Let X = Y uncountable, and let A = B be the σ algebra of countable-cocountable subsets of X, Y respectively. Let C be the countable-cocountable σ-algebra on X × Y.Is C = σ ( A × B )? ⊆ is obvious, but I am not sure about the other direction. I know that in general the two do not have to be equal, so I am trying to find a counterexample i.e find an element in C which is not in σ ( A × B ). Perhaps AC is involved, which makes it tricky? I am not sure.Also, is it true that σ ( A × B ) = A × B ? This may help simplify the question.

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Let   X  =  Y uncountable, and let   A  =  B be the   σ algebra of countable-cocountable subsets of   X,  Y respectively. Let   C be the countable-cocountable   σ-algebra on   X  ×  Y.Is   C  =  σ  (  A  ×  B  )?  ⊆ is obvious, but I am not sure about the other direction. I know that in general the two do not have to be equal, so I am trying to find a counterexample i.e find an element in C which is not in   σ  (  A  ×  B  ). Perhaps AC is involved, which makes it tricky? I am not sure.Also, is it true that   σ  (  A  ×  B  )  =  A  ×  B  ? This may help simplify the question.

upornompe · Accepted Answer

As you said it is obvious that   C  ⊆  σ  (  A  ×  B  ).However the inclusion in the other direction is not true. Here is a simple counter-example:Let   X  =  Y  =  [  0  ,  1  ] and let A=B be the   σ algebra of countable-cocountable subsets of X,Y respectively. Let C be the countable-cocountable   σ-algebra on   X  ×  Y. Let   S  =  {      1    2    }  ×  [  0  ,  1  ]. It is immediate that   S  ∈  σ  (  A  ×  B  ), but   S  ∉  C.Remark: The example above is actually a general example. In fact:Let   X  =  Y uncountable, and let   A  =  B be the   σ algebra of countable-cocountable subsets of X,Y respectively. Let C be the countable-cocountable   σ-algebra on   X  ×  Y.Let us choose   a  ∈  X. Let   S  =  {  a  }  ×  Y. It is immediate that   S  ∈  σ  (  A  ×  B  ), but   S  ∉  C.

preityk7t · Accepted Answer

Also we have a counterexample with general   X, guess i`m right

Let X = Y uncountable, and let A = B be the σ algebra

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