Is every $\sigma$-algebra of sets *abstractly* the Borel algebra of a topology on perhps some other set? 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:
Link
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.
 A: I believe the same example as in the the answer to the question you link should provide a counterexample. It is a particular case of a more general result, that I am stating down below.
Namely, set $X=\{0,1\}^{\mathbb R}$ endowed with the $\sigma$-algebra $\mathcal A$ generated by the elementary sets $\{x:x_t=\epsilon\}$ for $t\in\mathbb R$, $\epsilon\in\{0,1\}$. We call $\mathcal E$ the collection of these elementary sets.
We assume $\mathcal A$ is isomorphic to the Borel algebra $\mathcal B(Y)$ of some topological space $(Y,\mathcal T)$ (as posets), and we will arrive to a contradiction. Since the Borel poset of a topological space does not change upon taking the Kolmogorov quotient, we can assume $Y$ is $T_0$. We denote by $\phi:\mathcal A\to\mathcal B(Y)$ the isomorphism of posets.
Define $\omega\mathfrak U(\mathcal A)$ (ultrafilters stable under countable intersections) the collection of all $U\subset\mathcal A$ such that

*

*$\varnothing\notin U$;

*$U$ is stable under countable intersections;

*if $A\subset A'$ are elements of $\mathcal A$ and $A\in U$, then $A' \in U$;

*for all $A\in\mathcal A$, either $A\in U$ or $A^\complement\in U$.

There are natural maps from $X$ and $Y$ to $\omega\mathfrak U(\mathcal A)$, namely
$$\begin{aligned}x&\mapsto\{A\in \mathcal A:x\in A\}=:U_x,
&y&\mapsto\{A\in \mathcal A:y\in\phi(A)\}=:V_y.\end{aligned}$$
The fundamental observation is the following, which we shall prove later.

Lemma.

*

*The map $x\to U_x$ is a bijection.

*The map $y\to V_y$ is injective.

*The induced map $i:Y\to X$ is such that $\phi(A)=i^{-1}A$.


Since the preimage has much nicer properties than $\phi$, this gives a lot more information about how the latter behaves. For instance, it shows that if a (not necessarily countable) union of elements of $\mathcal A$ is itself in $\mathcal A$, then $\phi$ commutes with the union. It is also true for the inverse $\phi^{-1}$ over $\mathcal B(Y)$, and for arbitrary intersections.
We deduce that $\mathcal A$ must be induced by a topology; namely, setting $\mathcal S$ the collection of all $\phi^{-1}(B)$ for $B\in\mathcal T$, we see that

Corollary.
$\mathcal S$ is a topology over $X$, and it generates $\mathcal A$ as a $\sigma$-algebra.

But this is not possible, as discussed in this answer.
Proof of the corollary.
The fact that $\mathcal S$ is a topology comes from the fact that $\mathcal T$ is a topology and $\phi^{-1}$ preserves unions and intersections. The fact that it generates $\mathcal A$ can be seen by defining $\mathcal B'=\{ B\in\mathcal B(Y):\phi^{-1}B\in\sigma(\mathcal S)\}$. It is a $\sigma$-algebra (because $\phi^{-1}$ preserves unions and intersections) and it contains $\mathcal T$, so $\mathcal B(Y)\subset\mathcal B'$, and by definition it means $\mathcal A\subset\sigma(\mathcal S)\subset\mathcal A$.
It remains to prove the lemma.
Proof of the lemma: the map from $X$.
It should be clear that the map is well-defined, i.e. $U_x\in\omega\mathfrak U(\mathcal A)$. The fact that it is injective comes from the fact that $\mathcal E$ separates points,¹ and that such sets are in $\mathcal A$.
For the surjective part, write $\mathcal T(\mathcal E)$ for the topology generated by $\mathcal E$, which makes $X$ a compact Hausdorff space. Let $U$ be an element of $\omega\mathfrak U(\mathcal A)$. Note that by the finite intersection property, the collection of open sets $O\in(\mathcal A\setminus U)\cap\mathcal T(\mathcal E)$ cannot be an open cover. Choose some $x\in X$ not covered by this collection; we will show that $U=U_x$.
The fact that $U=U_x$ is equivalent to $U_x\subset U$ (by the third and fourth properties of elements of $\omega\mathfrak U(\mathcal A)$), which is equivalent to all elements $A$ of $\mathcal A$ satisfying the following property: “exactly one of $A$ or $A^\complement$ contains $x$, and the one that does belongs to $U$”. In other words, setting $\mathcal A'$ those elements, we want to have $\mathcal A'=\mathcal A$.
In fact it is not too difficult to see that $\mathcal A'$ is a $\sigma$-algebra, and it contains $\mathcal E$ because $x$ is not in the open cover discussed above. This concludes the proof that $U=U_x$, and with it that of the first part of the lemma.
Proof of the lemma: the map from $Y$.
Denote by $y\mapsto V_y$ the map under consideration. It is well-defined and injective for the same reason $x\mapsto U_x$ enjoys the same properties.
Proof of the lemma: the map $i$.
To be clear, this map sends $y\in Y$ to the only $x\in X$ such that $U_y=U_x$. It is clearly well-defined and injective by the first two points of the lemma. The fact that $\phi(A)$ is always the inverse image of $A$ under $i$ is a matter of unfolding the definitions, which I would personally write as a two-way inclusion.

In fact, using the exact same sketch of proof as above, one can show the following. Say that a poset is Borel-representable if it is order-isomorphic to the Borel algebra of some topological space (in particular, it will be a Boolean algebra). Recall that a Stone space is a totally separated compact topological space.
Theorem.
Let $X$ be a Stone space, $\mathcal E$ its algebra of clopen sets, and $\mathcal A$ the $\sigma$-algebra generated by $\mathcal E$. Then $\mathcal A$ is Borel-representable if and only if it is already the Borel algebra of a topology on $X$.
In the case of $X=\{0,1\}^{\mathbb R}$, the topology we considered was the product topology, and the algebra of clopen sets consisted of the closure of the elementary open sets under finite union and intersection.

¹ For every pair $x,x'\in X$ of distinct elements, there is some $E\in\mathcal E$ containing precisely one of the two.
