Hi,

it is determined that . Does holds?

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- Jun 3rd 2011, 08:32 AMwaytogoalgebra of sets law
Hi,

it is determined that . Does holds? - Jun 3rd 2011, 08:48 AMPlato
- Jun 3rd 2011, 08:50 AMemakarov
No. For example, let and suppose , for . Then . However, , while .

- Jun 3rd 2011, 09:27 AMwaytogo
you are right Plato, my mistake. made it correct now.

- Jun 3rd 2011, 09:38 AMPlato
- Jun 3rd 2011, 01:49 PMwaytogo
- Jun 3rd 2011, 01:56 PMPlato
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.** - Jun 3rd 2011, 03:30 PMMoeBlee
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.