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What exactly does $\bigcup_{n=1}^\infty\bigg(\bigcap_{k=n}^\infty E_k\bigg)$ mean? I am confused particularly by $k$ depending on $n$.

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I think I've posted an answer to essentially this same question here before, except that I think it was phrased in the language of lim-infs and lim-sups of sets. –  Michael Hardy Sep 8 '12 at 16:54

5 Answers 5

up vote 2 down vote accepted

For each positive integer $n$ you compute the intersection

$$\bigcap_{k=n}^\infty E_k=E_n\cap E_{n+1}\cap E_{n+2}\cap\dots\,;$$

call this set $T_n$. Now you compute the union of all of the sets $T_n$:

$$\bigcup_{n=1}^\infty\left(\bigcap_{k=n}^\infty E_k\right)=\bigcup_{n=1}^\infty T_n\;.$$

$T_n$ is the set of all points that are in every $E_k$ from $E_n$ on; we might say that these are the points that are in every member of the $n$-th tail of the sequence $\langle E_k:k\in\Bbb Z^+\rangle$ of sets. The bigger $n$ is, the easier it is to be in $T_n$: for instance, to be in $T_7$ you have to be in $E_7$ (and every $E_k$ with $k\ge 7$), but to be in $T_{17}$ you don’t have to be in $E_7$ (or in $E_8,E_9,\dots,E_{15}$, or $E_{16}$). Thus, $T_1\subseteq T_2\subseteq T_3\subseteq\dots$, and the union of these sets $T_n$ may be larger than any of the individual $T_n$’s.

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It is a way of taking a 'limit' of sorts of a collection of sets (the $\liminf$).

When you see $\bigcup$, think $\exists$, when you see $\bigcap$, think $\forall$.

So, the above would be $ x \in \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty E_k $ iff $\exists n \geq 1 $ such that $ \forall k \geq n$, $ x \in E_k$.

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For any individual number $n$, $$\bigcap_{k=n}^{\infty}E_k=E_n\cap E_{n+1}\cap\cdots$$ is some set we may call $F_n$. Then we want $\bigcup\limits_{n=1}^{\infty}F_n$.

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A higher-level interpretation is that $x \in \bigcup_{n=1}^\infty \left( \bigcap_{k=n}^\infty E_k \right)$ iff there are only finitely many $k$ such that $x \notin E_k$.

To see this, first note note that if $x \in \bigcup_{n=1}^\infty \left( \bigcap_{k=n}^\infty E_k \right)$ then there must be an $n \geq 1$ such that $x \in \bigcap_{k=n}^\infty E_k$, but then $x \in E_k$ for all $k \geq n$, and therefore if $x \notin E_k$ it must be that $k$ is among $1,2,\ldots,n-1$.

In the opposite direction, if $\{ k \geq 1 : x \notin E_k \}$ is finite, then it has a maximum element $n$. It then follows that $x \in E_k$ for all $k \geq n+1$, and so $x \in \bigcap_{k=n+1}^\infty E_k \subseteq \bigcup_{n=1}^\infty \left( \bigcap_{k=n}^\infty E_k \right)$.

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This is a limit-infimum of a collection of sets $E_k$. Think of $\cap_{k=n}^\infty E_k =E_n\cap E_{n+1}\cap\ldots$, so taking the union over $n$ defines those points $x$ which belong in some tail $E_n\cap E_{n+1}\cap\ldots$. More specifically, a point $x$ belongs to $\cup_{n=1}^\infty \left(\cap_{k=n}^\infty E_k\right)$ if there exists a $n=n(x)$ (that depends on $x$) with $x\in \cap_{k=n(x)}^\infty$. In other words, $x$ belongs to every $E_i$ for all $i\geq n(x)$.

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