# Limsup and liminf

1. Dec 17, 2007

### ice109

i'm reading this book called "measure and integral: an introduction to real analysis" by antoni zygmund and richard l wheeden and i'm stuck on page two .

really the question is what are limsup and liminf but i would appreciate if some other questions were answered and maybe i'll be able to figure it out on my own.

in this book they define F a family of subsets of $R^n$ then say if F is countable( finite or countably infinte) then it is a sequence of sets. now does this imply an ordering of F where it is either decreasing or increasing? where the book defines increasing as $x_k \subset x_{k+1}$ and decreasing as $x_{k+1} \subset x_k$ where obviously $x_k$ is set k in the sequence.

if this is not the case about ordering or sequences then i don't understand how we can discuss limsup and liminf of a sequence of sets.

2. Dec 17, 2007

The "subset ordering" is a partial ordering, any two given sets may not be comparable (for instance of they are nonempty disjoint). As for "what is lim sup E_n", etc. I presume the definition is in your book. lim sup and lim inf allow for a notion of a limit that is more general than the intersection of a nested decreasing sequence of sets or the union of a nested increasing sequence of sets, both of which are intuitive ideas. No epsilon-delta is involved.

3. Dec 17, 2007

### HallsofIvy

Staff Emeritus
It might be easier to think of numbers rather than sets first. If you have a countable collection of numbers, then you can think of it as a sequence with some given order (of course, it does not have to be increasing or decreasing[/b]. Obviously, different orders will give you different sequences but the limit (or subsequential limits if it has more than one) do not depend on the ordering! If you are studying measure, then you should already know that the "limsup" of a sequence of numbers is the least upper bound on the set of all subsequential limits and the "liminf" is the greatest lower bound. Those, again, do not depend on the particular order given to the sequence! If the sequence of numbers is not bounded, the limsup and liminf may not be real numbers in which case it is standard to say they are "$+\infty$" or "$-\infty$".

Now, for sets, we use "subset" as an ordering. Again, a countable collection of sets can be thought of as a sequence: "countable" means there is a one-to-one function from the natural numbers onto the collection- that's an automatic ordering. Of course the same collection of sets might have many different such functions and so many different ordering. Again, they do not have to be either increasing or decreasing. Given an increasing or decreasing sequence of sets, its limit is (for increasing) the smallest set containing all sets in the sequence or (for decreasing) the largest set contained in all sets in the sequence. If a sequence of sets is neither increasing nor decreasing, it may have many such sequences with many different limits. The "limsup" and "liminf" are again the "least upper bound" (the smallest set containing all the subsequential limits) and the "greatest lower bound" (the largest set contained in all the subsequential limits). The difficulty is that since the order defined by "subset" is not a linear order (there exist set A and B such that neither is a subset of the other), there may not exist such limsup and liminf.

4. Dec 17, 2007

### ice109

please more on this. if partially ordered they can or they cannot be "nonempty disjoint"

halls:

here is how they are defined in my book

$$limsup ~ E_k = \bigcap^{\infty}_{j=1}\left(\bigcup^{\infty}_{k=j} E_k\right)$$

$$liminf ~ E_k = \bigcup^{\infty}_{j=1}\left(\bigcap^{\infty}_{k=j} E_k\right)$$

Last edited: Dec 17, 2007
5. Dec 17, 2007

The set {$$\emptyset$$,{5},{6}} is only partially ordered. You have $$\emptyset \subset \{ 5 \} \, \emptyset \subset \{ 6 \}$$, but you don't have $$\{ 5 \} \subset \{ 6 \}$$ or $$\{ 6 \} \subset \{ 5 \}$$. ({5},{6} are nonempty and disjoint.)

Given the definitions in your book, you are probably asked to prove in a problem that lim inf {E_n} = {x: x belongs to all but finitely many E_n}, and lim sup E_n = {x: x belongs to infinitely many E_n}.. and as an exercise you can observe the other important relation that lim inf E_n $$\subseteq$$ lim sup E_n, but the opposite inequality might not hold.

