# Thread: algebra of sets law

1. ## algebra of sets law

Hi,

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

2. Originally Posted by waytogo it is determined that $\displaystyle (\bigcup_{n=1}^m A_n)^c = \bigcup_{n=1}^m (A_n)^c$.
That is not true.

It is $\displaystyle (\bigcup_{n=1}^m A_n)^c = \bigcap_{n=1}^m (A_n)^c$

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

4. you are right Plato, my mistake. made it correct now.

5. Originally Posted by waytogo it is determined that $\displaystyle (\bigcup_{n=1}^m A_n)^c = \bigcap_{n=1}^m (A_n)^c$. Does $\displaystyle (\bigcup_{n=1}^\infty A_n)^c = \bigcap_{n=1}^\infty (A_n)^c$ holds?
In the set $\displaystyle \mathbb{N}=\{0,1,2,3,\cdots\}$ define $\displaystyle A_n=\{n\}$.
For each $\displaystyle m$ does $\displaystyle (\bigcup_{n=1}^m A_n)^c = \bigcap_{n=1}^m (A_n)^c$ hold ?

6. Originally Posted by Plato In the set $\displaystyle \mathbb{N}=\{0,1,2,3,\cdots\}$ define $\displaystyle A_n=\{n\}$.
For each $\displaystyle m$ does $\displaystyle (\bigcup_{n=1}^m A_n)^c = \bigcap_{n=1}^m (A_n)^c$ hold ?
yes, for me it seems correct in this case.

7. 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.

8. 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.

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