algebra of sets law

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June 3rd 2011, 07:32 AM
waytogo

algebra of sets law

Hi,

it is determined that $(\bigcup_{n=1}^m A_n)^c = \bigcap_{n=1}^m (A_n)^c$ . Does $(\bigcup_{n=1}^\infty A_n)^c = \bigcap_{n=1}^\infty (A_n)^c$ holds?
June 3rd 2011, 07:48 AM
Plato

Quote:

Originally Posted by waytogo

it is determined that $(\bigcup_{n=1}^m A_n)^c = \bigcup_{n=1}^m (A_n)^c$ .

That is not true.

It is $(\bigcup_{n=1}^m A_n)^c = \bigcap_{n=1}^m (A_n)^c$
June 3rd 2011, 07:50 AM
emakarov

No. For example, let $m = 2$ and suppose $A_1=A_2 = \{0\}$ , $A_n=\{i\in\mathbb{N}\mid i>0\}$ for $n>2$ . Then $(A_1\cup A_2)^c=A_1^c=A_2^c=A_3=A_1^c\cup A_2^c$ . However, $\big(\bigcup_{n=1}^\infty A_n\big)^c=\mathbb{N}^c=\emptyset$ , while $\bigcup_{n=1}^\infty A_n^c=\mathbb{N}$ .
June 3rd 2011, 08:27 AM
waytogo

you are right Plato, my mistake. made it correct now.
June 3rd 2011, 08:38 AM
Plato

Quote:

Originally Posted by waytogo

it is determined that $(\bigcup_{n=1}^m A_n)^c = \bigcap_{n=1}^m (A_n)^c$ . Does $(\bigcup_{n=1}^\infty A_n)^c = \bigcap_{n=1}^\infty (A_n)^c$ holds?

In the set $\mathbb{N}=\{0,1,2,3,\cdots\}$ define $A_n=\{n\}$ .
For each $m$ does $(\bigcup_{n=1}^m A_n)^c = \bigcap_{n=1}^m (A_n)^c$ hold ?
June 3rd 2011, 12:49 PM
waytogo

Quote:

Originally Posted by Plato

In the set $\mathbb{N}=\{0,1,2,3,\cdots\}$ define $A_n=\{n\}$ .
For each $m$ does $(\bigcup_{n=1}^m A_n)^c = \bigcap_{n=1}^m (A_n)^c$ hold ?

yes, for me it seems correct in this case.
June 3rd 2011, 12:56 PM
Plato

I have a bad feeling that we are missing some fine points here.
Are you asking about a particular collection of sets with the same index set? Or are you asking about a collection with a finite index set that can be expanded into an infinite collect?

It would be nice if you posted an exact, fully worded question.
June 3rd 2011, 02:30 PM
MoeBlee

In what theory? In Z set theories, there is no absolute complement of a set. So suppose this is relative to some universe S:

Suppose x in S\U[n = 1 to oo]An. Show that for all n, we have x in S\An. Well, x is in S. And if there were an n such that x in An, then x would be in U[1 = n to oo]An, but x is not in U[n = 1 to oo]An.

Suppose that for all n, we have x in S\An. Show x in S\U[n = 1 to oo]An. Well, x is in S. And if x where in U[n = 1 to oo]An, then x would be in An for some n, but for no n is x in An.

Indeed, in even the most general case, the relative complement of the union of B equals the intersection of the relative complements of the members of B.

This is generalized De Morgan's law for sets.