# How to Quantify $\limsup \limits_{i \to \infty} \; E_i = \bigcap \limits_{k=1}^{\infty} \bigcup \limits_{i=k}^{\infty}\; E_i$?

$$\limsup_{i \to \infty} \; E_i = \bigcap_{k=1}^{\infty} \; \bigcup_{i=k}^{\infty} \; E_i$$ So $$\bigcup_{i=k}^{\infty} \; E_{i} = \{x \in E_{i} \mid i \in I\}= S$$ where $I$ is some index set. When I try try to understand next part, I get confused, seems too trivial: $$\bigcap_{k=1}^{\infty}S = S$$ since there is only one set $S$ as the sets were already evaluated with the or quantifier in the operation $\bigcup$. I am probably wrong so how do you quantify inner structure in $\limsup \limits _{i\to \infty} \; E_i$?

[Updated]

$\bigcap_{k=1}^{\infty}S$ = S clearly cannot be true in every cases, only trivially. Thank you for the notice. Perhaps, I am not clear what I mean by quantification. Let's consider simpler case with finite sets.

1. Suppose an arbitrary amount of sets. What is the maximun number of different types of sets in $A = \bigcap_{k=1}^{a}\bigcup_{i=k}^{b} E_{i}$?
2. What is the maximum amount of different sets in $\bigcup_{i=k}^{\infty} E_{i}$?
3. I pretty sure you know better measures to analyze it, how?

... and so on, I want to break the problem into parts to analyze it.

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The mistake occurs right at the beginning: Note that the set $S_{k} = \bigcup_{i=k}^{\infty} E_{i} = \bigcup_{i=k}^{\infty} \{x\,|\,\exists\,i \geq k\,:\,x \in E_{i}\}$ depends on $k$. – t.b. Jan 21 '11 at 2:06
Your new questions are unclear. Does "arbitrary" mean "countable", "possibly uncountable", "finite but arbitrary"? Question 2 is likewise unclear; the union is a single set, so it makes no sense to talk about "maximum amoung of different sets". – Arturo Magidin Jan 25 '11 at 16:38
Arturo Magidin: Yes it does in the context of finite sets. Suppose $A={1,2}$ and $B={1}$. What is the maximum number of different combinations you can get with them? Trivially union just $1$ but when you consider you have also intersection, you can get ${1}$. So 2 pcs (without empty set). – hhh Jan 25 '11 at 16:45
err $A=\{1,2\}$, $B=\{1\}$ – hhh Jan 25 '11 at 16:46
@hhh: The set $A$ is just a single set; it makes no sense to talk about "different sets in A". Perhaps you meant something like $A(k,b)$ and asking about possible difference as $k$ and $b$ change, but that is not what you wrote. Likewise, $\cup_{i=1}^{\infty}E_i$ is a single set, it makes no sense to ask about "maximum amount of sets in" a single set; clearly you meant something else, but what you wrote is not what you mean (apparently). This has nothing to do with whether your sets are finite or not, but with what you are writing. – Arturo Magidin Jan 25 '11 at 19:37

I assume your collection of sets is countable (and that this is what you mean by "arbtirary"); that is, you have one set for each natural number (or perhaps only up to a certain bound), $E_1$, $E_2$, $E_3$, $E_4,\ldots$.

$\limsup E_i$ consists of all elements in $\cup E_i$ that are in infinitely many of the $E_i$. $\liminf E_i$ consist of all elements in $E_i$ that are in all except perhaps for finitely many of the $E_i$.

If you imagine the $E_i$ lined up in front of you, then $\limsup E_i$ is the collection of all things which, no matter where you stand on the line, there will always be an instance of that thing behind you (I think of $E_{i+1}$ as "behind" $E_i$). On the other hand, $\liminf E_i$ consists of all thing which are in each position of the queue after a certain point.

In general, if you think of $\limsup E_i$ as a "sequence" of sets, namely the sets $F_k = \cup_{i=k}^{\infty} E_i$, which you are then intersecting, then the $F_k$ can be all different, or they can change as often as you want or as little as you want, even if all your sets are finite.

For an example where all the $F_i$ are different, take $E_i = \{i\}$. Then $F_k = \{k,k+1,\ldots\}$. Or you can tweak it to change every so often; $E_1 = \{1\}$, $E_2=\{1,2\}$, $E_3=\{1,3\}$, $E_4=\{4\}$, $E_5=\{4,5\}$, $E_6=\{4,6\},\ldots$. Then $F_k$ will be $\{r,r+1,r+2,\ldots\}$ if $3r+1\leq k\leq 3r+r$. You can make the changes happen at distinct intervals. Etc.

If all your sets are contained in a finite set $A$ with $n$ elements, then you can set the $F_i$ to be as many as $n+1$ distinct sets (notice that you will have $F_1\supseteq F_2\supseteq F_3\supseteq F_4\supseteq\cdots$, and it is easy to check that if they are all contained in a set with $n+1$ elements, then such a chain can have at most $n$ distinct elements in it, since every time you "change" sets you lose at least one element, and you can go from $n$ elements down to $0$ elements).

Symmetrically with the limit inferior. if you set $G_i = \mathop{\cap}_{i=1}^{\infty} E_i$, then you have $G_1\subseteq G_2\subseteq G_3\subseteq\cdots$; if all sets are contained in an infinite set $A$, then you can have the $G_i$ change at each step, every other step, or you can designate exactly where you want them to change and you can set it up so that they change exactly there. If your $E_i$ are all subsets of a finite set with $n$ elements, then there can be at most $n+1$ distinct $G_i$, for the same reason as there can be at most $n+1$ distinct $F$.

