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Given a topological space $(X,\cal T)$ a subcollection $\cal A$ of $\cal T$ is a base of open sets for $\cal T$ if any open set $O$ of $\cal T$ is union of any subcollection $\mathcal A_O$ of $\cal A$ so that it is usuasl to say that a collection $\mathcal S_O$ is a subbase of open sets if the collection $$ \mathcal F_O:=\Big\{Y\in \mathcal P(X):Y=\bigcap \mathcal A_n\,\text{where }\mathcal A_n\subseteq\mathcal A\,\text{is such that }|\mathcal A_n|=n\,\text{for }n\in\omega\Big\} $$ of finite intersection of $\mathcal S_O$ is a base of open sets.

However it is a well know result that a topology can be defined using closed sets with interchanging finite intersection with arbitrary intersection and arbitrary union with finite union so that a collection $\cal C$ of $\cal T$ is a base of closed sets for $\cal T$ if any closed set $C$ is intersection of a subcollection $\cal C_C$ of $\cal C$ (see here for details) so that I would like to know if it is possibile to define a subbase $\cal S$ of closed sets if the collection $$ \mathcal F_C:=\Big\{Y\in \mathcal P(X):Y=\bigcup \mathcal C_n\,\text{where }\mathcal C_n\subseteq\mathcal C\,\text{is such that }|\mathcal C_n|=n\,\text{for }n\in\omega\Big\} $$ of finite union of $\mathcal S_C$ is a base of closed sets: indeed, it is a well know resul that if $\mathcal A_O$ and $\mathcal C_C$ are base of open and closed sets then the collections $$ \mathcal A_O^*:=\{X\setminus A_O:A_O\in\mathcal A\}\quad\text{and}\quad\mathcal C^*_C:=\{X\setminus C_C:C_C\in\mathcal C_C\} $$ are a base of closed and open sets respectively so that I thought that by symmetry it is possibile to define a subbase of closed sets obtaining analogous results for open and closed subbases as the following observation shows.

So we observe that if $\mathcal F_O$ is a subbase of open sets then for any closed set $C$ the identity $$ X\setminus C=\bigcup_{i\in I}\Biggl(\bigcap_{h_i\in n_i}(F_O)_{h_i}\Biggl) $$ holds where $n_i\in\omega$ and $(F_O)_{h_i}\in\mathcal F_O$ for any $h_i\in n_i$: thus observing that $$ C=X\setminus\bigcup_{i\in I}\Biggl(\bigcap_{h_i\in n_i}(F_O)_{h_i}\Biggl)=\bigcap_{i\in I}\Biggl(X\setminus\bigcap_{h_i\in n_i}(F_O)_{h_i}\Bigg)=\bigcap_{i\in I}\Biggl(\bigcup_{h_i\in n_i}\Big(X\setminus(F_O)_{h_i}\Big)\Biggl) $$ we conclude that the collection $$ \mathcal F_O^*:=\{X\setminus F_O:F_O\in\mathcal F_O\} $$ is a subbase of closed sets. Moreover by analogous arguments it is possible to prove that if $\mathcal F_C$ is a subbase of closed set then the collection $$ \mathcal F_C^*:=\{X\setminus F_C:F_C\in\mathcal F_C\} $$ is a subbase for open sets.

So with respect this facts could it make sense define a subbase for closed sets? Could someone help me, please?

