Given a topological space $\langle X, T \rangle $, consider the Borel algebra $B$ generated by $T$. The question is, can you write any $b \in B$ using only finitely many open (or closed) sets using set operations $\cup$ and $\sim$ (and $\cap$) ?

For closed and open sets, this is obviously the case. So the characterization is only interesting for sets which are neither open nor closed.

For example, consider $\mathbb{R}$ with the standard topology and the following set: $$A = \{\frac{1}{n} : n \in \mathbb{N}\}.$$

$A$ is a Borel set and obviously, $A$ is neither open nor closed. But $A' = A \cup \{ 0 \}$ is closed and one can rewrite $A$ as $A' \setminus \{ 0 \}$. Since both sets are closed, $A$ is "finitely decomposable into open (or closed) sets."

If not, can you give a counter example?


Here is an elementary proof that $\mathbb{Q} \subset \mathbb{R}$ can not be constructed as you describe.

Consider $\mathbb{R}$ with the standard topology. We will prove that all sets $A$ generated in the way you describe have the following property (let's call it property P):

Property P: For any open interval $I$, either $I \cap A$ or $I - A$ contains an open interval.

This is clearly true of closed and open sets: Either $I \cap A$ or $I - A$ is open, thus either contains an open interval or is empty, but if $I \cap A$ is empty, $I - A = I$ and vice versa.

Thus we need only show this property is preserved under complements and finite unions. Complements are clear, since they merely switch $I \cap A$ and $I - A$.

So we now need only show that if $A_1,A_2 \subset R$ have property $P$, then so does $A_1 \cup A_2$. So fix an $I$. Clearly if $I \cap A_1$ or $I \cap A_2$ contains an open interval then we're done, so we can assume $I - A_1 $ and $I - A_2$ both contain open subintervals of $I$, call them $I_1$ and $I_2$, respectively.

By property $P$, either $I_1 \cap A_2$ or $I_1 - A_2$ contains an open interval $I'$. If the former, then $I' \subset I \cap (A_1 \cup A_2)$. If the latter, then $I'$ is disjoint from both $A_1$ and $A_2$, thus $I' \subset I - (A_1 \cup A_2)$. This shows that $A_1 \cup A_2$ has property P.

Since $\mathbb{Q}$ is dense in $\mathbb{R}$ with dense complement, it clearly does not satisfy property $P$. But it is Borel (as it is a countable union of closed points), and thus is a counterexample.

  • $\begingroup$ Thank you for your answer. My follow up question is, do you have any reference for the proposition that "finitely decomposable sets" have the property P? On the net or not doesn't matter... $\endgroup$ – user97110 Sep 28 '13 at 13:36
  • $\begingroup$ No, I just came up with this property as a way to prove the statement (though I'm sure others have come up with it before me). Is there something wrong with the proof I provided that finitely decomposable sets have this property? $\endgroup$ – MartianInvader Sep 29 '13 at 15:05
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    $\begingroup$ When you formulate property P, I think you meant $I \cap A$ and not $I \cup A$, right? This mix-up seem to continue in the rest of the proof as well, or I am missing something? $\endgroup$ – user97110 Sep 30 '13 at 12:39
  • $\begingroup$ Oh man, no wonder it was causing confusion! Thanks for figuring that out; I've corrected it to use $I \cap A$. $\endgroup$ – MartianInvader Sep 30 '13 at 17:22
  • $\begingroup$ Okay, one final (hopefully) question. I'm now looking into a "practical" characterisation of these finitely decomposable sets (FDS). How about the claim that sets x such that int(x) = int (Cl(x)) and Cl(x)=Cl(int(x)) are always FDS? $\endgroup$ – user97110 Oct 2 '13 at 9:48

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