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I want to show that A = $\bigcap\limits_{n=1}^\infty \bigcup\limits_{k=n}^\infty E_k$ is the set of elements that lie in infinitely many $E_k$. However, I am a little bit confused because:

For all x in infinitely many $E_k$, x is in $\bigcup\limits_{k=n}^\infty E_k$, however, from here I'm not sure that I can conclude x is in $\bigcap\limits_{n=1}^\infty \bigcup\limits_{k=n}^\infty E_k$, because it's a countable intersection. For example 0 is in $(-1/n, 1/n)$ for all n, but it is not in $\bigcap\limits_{n=1}^\infty (-1/n, 1/n)$

Could anyone help me please?

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No. We certainly have $0\in\bigcap\limits_{n=1}^\infty (-1/n, 1/n)$. You seem to have a misunderstanding of how intersections are defined. The meaning of $0\in\bigcap\limits_{n=1}^\infty (-1/n, 1/n)$ is exactly that $0\in(-1/n,1/n)$ for all $n$. –  Michael Greinecker Mar 11 '12 at 11:11
    
If $x\in\bigcup_{k=n}^\infty E_k$ for all $n\geq 1$, then by definition you can conclude that $x\in\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty E_k$. Your last example is wrong, since $\bigcap_{n=1}^\infty (-\tfrac{1}{n},\tfrac{1}{n}) = \{0\}$. –  Stefan Hansen Mar 11 '12 at 11:13
    
It probably easier to show that $x \notin \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty E_k$ iff $x$ is an element of (at most) finitely many $E_k$. –  Arthur Fischer Mar 11 '12 at 11:26
    
Thank you, yes I think I misunderstood the definition of the intersection. –  user1412 Mar 11 '12 at 11:43

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up vote 5 down vote accepted

We wish to show that $A = \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} E_{k}$ is the set of elements that lie in infinitely many $E_{k}$.

To see this, we first note what this notation is saying. We first take the union $\bigcup _{k=1}^{\infty} E_{k}$, then we intersect it with the union $\bigcup_{k=2}^{\infty} E_{k}$, which is equivalent to throwing away $E_{1}$. Then, we intersect with $\bigcup_{k=2}^{\infty} E_{k}$, which means we are throwing away $E_{2}$. If we keep on throwing away $E_{n}$, the only elements that remain are those that are in $E_{k}$ for arbitrarily large $k$, which is just the set of elements that lie in infinitely many $E_{k}$. With this intuition, we attempt a formal proof:

Let $x$ lie in infinitely many $E_{k}$. Then, for every $n$, there $j > n$ such that $x \in E_{j}$. Thus, $x \in \bigcup_{k=n}^{\infty}$ for every $n$, and hence it is in the intersection $\bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} E_{k}$. For the converse, we do a proof by contrapositive. Suppose that $x$ does not lie in infinitely many $E_{k}$. Then there is some largest $E_{K}$ in which $x$ lies. Hence, for $n > K$, we have $x \notin \bigcup_{k=n}^{\infty} E_{k}$. Thus $x$ will not be in the intersection $\bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} E_{k}$, as it is not in every $\bigcup_{k=n}^{\infty} E_{k}$.

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