Halls was pretty thorough in the relation between the real case.. One of the points is that you will get used to these types of definitions as long as you keep on studying math. Hopefully that will help get you past page 2.

6. Dec 17, 2007

### ice109

look i can easily think of a set whose limsup and liminf change depending on the assignment of the index so i think i'm misunderstanding something about that. call E_1={1,2,3} E_2={1,2} E_3={1}. limsup equals 1 and liminf equals 1. now reverse the order with E_1={1}
E_2={1,2} and E_3={1,2,3} and now limsup equals {1,2,3} and liminf equals {1,2,3}. so i don't understand how these limits can be independent of ordering.

7. Dec 18, 2007

What you said is like saying x_1 = 1, x_2 = 2, x_3 = 3, therefore lim x_n = 3. But the limit (and also lim inf, lim sup) definition applies to an infinite sequence, not 3 terms. Same for the system of sets. I think if you define E_n for n > 3, then you will not find any such contradiction.

8. Dec 18, 2007

### mathwonk

any set has subsequences. i am going out on a limb and guess that the limsup is the largest limit of a subsequence consisting of distinct points of the set.

equivalently it should be the largest number such that any smaller number is exceeded by an infinite number of elements of the set. ??? does that seem right?

intuitively it is the largest "cluster point" of the set, i.e. the largest limit point, i.e. the supremum of all the limit points.

9. Dec 18, 2007

### ice109

i don't see the difference between the finite and infinite case? define the sequence for infinite terms and you will still get different limits if you arrange it increasing or decreasing as previously defined.

Last edited: Dec 18, 2007
10. Dec 19, 2007

It is easier to look at
lim inf {E_n} = {x:x belongs to all but finitely many E_n}, and
lim sup {E_n} = {x:x belongs to infinitely many E_n}.

From this identity, it's not that hard to see no matter how you rearrange the E_n's you will get the same answer. Again, lim sup, lim inf doesn't make sense for a finite list of sets.

You can actually define lim inf, lim sup this way. But the set identity makes it easier to prove (using De Morgan) that
$$(lim \ inf \{ E_n \} )^c \ = \ lim \ sup \{ E_n^c\}$$, etc. This formula kind of looks similar to
- lim inf a_n = lim sup -a_n.

One thing that sticks out to me is that x is in lim sup {E_n} if and only if there is some subsequence E_nk that contains x for all k. x is in lim inf {E_n} if and only if every subsequence E_nk contians x for all k >= N, for some N.

11. Mar 20, 2008

### Sesp

Hi, I was just reading through your posts and I thought I understood it, but the last sentence confused me. x is in liminf {E_n} iff every subsequence contains x for all k>=N --- doesn't that mean that you need one ordering for the sequence of sets? Otherwise what does k>=N mean?

12. Mar 20, 2008

### Sesp

Actually I think I get it. Had a look at wikipedia's definition of liminf and limsup for sets http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior
In the example there's a sequence of sets {0},{1},{0},{1},{0},{1}.....
It doesn't matter which way these are ordered because there's always going to be an infinite number of sets {0} and infinite number of sets {1} remaining in the sequence.

So limsup will always be {0,1} and liminf will always be empty set no matter how they are ordered.

13. Mar 20, 2008

### e(ho0n3

I agree with mathwonk's post: the lim sup of a sequence is the largest limit of all the subsequences that have a limit. Similarly, the lim inf of a sequence is the smallest limit of all the subsequences that have a limit.

Example: Let $$a_n = \cos \pi n$$. This is just the sequence 1, -1, 1, -1, ... The lim sup is 1, the lim inf is -1.

14. Mar 20, 2008

### bomba923

For fun:
Show that
$$\mathop {\lim }\limits_{n \to \infty } \bigcup\limits_{i = 0}^n {\left\{ {\frac{i} {n}} \right\}} \ne \mathop {\lim }\limits_{n \to \infty } \bigcup\limits_{i = 0}^{n!} {\left\{ {\frac{i}{{n!}}} \right\}} = \mathbb{Q} \cap \left[ {0,1} \right]$$

Last edited: Mar 20, 2008