# intersection of union

• Sep 12th 2009, 09:14 PM
Kat-M
intersection of union
i do not understand how this notation works. $\bigcap_{n=1}^\infty$ $\bigcup_{m \geq n}$ $A_m$
• Sep 13th 2009, 01:08 AM
Taluivren
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 $A_k$.
• Sep 13th 2009, 02:10 AM
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 $sup$ $
A_5$
?
• Sep 13th 2009, 02:25 AM
Taluivren
Quote:

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 $sup$ $
A_5$
?

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