Given a topological space $(X , \mathcal{O})$, along with the "topology"/"family of open sets" $\mathcal{O}$ one gets (implicitly) the family of closed sets $\mathcal{C}$. Then $\mathcal{O} \cup \mathcal{C}$ is a subset of the powerset $\mathcal{P}(X)$ of $X$, and we can then ask questions about the "closure" of $\mathcal{O} \cup \mathcal{C}$ under both finite intersections and finite unions, call it e.g. $\mathcal{W}$ or something.
So for example, given $C_n$ closed and $O_n$ open, $((C_1 \cap (C_2 \cup O_3)) \cup (O_4 \cap C_5)) \cap O_6 \in \mathcal{W}$.
Question: Does this family of subsets ("$\mathcal{W}$") have a name? Is it equal to either:
- the locally closed subsets of $X$?
- the $\boldsymbol{\Delta}_2^0$ subsets of $X$ in the Borel Hierarchy, i.e. those sets that are both $G_{\delta}$ and $F_{\sigma}$?
Comments:
For $\boldsymbol{\Delta}_2^0$ subsets, I know we have to require $X$ to be Polish or something similar, because otherwise we're not guaranteed that open sets are $F_{\sigma}$ and closed sets are $G_{\delta}$. That being said, even in the Polish case my guess is that it is too broad of a class of subsets.
Because every locally closed set can be written as the intersection of one open set and one closed set, they seem like a possible candidate. However, this class of subsets might be too restrictive, because it is not clear to me whether they are closed under finite unions and intersections.
Motivation: Basically, e.g. in the case of $\mathbb{R}$, a family of subsets that includes all of $(a,b), [a,b], (a,b], [b,a)$ while still excluding $\mathbb{Q}$. I can explain the motivation further but it might be off-topic.
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Best Answer
Community Wiki
Apparently in algebraic geometry there is some notion called "constructible set" which is (at least for the Zariski topology) the set of all finite unions of locally closed sets. So in particular it would seem that locally closed sets are at least not closed under finite union.
According to Wikipedia, constructible sets are closed under both finite unions and finite intersections (at least because that should follow from being closed under finite unions and complements If a set is closed under unions and intersections is it closed under complements?) and apparently constructible sets are the Boolean algebra generated by the open sets and closed sets together, which seems like it is the same as "$\mathcal{W}$" above.
So the answer seems to be: constructible sets.
In particular it is trivial that every locally closed set is constructible, and in Polish spaces it seems that constructible sets are all $\boldsymbol{\Delta}_2^0$, i.e. $G_{\delta}$ and $F_{\sigma}$. Cf. this question: Boolean Closure and Borel sets
Sorry for answering my own question -- I didn't know the answer when I started writing this, but found it in the process of searching for an answer while writing.
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