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This question arises from the excellent question posed on math.SE by Salvo Tringali, namely, Correspondence between Borel algebras and topology. Since the question was not answered there after some time, I am bringing it up here on mathoverflow in the hopes that it may find an answer here.

For any topological space, one may consider the Borel sets of the space, the $\sigma$-algebra generated by the open sets of that topology. The question is whether every $\sigma$-algebra arises in this way.

Question. Is every $\sigma$-algebra the Borel algebra of a topology?

In other words, does every $\sigma$-algebra $\Sigma$ on a set $X$ contain a topology $\tau$ on $X$ such that $\Sigma$ is the $\sigma$ algebra generated by the sets in $\tau$?

Some candidate counterexamples were proposed on the math.SE question, but ultimately shown not to be counterexamples. For example, my answer there shows that the collection of Lebesgue measurable sets, which seemed at first as though it might be a counterexample, is nevertheless the Borel algebra of the topology consisting of sets of the form $O-N$, where $O$ is open in the usual topology and $N$ is measure zero. A proposed counterexample of Gerald Edgar's there, however, remains unresolved. And I'm not clear on the status of a related proposed counterexample of George Lowther's.

Meanwhile, allow me to propose here a few additional candidate counterexamples:

• Consider the collection $\Sigma_0$ of eventually periodic subsets of $\omega_1$. A set $S\subset\omega_1$ is eventually periodic if above some countable ordinal $\gamma$, there is a countable length pattern which is simply repeated up to $\omega_1$ to form $S$. This is a $\sigma$-algebra, since it is closed under complements and countable intersections (one may find a common period among countably many eventually periodic sets by intersecting the club sets consisting of starting points of the repeated pattern).

• Consider the collection $\Sigma_1$ of eventually-agreeing subsets of the disjoint union $\omega_1\sqcup\omega_1$ of two copies of $\omega_1$. That is, sets $S\subset \omega_1\sqcup\omega_1$, such that except for countably many exceptions, $S$ looks the same on the first copy as it does on the second. Another way to say it is that the symmetric difference of $S$ on the first copy with $S$ on the second copy is bounded. This is a $\sigma$-algebra, since it is closed under complement and also under countable intersection, as the countably many exceptional sets will union up to a countable set.

Please enlighten me by showing either that these are not actually counterexamples or that they are, or by giving another counterexample or a proof that there is no counterexample.

If the answer to the question should prove to be affirmative, but only via strange or unattractive topologies, then consider it to go without saying that we also want to know how good a topology can be found (Hausdorff, compact and so on) to generate the given $\sigma$-algebra.

This question arises from the excellent question posed on math.SE by Salvo Tringali, namely, Correspondence between Borel algebras and topology. Since the question was not answered there after some time, I am bringing it up here on mathoverflow in the hopes that it may find an answer here.

For any topological space, one may consider the Borel sets of the space, the $\sigma$-algebra generated by the open sets of that topology. The question is whether every $\sigma$-algebra arises in this way.

Question. Is every $\sigma$-algebra the Borel algebra of a topology?

In other words, does every $\sigma$-algebra $\Sigma$ on a set $X$ contain a topology $\tau$ on $X$ such that $\Sigma$ is the $\sigma$ algebra generated by the sets in $\tau$?

Some candidate counterexamples were proposed on the math.SE question, but ultimately shown not to be counterexamples. For example, my answer there shows that the collection of Lebesgue measurable sets, which seemed at first as though it might be a counterexample, is nevertheless the Borel algebra of the topology consisting of sets of the form $O-N$, where $O$ is open in the usual topology and $N$ is measure zero. A proposed counterexample of Gerald Edgar's there, however, remains unresolved.

Meanwhile, allow me to propose here a few additional candidate counterexamples:

• Consider the collection $\Sigma_0$ of eventually periodic subsets of $\omega_1$. A set $S\subset\omega_1$ is eventually periodic if above some countable ordinal $\gamma$, there is a countable length pattern which is simply repeated up to $\omega_1$ to form $S$. This is a $\sigma$-algebra, since it is closed under complements and countable intersections (one may find a common period among countably many eventually periodic sets by intersecting the club sets consisting of starting points of the repeated pattern).

• Consider the collection $\Sigma_1$ of eventually-agreeing subsets of the disjoint union $\omega_1\sqcup\omega_1$ of two copies of $\omega_1$. That is, sets $S\subset \omega_1\sqcup\omega_1$, such that except for countably many exceptions, $S$ looks the same on the first copy as it does on the second. Another way to say it is that the symmetric difference of $S$ on the first copy with $S$ on the second copy is bounded. This is a $\sigma$-algebra, since it is closed under complement and also under countable intersection, as the countably many exceptional sets will union up to a countable set.

Please enlighten me by showing either that these are not actually counterexamples or that they are, or by giving another counterexample or a proof that there is no counterexample.

If the answer to the question should prove to be affirmative, but only via strange or unattractive topologies, then consider it to go without saying that we also want to know how good a topology can be found (Hausdorff, compact and so on) to generate the given $\sigma$-algebra.

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

This question arises from the excellent question posed on math.SE by Salvo Tringali, namely, Correspondence between Borel algebras and topology. Since the question was not answered there after some time, I am bringing it up here on mathoverflow in the hopes that it may find an answer here.

For any topological space, one may consider the Borel sets of the space, the $\sigma$-algebra generated by the open sets of that topology. The question is whether every $\sigma$-algebra arises in this way.

Question. Is every $\sigma$-algebra the Borel algebra of a topology?

In other words, does every $\sigma$-algebra $\Sigma$ on a set $X$ contain a topology $\tau$ on $X$ such that $\Sigma$ is the $\sigma$ algebra generated by the sets in $\tau$?

Some candidate counterexamples were proposed on the math.SE question, but ultimately shown not to be counterexamples. For example, my answer there shows that the collection of Lebesgue measurable sets, which seemed at first as though it might be a counterexample, is nevertheless the Borel algebra of the topology consisting of sets of the form $O-N$, where $O$ is open in the usual topology and $N$ is measure zero. A proposed counterexample of Gerald Edgar's there, however, remains unresolved.

Meanwhile, allow me to propose here a few additional candidate counterexamples:

• Consider the collection $\Sigma_0$ of eventually periodic subsets of $\omega_1$. A set $S\subset\omega_1$ is eventually periodic if above some countable ordinal $\gamma$, there is a countable length pattern which is simply repeated up to $\omega_1$ to form $S$. This is a $\sigma$-algebra, since it is closed under complements and countable intersections (one may find a common period among countably many eventually periodic sets by intersecting the club sets consisting of starting points of the repeated pattern).

• Consider the collection $\Sigma_1$ of eventually-agreeing subsets of the disjoint union $\omega_1\sqcup\omega_1$ of two copies of $\omega_1$. That is, sets $S\subset \omega_1\sqcup\omega_1$, such that except for countably many exceptions, $S$ looks the same on the first copy as it does on the second. Another way to say it is that the symmetric difference of $S$ on the first copy with $S$ on the second copy is bounded. This is a $\sigma$-algebra, since it is closed under complement and also under countable intersection, as the countably many exceptional sets will union up to a countable set.

Please enlighten me by showing either that these are not actually counterexamples or that they are, or by giving another counterexample or a proof that there is no counterexample.