Reconciling two different definitions of constructible sets

This question is really about sets and topology, but it is motivated from commutative algebra, hence the tag.

Setup: Let $X$ be a set and let $\{U_\lambda\}_{\lambda\in\Lambda}\subset 2^X$ be a family of subsets of $X$ ($\Lambda$ is just an index set for the family) closed under finite intersection. Then the $U_\lambda$'s form the base of a topology; call it $\mathscr{T}$.

Let $\mathscr{F}\subset 2^X$ be the smallest family of subsets of $X$ containing $\mathscr{T}$ and closed under finite intersection and complementation.

Meanwhile, let $\mathscr{G}\subset 2^X$ be the coarsest topology in which every $U_\lambda$ is clopen.

It seems to me, though I haven't written down the proof to my satisfaction yet, that if it happens that $(X,\mathscr{T})$ is a noetherian space, then $\mathscr{F}=\mathscr{G}$. However, it seems to me that in general, without the noetherian hypothesis, they should not be equal and neither can be guaranteed to contain the other. E.g. it seems to me that $\mathscr{F}$ needn't be a topology, and that $\mathscr{G}$ needn't be closed under complementation. Also, in principle, while $\mathscr{F}$ clearly depends only on $\mathscr{T}$, $\mathscr{G}$ might actually depend on the base $\{U_\lambda\}_{\lambda\in\Lambda}$ chosen for $\mathscr{T}$. But my attempts to give examples of all this haven't been successful so far.

So my questions are:

1. Is it true that $\mathscr{F}$ needn't contain $\mathscr{G}$? If so, what's an example? If $\mathscr{F}$ must contain $\mathscr{G}$, what's the proof?

2. Same question with the roles of $\mathscr{G}$ and $\mathscr{F}$ reversed.

3. Is it true that $\mathscr{G}$ may depend on the base $\{U_\lambda\}_{\lambda\in\Lambda}$ chosen for $\mathscr{T}$? If so, what's an example? If not, what's the proof that it is determined entirely by $\mathscr{T}$?

Context: $\mathscr{F}$ and $\mathscr{G}$ are two different definitions of the constructible sets given in Atiyah-MacDonald. ($\mathscr{F}$ is in exercises 20-23 of chapter 7; $\mathscr{G}$ is in exercises 27-30 of chapter 3.) It seems to me that in the case of the Zariski topology on the Spec of a noetherian ring, they will coincide, but not in general. I could be totally wrong; this is what I'm trying to probe here.

• Let $\mathcal{B} = \{U_{\lambda} : \lambda \in \Lambda \}$ be the usual clopen basis on $X = 2^{\omega}$. Then, $\mathcal{G} = \mathcal{T}$ is just the usual topology so $\mathcal{G}$ is smaller than $\mathcal{F}$ as the latter contains all closed sets. – hot_queen Jun 24 '15 at 19:08
• For the other non inclusion, Let $X = [0, 1]$, $\mathcal{B} = \{A \subseteq [0, 1] : \mu(A) = 1\}$. Then $\mathcal{G} = \mathcal{P}([0, 1])$ while $\mathcal{F} = \{A \subseteq [0, 1]: \mu(A) = 0$ or $1 \}$. So $\mathcal{G}$ is not contained in $\mathcal{F}$. – hot_queen Jun 24 '15 at 19:21
• @hot_queen - Thanks so much! If you expanded this pair of comments into an answer I would accept it! Also: I am unfamiliar with "the usual clopen basis / usual topology on $2^\omega$" - what does this refer to? – Ben Blum-Smith Jul 1 '15 at 17:58

Let $Fn(\omega, 2)$ be the set of all finite partial functions from $\omega$ to $2$. For $s \in Fn(\omega, 2)$, let $[s] = \{x \in 2^{\omega} : s \subseteq x\}$. Then $\mathcal{B} = \{[s] : s \in Fn(\omega, 2)\}$ is a basis for the product topology $\mathcal{T}$ on $2^{\omega}$ where $2$ has discrete topology. Note that each set in $\mathcal{B}$ is also closed. Now $\mathcal{G} = \mathcal{T}$ and hence $\mathcal{F}$ is not contained in $\mathcal{G}$ as there are non open closed sets.

Let $\mathcal{B} = \{A \subseteq [0, 1]: \mu(A) = 1\}$. Then $\mathcal{G} = \mathcal{P}([0, 1])$ because every singleton is in $\mathcal{G}$. Also $\mathcal{F} = \{A \subseteq [0, 1] : \mu(A) = 0$ or $1\}$. Hence $\mathcal{G}$ is not contained in $\mathcal{F}$.

To see that $\mathcal{G}$ depends on the basis, in the first example, change $\mathcal{B}$ to the family of all open sets in $2^{\omega}$. Then $\mathcal{G}$ is no longer $\mathcal{T}$.

Take $X = \mathbb{N}$ with the cofinite topology. Let us say the $U_\lambda$ are all complements of singletons (though I think what follows is true for any other choice of $\{U_\lambda\}$). Then if I am not mistaken, $\mathscr{F}$ is the set of all subsets that are either finite or cofinite, while $\mathscr{G}$ is the discrete topology.

Note that this space is actually Noetherian!!

(In fact $X = \operatorname{Spec}(\mathbb{Z})$ is also a counterexample. If we forget about the generic point, this is basically the same thing as $\mathbb{N}$ with the cofinite topology.)

• +1 Beautiful example, showing my assumption about the $\operatorname{Spec}$-of-a-Noetherian-ring case was wrong! – Ben Blum-Smith Feb 10 '17 at 12:32