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We know that the Borel $\sigma$-algebra of the Cartesian product space (with the product topology) of two topological spaces is equal to the product of the Borel $\sigma$-algebras of the factor spaces. (The product $\sigma$-algebra can be defined via pullbacks of projection maps...)

When one upgrades the above statement to a product of a countable family of topological spaces, the analagous result, namely that the Borel $\sigma$-algebra is the the product Borel $\sigma$-algebra, is conditioned by the topological spaces being second countable. Why?

My question is this: how and why does second countability make its appearance when we upgrade the finite product to countable product? (Second countability means that the base is countable, not just locally...)

My difficulty is that I do not see how suddenly second countability is important when we pass from finite to countable products.

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up vote 3 down vote accepted

This is not even true for a product of two spaces: see this Math Overflow question. To rephrase, second countability can be important even for products of two topological spaces.

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So secound countability is needed even for two spaces? –  Gandhi Viswanathan Oct 25 '11 at 21:37
    
It suffices that both spaces are Hausdorff, and one of them has a countable base. Bogachev's book on measure theory has the proof. –  Byron Schmuland Oct 25 '11 at 21:39

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