If you consider pairs $(a,b)$ with $a\leq b$, and you set $H_{a,b} = \cup_{k=1}^a\cap_{i=k}^{b} E_i$, then you have $H_{1,1}\supseteq H_{1,2}\supseteq H_{1,3}\supseteq\cdots$, $H_{2,2}\supseteq H_{2,3}\supseteq \cdots$ (and in general $H_{a,b}\supseteq H_{a,b+1}$ for all $a\leq b$, and $H_{a,b}\subseteq H_{a+1,b}$ for all $a\lt b$). You can have lots of distinct things, or they may be limited depending on where your sets $E_i$ are taken from.

The questions you pose are too vague to give anything but very general answers absent more information.

Added: Clearly, we are having an impasse of undertanding. Below you say that "quantify" to you means "break up and construct interesting stuff". I'll wager the reason people (me included) are so confused about your question is that nobody took that word to mean anything like that (I certainly did not). The dictionary gives two meanings: "to determine or express the quantity of", and the logic related one of "to limit the variables of a proposition by prefixing an operator such as for all or some". I took it to mean that you wanted to give it a kind of "quantity" or "size"; perhaps you are taking your queue from "quantum mechanics" when you use "quantify", but I don't think anybody understood you. I confess to still being rather mystified.

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Arturo Magidin: Can you elaborate the last part. How did you see the lower indices with $H_{a,b}\subseteq H_{a+1,b}$ for all $a<b$? Where is the contradiction if $H_{a+1,b} \subseteq H_{a,b}$ for all $a<b$? – hhh Feb 1 '11 at 12:14
@hhh: No contradiction arises from simply that assumption, because you could have equality. As to why the inclusion I indicated holds, note that $$H_{a+1,b} = \cup_{k=1}^{a+1}(\cap_{i=k}^bE_i) = (\cup_{k=1}^a(\cap_{i=k}^bE_i))\cup (\cap_{i={a+1}}^bE_i) = H_{a,b}\cup (\cap_{i=a+1}^bE_i)\supseteq H_{a,b}.$$ – Arturo Magidin Feb 1 '11 at 14:09
Arturo Magidin: so for $a\leq b$, $H_{a,b} \supseteq H_{a,b+1}= \cup_{k=1}^{a}\cap_{i=k}^{b+1}E_{i} = \cup_{k=1}^{a} ( (\cap_{i=k}^{b}E_{i}) \cap E_{b+1} ) \subseteq \cup_{k=1}^{a}\cap_{i=k}^{b}E_{i} = H_{a,b}$. Is there a way to break $(\cap_{i=k}^{b}E_{i}) \cap E_{b+1} )$ somehow? I feel it quantifies it but perhaps someone can go even deeper. – hhh Feb 1 '11 at 16:06
@hhh: Yes, that is how I deduced the inclusions in that direction. As to "break up" the intersection, it is associative and commutative, so you can break it up into any number of intersections you wish, but that's about it. – Arturo Magidin Feb 1 '11 at 16:25
@hhh: That is not De Morgan's Law! De Morgan's Law is that $(A\cap B)^c = A^c\cup B^c$ (or more generally, $(\cap A_i)^c = \cup(A^c)$) and $(A\cup B)^c = A^c\cap B^c$ (or that $(\cup A_i)^c = \cap A_i^c$). What you describe are distributive laws of intersection and union. – Arturo Magidin Feb 2 '11 at 2:18

I am not quite sure I understand your question. But in order to understand something like $\bigcap_{k=1}^{\infty}\bigcup_{i=k}^{\infty} E_{i}$, we always translate $\bigcap$ as "For Any" and translate $\bigcup$ as "there exists". Then $\bigcap_{k=1}^{\infty}\bigcup_{i=k}^{\infty} E_{i}$ means "For any k, there exists some $i\geq k$, such that $x\in E_i$".

Thus, let $S_I = \bigcup _{i\in I} E_{i}$. If $I$ only contains finite number of indices, there is actually not much to say about the relation between $S_I$ and $S$. But if I contain infinite many elements, $S\subset S_I$. In fact, $S$ is the joint of all such $S_I$'s.

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In your third sentence, did you mean to write "For any k there exists i $\geq$ k ..."? – Rudy the Reindeer Jan 20 '11 at 12:27
thanks I corrected it. – Xiaochuan Jan 20 '11 at 12:36
Xiaochuan: what about vice versa, is the terminology still the same for $\cup_{k=1}^{\infty} \cap_{k=1}^{\infty} E_{i}$ i.e. "there exist --- for any --". – hhh Feb 1 '11 at 16:19
what about $\cap_{k=1}^{\infty} \cap_{k=1}^{\infty} E_{i}$ or $\cup_{k=1}^{\infty} \cup_{k=1}^{\infty} E_{i}$? – hhh Feb 1 '11 at 16:21

Another way to think of this is to notice that

$$\limsup_{n\to\infty} E_n = \lbrace x| x\in E_n \hbox{for infinitely many n}\rbrace.$$

In other words, $x\in E_n$ occurs "frequently."

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$$\bigcup_{i=k}^{\infty} E_i = \{ x \mid \exists i \geq k : x \in E_i\}$$

Did you mean this by inner structure of $\limsup$?

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