Proof that : about sigma-algebra

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let C be acollection . of subsets of a set X , then there is asmallest sigma-algebra H containing C proof that:
H = the intersection of Bs ,B from F and F = familly of all sigma algebra that contains C
 
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You need to demonstrate that H is a subset of any sigma algebra which contains C and that H is a sigma algebra.
 
how? can u help me?
 
If you don't know how to demonstrate that the intersection of all sigma algebras containing C is a subset of any sigma algebra containing C, then no, I don't think I can.
 
Can you show: the intersection of any family of sigma-algebras (on X) is a sigma-algebra?
 
1) H, being the intersection of all sigma algebras containing C, also contains C.

2) Need to show H is a sigma algebra. Show this by showing that it's closed under union and complimentation. To do this remember that an element of H is an element of all B's.

3) The intersect is always smaller than each of the elements. Therefore with the intersect you'll find the sigma algebra that's the minimum. H is the minimum.
 
RedX said:
1) H, being the intersection of all sigma algebras containing C, also contains C.

H is defined to contain C. Otherwise the statement "H, being the intersection of all sigma algebras containing C, also contains C" need not be true. No?
 
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SW VandeCarr said:
H is defined to contain C. Otherwise the statement "H, being the intersection of all sigma algebras containing C, also contains C" need not be true. No?

H is defined to contain C.

The intersect of all sigma algebras containing C should also contain C.

Therefore it is possible for H to be the intersect of all sigma algebras containing C.

I just wanted to show that the intersection of all those sigma algebras satisfies everything that H satisfies, and later on argue that it's the smallest.
 
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