven't really defined limits of sets in this course, but if we had...”. So it's pretty intuitive that for an increasing sequence of sets, the limit should be defined as the union of all sets in the sequence.

Similarly, for a decreasing sequence $E_1 supseteq E_2 supseteq E_3 supseteq dotsb$, it's natural to define the limit as the intersection
: $E_n to E$ as $ntoinfty$, where $E=bigcap_{n=1}^infty E_n$.

Now, for an arbitrary sequence of sets $(A_1,A_2,A_3,dots)$, we can squeeze it between an increasing sequence $(D_n)$ and a decreasing sequence $(E_n)$, like this:
$$
begin{array}{lcl}
D_1 = A_1 cap A_2 cap A_3 cap dotsb
&quadsubseteqqquad
A_1
qquad
&subseteq qquad
E_1 = A_1 cup A_2 cup A_3 cup dotsb
\
D_2 = phantom{A_1 cap{}} A_2 cap A_3 cap dotsb
&quadsubseteqqquad
A_2
qquad
&subseteq qquad
E_2 = phantom{A_1 cup{}} A_2 cup A_3 cup dotsb
\
D_3 = phantom{A_1 cap A_2 cap{}} A_3 cap dotsb
&quadsubseteqqquad
A_3
qquad
&subseteq qquad
E_3 = phantom{A_1 cup A_2 cup{}} A_3 cup dotsb
\
& qquadvdots &
end{array}
$$
Moreover, $(D_n)$ is the largest increasing sequence such that $D_n subseteq A_n$ for all $n$,
and $(E_n)$ is the smallest decreasing sequence such that $A_n subseteq E_n$ for all $n$,
so it makes sense to define
$$
liminf_{ntoinfty} A_n = lim_{ntoinfty} D_n
,qquad
limsup_{ntoinfty} A_n = lim_{ntoinfty} E_n
.
$$
And if these lower and upper limits are equal, then we say that that's $limlimits_{ntoinfty} A_n$.

(What Willie Wong's answer says it that the same construction makes sense in any partially ordered set where you can take the least upper bound and greatest lower bound of infinitely many elements, if you replace $(subseteq,cap,cup)$ by $(le,wedge,vee)$.)

In terms of sets, we have the following interpretations:

• $displaystyle xinbigcup_{iin I} A_i$ means that $x$ is in at least one of the $A_i$ sets.


• $displaystyle xinbigcap_{iin I} A_i$ means that $x$ is in all of the $A_i$ sets.

So this means that

1. $bigcap_{Nge1}bigcup_{nge N} A_n$ are all elements somewhere in $A_N,A_{N+1},A_{N+2},dots$, no matter how large N is. Being a member of this set is logically equivalent to being "in infinitely many of the $A_i$ sets".


2. $bigcup_{Nge1}bigcap_{nge N} A_n$ are all elements in every single one of $A_{N},A_{N+1},A_{N+2},dots$ for some $N$. Being a member of this set is logically equivalent to being "in all but finitely many the $A_i$ sets".