intersection of union

**Kat-M** · September 12th 2009, 09:14 PM

i do not understand how this notation works. $\bigcap_{n=1}^\infty$ $\bigcup_{m \geq n}$ $A_m$
help me please.

**Taluivren** · September 13th 2009, 01:08 AM

Hi, i don't know if the following visualisation is helpful:
$\bigcap_{n=1}^\infty \bigcup_{m \geq n} A_m = \bigcap_{n=1}^\infty (A_n \cup A_{n+1}\cup A_{n+2}\cup A_{n+3}\ldots) =$

$= (A_1 \cup A_{2}\cup A_{3}\cup A_{4}\cup A_{5}\cup A_{6}\ldots)\cap (A_{2}\cup A_{3}\cup A_{4}\cup A_{5}\cup A_{6}\ldots)\cap (A_{3}\cup A_{4}\cup A_{5}\cup A_{6}\ldots)\cap\ldots$

but maybe this little proposition will make things clear:

$x \in \bigcap_{n=1}^\infty \bigcup_{m \geq n} A_m \mbox{ if and only if } x \in A_k \mbox{ for infinitely many } k \in \mathbb{N}.$

Proof: "=>" assume $x\in A_k$ for only finitely many $k\in\mathbb{N}$ , let $l$ be the biggest such $k$ , then for all $m\ge l+1$ we have $x\not \in A_m$ . This means $x \not \in \bigcup_{m\ge l+1}A_m$ , so $x\not\in \bigcap_{n=1}^\infty \bigcup_{m \geq n} A_m$ .

"<=" assume $x\not\in \bigcap_{n=1}^\infty \bigcup_{m \geq n} A_m$ , this means there's some $n\in\mathbb{N}$ such that $x \not \in \bigcup_{m\ge n}A_m$ , that is, $x\not\in A_m$ for all $m\ge n$ , so $x$ belongs to only finitely many $A_k$ .

**Kat-M** · September 13th 2009, 02:10 AM

So if we define $lim_{n \rightarrow \infty}sup$ $A_m$ = $\bigcap_{n=1}^\infty \bigcup_{m \geq n} A_m$ , what is $sup$ $<br /> A_5$ ?

**Taluivren** · September 13th 2009, 02:25 AM

Originally Posted by Kat-M

So if we define $lim_{n \rightarrow \infty}sup$ $A_m$ = $\bigcap_{n=1}^\infty \bigcup_{m \geq n} A_m$ , what is $sup$ $<br /> A_5$ ?

this doesn't make sense.
it is "limes superior of a sequence of sets $\{A_n:\, n\in \mathbb{N}\}$ " , by definition:
$\limsup_{n\rightarrow \infty}A_n =\bigcap_{n=1}^\infty \bigcup_{m \geq n} A_m$

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