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Is every sigma-algebra the Borel algebra of a topology?

inspires the present question which asks for less.

Question: Given a $\sigma$-algebra ${\cal A}$ on a set $X$, does there exist a topology ${\cal T}$ on perhaps some other set $Y$ such that ${\cal A}$ is isomorphic to the Borel sets determined by ${\cal T}$?

Examples contained in the answers to the quoted question indicate an answer of "not necessarily" if one also requires $X=Y$. I may be wrong, but it seems to me that a negative answer here (if appropriate) will require a new idea.

I've changed the title of my question to account for Gerald Edgar's comment. One could still ask to represent any abstract $\sigma$-algebra as a Borel field, but this isn't possible, as noted by Loomis here: http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.bams/1183510979 That said, the theorem Loomis proves indeed realizes abstract $\sigma$-algebras as Borel fields modulo a $\sigma$-ideal. I don't believe this settles my intended question though.

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What is an isomorphism of $\sigma$-algebras? An isomorphism of Boolean algebras that preserves countable sups? –  Stefan Geschke Oct 14 '12 at 8:33
An isomorphism $f$ between any two posets obviously preserves sups (i.e. $I$ has a sup iff $f(I)$ has a sup in which case $f(\sup(I))=\sup(f(I))$. So an isomorphism between $\sigma$-algebras just means an isomorphism of the underlying Boolean algebras. –  YCor Oct 14 '12 at 9:24
Plus you added the problem of starting with an abstract sigma-algebra, and realizing it as a sigma-algebra of sets: where the countable sups correspond to countable unions. –  Gerald Edgar Oct 14 '12 at 12:19

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