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Let $X$ be a set on which we wish to define a topology. Let $\mathcal{A}$ be a subset of the power set of $X$, i.e. elements of $\mathcal{A}$ are subsets of $X$. This question arose when studying the Zariski topology, so I'm approaching it by closed subsets, not open ones. Consider the intersection of all topologies for which elements of $\mathcal{A}$ are all closed : it is itself a topology. Equip $X$ with that topology and call $\mathcal{T}$ the set of $X$'s closed elements.

From this Wikipedia page I get the impression that it is easy to explicitly describe $\mathcal{T}$ and to do so I introduce the notation $\mathcal{P}_f(\mathcal{A})$ to denote the finite elements of the power set of $\mathcal{A}$. Then, my understanding is that :

$$\mathcal{T} = \lbrace{\bigcap_{\omega \in \Omega} \bigcup_{F \in \omega} F \quad | \quad \Omega \subset \mathcal{P}_f(\mathcal{A}) \quad \rbrace} $$

I tried to prove this, convinced that it would be straightforward but I'm actually really stuck. To make notation simpler, notice that a supposedly closed subset of $X$ is entirely determined by $\Omega$ so I defined :

$$ \Phi : \mathcal{P}(\mathcal{P}_f(\mathcal{A})) \rightarrow \mathcal{P}(X) $$ by $$ \Phi(\Omega)=\bigcap_{\omega \in \Omega} \bigcup_{F \in \omega} F$$

My claim now simply turns into $\quad\mathcal{T}=\text{Im}\Phi$.

It is clear that $\text{Im}\Phi \subset \mathcal{T}$ from the properties of closed subsets.

I need to prove that also $\mathcal{T} \subset \text{Im}\Phi$, for which it suffices to prove that $\text{Im}\Phi$ is a topology (or rather, the set of closed sets of a topology).

  • $X=\Phi(\emptyset)$ and $\emptyset=\Phi(\lbrace \emptyset \rbrace)$ so $X$ and $\emptyset$ are given as closed subsets by $\Phi$
  • The intersection property should be easy to prove
  • Let $I \subset \mathcal{P}(\mathcal{P}_f(\mathcal{A}))$ be finite. We need to prove that $\quad \bigcup_{\Omega \in I} \Phi(\Omega)\in \text{Im}\Phi$ that is exactly : $$ \bigcup_{\Omega \in I}\bigcap_{\omega \in \Omega} \bigcup_{F \in \omega}F $$ can be written in the form $$\bigcap_{\omega \in \Omega'} \bigcup_{F \in \omega} F $$ and I've drawn diagrams and all sorts but I can't swap that left-hand side cup with the cap.
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  • $\begingroup$ The Wikipedia page in english: Subbase. Generally, if you are in a wikipedia page in some language, you can find a link for that page in some other languages by choosing the language in the left pane. $\endgroup$ – amrsa Apr 28 '17 at 15:25
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We need to prove that $H=\bigcup_{\Omega\in I}\bigcap_{\omega\in\Omega}\bigcup_{F\in\omega}F$ can be written in the form $\bigcap_{\omega\in\Omega'}\bigcup_{F\in\omega}F$ for some $\Omega'\in\mathcal{P}(\mathcal{P}_f(\mathcal{A}))$.

Denote $\mathcal{F}_I$ the family of all functions $\varphi\colon I\to\mathcal{P}_f(\mathcal{A})$ satisfying the property $(\forall \Omega\in I)\,\varphi(\Omega)\in\Omega$. For every $\varphi\in\mathcal{F}_I$ denote $\omega_\varphi=\bigcup_{\Omega\in I}\varphi(\Omega)$. Since $I$ is finite and $\varphi(\Omega)$ is a finite subset of $\mathcal{A}$ for every $\Omega\in I$, we have $\omega_\varphi\in\mathcal{P}_f(\mathcal{A})$. Put $\Omega'=\{\omega_\varphi:\varphi\in\mathcal{F}_I\}$.

Now, we have $x\in X\setminus H$ iff $$x\in\bigcap_{\Omega\in I}\bigcup_{\omega\in\Omega}\bigcap_{F\in\omega}(X\setminus F)$$ iff $$(\forall\Omega\in I)(\exists\omega\in\Omega)\ x\in\bigcap_{F\in\omega}(X\setminus F)$$ iff $$(\exists\varphi\in\mathcal{F}_I)(\forall\Omega\in I)\ x\in\bigcap_{F\in\varphi(\Omega)}(X\setminus F)$$ iff $$x\in\bigcup_{\varphi\in\mathcal{F}_I}\bigcap_{\Omega\in I}\bigcap_{F\in\varphi(\Omega)}(X\setminus F),$$ hence $$H=\bigcap_{\varphi\in\mathcal{F}_I}\bigcup_{\Omega\in I}\bigcup_{F\in\varphi(\Omega)}F=\bigcap_{\omega\in\Omega'}\bigcup_{F\in\omega}F.$$

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  • $\begingroup$ You did really well on taking it exactly where I'd left it. Now since I could not have come up with this myself, I'd like to know if you think there's a simpler approach to this. Because it seems to me that anyone who has been introduced to general topology would want to know if a generated topology can be conveniently described (just as with $\sigma$-algebras, for which I was disadvised from trying). I'm surprised at how far this goes into power sets of power sets and all that $\endgroup$ – James Well Apr 28 '17 at 21:44
  • $\begingroup$ I think that steps I did are unavoidable. Rewriting an intersection of unions by a union of intersections using a family of functions is a standard trick, here we also had a good luck that it preserved the finiteness. I believe that you would have a better chance to come up with it, if you started with open sets instead of closed. And yes, good skills in set theory are needed for topology. $\endgroup$ – Peter Elias Apr 28 '17 at 22:50
  • $\begingroup$ Given what I'd done I never doubted that your steps were necessary to finish it off. I was questioning my own start actually, but with what you just said I guess not. $\endgroup$ – James Well Apr 28 '17 at 23:18
  • $\begingroup$ I think that what you did was optimal. $\endgroup$ – Peter Elias Apr 28 '17 at 23:36
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    $\begingroup$ I chose that direction because the step involving switching the order of quantification explains how each function $\varphi$ encodes the choice of $\omega\in\Omega$ in dependence on given $\Omega\in I$. It would be possible to go the other way but then I would have to reason negatively: ‘if there is $\Omega$ working for each $\omega\in\Omega$ then there is no way to make a choice of $\omega\in\Omega$ in dependence on a given $\Omega$ that is breaking each $\Omega$', or something like that. $\endgroup$ – Peter Elias Apr 30 '17 at 1:45
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In this answer I describe a proposition for the open sets.

This can of course easily be dualised, using de Morgan and complements.

If we have a collection of sets $\mathcal{C}$ that we want to be closed, the minimal collection of a topology (described by closed sets) is

$$\hat{\mathcal{C}} = \left(\mathcal{C}^{\cup ,< \infty}\right)^{\cap}$$

and the open sets are their complements. The notation is analogous to the one used in the linked answer. It boils down to the same thing as your $\mathcal{T}$.

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