  • Closed set doesn't form a topology. – Sourav Ghosh Jul 03 '22 at 19:56
  • @LostinSpace Sorry, perhaps I use bad words: I wanted mean that if $\mathcal C$ is a collection of subsets of $X$ such that $$1.\quad\emptyset,X\in\mathcal C\2.\quad\mathcal C'\subseteq C\longrightarrow\bigcap\mathcal C'\in\mathcal C\ 3.\quad C_1,C_2\in\mathcal C\longrightarrow C_1\cap C_2\in\mathcal C$$ then the collection $$\mathcal A:={X\setminus C:C\in\mathcal C}$$ is a topology on $X$. – Antonio Maria Di Mauro Jul 03 '22 at 20:06
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    You want $C_1\cup C_2\in\cal C$ not $C_1\cap C2\in\cal C$ – FShrike Jul 03 '22 at 20:08
  • And as opposed to a basis where every open set contains a basic set, you want a basis where every closed set is contained in one of these basic sets, some caution required (I think) – FShrike Jul 03 '22 at 20:09
  • @Fshrike Oh, yeah: I had to wrote $C_1\cup C_2$ and not $C_1\cap C_2$ in the third point. However if you do not know the existence of closed base I can edit the question showing that a closed base generate an open base and vice versa. – Antonio Maria Di Mauro Jul 03 '22 at 20:12
  • @FShrike Hi, I edited the question: so what do you think about my last observation? – Antonio Maria Di Mauro Jul 04 '22 at 09:45
  • To be clear, which observation do you mean? Eg is the answer by hartkp satisfactory? – FShrike Jul 04 '22 at 11:11
  • I mean the observation about that a closed subbase determines an open subbase and viceversa. Moreover I surely appreciate Hartkp's answer and also surely I will appreciate any other answer that can add other informations. – Antonio Maria Di Mauro Jul 04 '22 at 13:23

1 Answers1

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There is no problem to define (sub)base for the closed sets, just use De Morgan's laws.

In a topological space $(X,\mathcal{O})$ let $\mathcal{F}$ denote the family of closed sets A subfamily $\mathcal{B}$ of $\mathcal{F}$ is a base for the closed sets if every member of $\mathcal{F}$ is the intersection of a subfamily of $\mathcal{B}$. A subfamily $\mathcal{S}$ of $\mathcal{F}$ is a subbase for the closed sets if the family $\mathcal{S}^+$ of finite unions of members of $\mathcal{S}$ is a base for the closed sets; in extreme cases you have to add $\emptyset$ and $X$ explicitly to $\mathcal{S}^+$.

An alternative definition/characterization: $\mathcal{S}$ is a subbase for the closed sets of $(X,\mathcal{O})$ if $\mathcal{O}$ is the smallest topology whose family of closed sets contains $\mathcal{S}$.

hartkp
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  • Okay, however I do not understand which are the extreme cases: could you indicate it? Perhaps the empty set case or the singleton case is an extreme case? – Antonio Maria Di Mauro Jul 04 '22 at 13:21
  • Moreover, is it true that a closed base is also a close subbase, right? – Antonio Maria Di Mauro Jul 04 '22 at 14:20
  • The family of singleton sets is a subbase for the co-finite topology on $\mathbb{N}$, say, but just taking finite unions and then intersections will not give you the whole set $\mathbb{N}$ as a closed set; some (very few) people don't like to use $\bigcup\emptyset$ to get $\emptyset$ in the collection. And, yes, a closed base is also a closed subbases. – hartkp Jul 04 '22 at 16:22
  • Okay, however effectively the "universe" set $X$ must be element of any closed base, right? – Antonio Maria Di Mauro Jul 05 '22 at 07:48
  • Yes, that's the only way you can get it as an intersection of a subfamily (unless you stipulate that $\bigcap\emptyset=X$, but some people don't like that). – hartkp Jul 05 '22 at 07:59
  • Okay, all clear now: question upvoted and approved. Regarding the position $$\bigcap\emptyset:=X$$ yesterday I posed a specific question about after reading this Brian M. Scott's answer: unfortunately, none had clarified the issue so could I ask your opinion? Anyway, thanks very much for your assistance!!! :-) – Antonio Maria Di Mauro Jul 05 '22 at 10:27
  • @AntonioMariaDiMauro I posted an answer, of sorts. – hartkp Jul 05 '22 at 19:48
  • @hrtkp I saw. Now it is all clear: answer upvoted and approved. Thanks very much for your assistance: you was very kind with me so that thanks yet. See you soon, goodbye. :-) – Antonio Maria Di Mauro Jul 06 '22 at 08:39