Re: Hulls in sigma algeras

On Fri, 3 Aug 2007, G. A. Edgar wrote:
<maninalift@xxxxxxxxxxxxxx> wrote:

Is it the case that the intersection of all members of a sigma-algebra
containing a given element of the underlying space must its self be a
member of that sigma algebra?

In general, no. If the sigma-algebra is countably generated, yes.

I'm not understanding the problem. Would you detail or formulize what is
meant by "the intersection (/\) of all members of a sigma algebra"? If S
is the algebra, does it mean /\{ A | A in S, x in A } for some given x in
the underlying space \/S?

It seems a very simple question but I'm not seeing a proof or
counterexample at the moment.

E = {0,1} with all four subsets as sigma-algebra.
X = uncountable product of copies of E, with product sigma-algebra.
Then points of X are not measurable sets.

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/

.

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