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Let $E = \{E_k\}_{k \in \mathbb{N}}$ be an infinite sequence of sets. Then, the following inclusion holds:

$\bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} E_k \quad\subseteq\quad \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} E_k$

I know the left-hand side (LHS) represents elements which belong to all but finitely many sets in the sequence $E$, and the right-hand side (RHS) represents elements which belong to infinitely many sets in the sequence $E$. This concludes the proof that the LHS is contained in the RHS. Yet it is not intuitive for me, since the above interpretation is not intuitive per se.

Is there an intuitive reason why the above inclusion holds? That is, (very informally) why union of intersections is contained in intersection of unions?

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4 Answers 4

up vote 6 down vote accepted

If $A$ and $B$ are sets, and $A\subseteq B_n$ for all $n$, then $A\subseteq\bigcap_n B_n$. Conversely, $A\subseteq\bigcap_n B_n$ means that $A\subseteq B_n$ for all $n$.

This tells us that the inclusion you are interested in reduces then to checking $(+)$: $$ \bigcup_n \bigcap_{k\ge n}E_k\subseteq \bigcup_{k\ge m}E_k $$ for all $m$.

(This is progress. It is similar to a common move in analysis: To prove that $a\le b$, it is not unusual to check instead that $a\le c$ for any $c>b$. But these inequalities $a\le c$ tend to be easier than the one you really want, $a\le b$.)

Now, the same idea gives us that to prove the new inclusion $(+)$, it is enough to prove $(++)$: $$\bigcap_{k\ge n}E_k \subseteq \bigcup_{k\ge m}E_k $$ for all $n,m$.

(The corresponding move in analysis is that to prove $a\le b$, it is enough to show $c\le b$ for all $c\le a$. Unions correspond to suprema, intersections to infima in these analogies.)

Now, $(++)$ is really obvious: If somebody is in the left hand side, then it is in all sufficiently large $E_k$, but then it is in the right hand side.

The point is: It is useful to train oneself to "decode" certain expressions the way we did, by focusing on their "atomic" or at least "more basic" components.

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Excellent! Just one point: the inclusion (+) is true for "all m", rather than "for all n,m". Am I right? –  Sadeq Dousti Jan 19 '11 at 15:37
1  
Right, thanks. Edited. –  Andres Caicedo Jan 19 '11 at 15:38

As you said the LHS is the collection of elements which appear in all but finitely many $E_k$, while the elements on the RHS are those which appear in infinitely many $E_k$.

If you appear in all but finitely many, then you appear in infinitely many; the converse is not always true as you can appear in all the even indices but not once in the odd indices.

The LHS is called sometimes denoted by $\liminf E$, and the RHS is $\limsup E$.

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I understand this, but as I pointed out, this interpretation is not intuitive per se. –  Sadeq Dousti Jan 19 '11 at 15:27
    
@Sadeq: In mathematics many times intuition is built after a while, and cannot be explained in simple terms. I think this is very much the case here. –  Asaf Karagila Jan 19 '11 at 15:31

So on the left side, an element must belong to every $E_k$ starting from some $k$. On the right side, an element only has to belong to sets that are arbitrarily large. So if it belongs to every other one, or every prime number one, or something like that, it will be included in the right side.

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There are good explanations of liminf and limsup of sets here. You will probably see these constructions in measure theory.

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