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Suppose that $(X,\tau)$ is a pair in which $X$ is a non-empty set and $\tau \subset 2^X$ with these conditions:

(1) $X,\emptyset \in \tau$,

(2) $\tau $ is closed under arbitrary unions, and

(3) $U_1 \cap U_2 \neq \emptyset$ implies $\operatorname{int}(U_1 \cap U_2 )\neq \emptyset$ for every $U_1 ,U_2 \in \tau$.

Is there a pair $(X,\tau)$ satisfying conditinons (1), (2) and (3) but it is not a topological space.

Here I define "$\operatorname{int}$" with respect to $\tau$ itself. $\operatorname{int}(A)$, $A\subset X$, is defined as the union of all sets $B\in \tau$ with $B\subset A$.

A pair $(X,\tau)$ satisfying (1) and (2) is called a generalized topology in the literature. So in other words, my question is that: Is there a generalized topological space (which is not a topological space) satisfying the condition (3)?

Xander Henderson
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M.Ramana
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    How is "int" defined? Some other topology? – Akiva Weinberger Nov 11 '22 at 13:30
  • @AkivaWeinberger A pair $(X,\tau )$ that satisfies condition (1) and (2) is called a generalized topological space in the literature. I just want to know about an example of a generalized topological space (which is not a topological space) satisfying condition (3). – M.Ramana Nov 11 '22 at 13:34
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    I understand, but that doesn't answer my question. What is "int"? If you are talking about interiors it means you already have a topology. – Akiva Weinberger Nov 11 '22 at 13:38
  • I suppose one possible example would be $X=\Bbb R^n$ and $\tau$ being the set of all open sets of volume at least $1$. Then it's closed under union but not intersection. EDIT: I don't think this satisfies 3 if you define int with respect to $\tau$ itself. – Akiva Weinberger Nov 11 '22 at 13:40
  • @AkivaWeinberger Sorry, I just got it. The definition of interior is defined in generalized topological spaces as well as topological spaces. Here we have a generalized topology (not topology). – M.Ramana Nov 11 '22 at 13:40
  • Or perhaps you could consider $X=\Bbb R^2$ and $\tau$ to be the set of open circles whose center lies on the x-axis, or any union of those circles. – Akiva Weinberger Nov 11 '22 at 13:43
  • @AkivaWeinberger Thank you so much for your good examples. I need time to check if these examples satisfy in the conditions. Thanks a lot. – M.Ramana Nov 11 '22 at 13:47
  • @AkivaWeinberger What do you mean by "open" sets in the first example? Here we call the elements of generalized topology "open" sets. Do you mean we consider $\tau$ to be set of all sets of volume at least 1 in $\mathbb{R}^n$ or the set of all open sets of volume at least 1 in $\mathbb{R}^n$ with the usual topology? – M.Ramana Nov 11 '22 at 13:52
  • I meant from the usual topology… but in that case, while their intersections are open and have nonempty interior in the usual topology, I don't think they do in the general topology. So that example probably doesn't work – Akiva Weinberger Nov 11 '22 at 13:54
  • @AkivaWeinberger Thanks so much for the comment. – M.Ramana Nov 11 '22 at 14:01
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    @M.Ramana can you write down the definition of int in the question? – hunter Nov 11 '22 at 14:06
  • @hunter Sure. I've just add it. – M.Ramana Nov 11 '22 at 14:29
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    your #3 would be simplified by just saying $U_1\cap U_2\not=\emptyset$ implies there exists nonempty $V\in\tau$ with $V\subseteq U_1\cap U_2$. – Steven Clontz Nov 11 '22 at 14:33

1 Answers1

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Take $X=\{0,1,2,3\}$ and $$ \tau=\{\emptyset,\{0\},\{0,1,2\},\{0,1,3\},\{0,1,2,3\}\} $$

Giulio R
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  • Thanks so much for your answer. Are generalized topological spaces satisfying condition (3) (like your example) always Baire spaces? – M.Ramana Nov 11 '22 at 16:17
  • Thanks for accepting the answer. I would say no, take a disjoint union of $(X,\tau)$ in my answer with any other topological space $(X',\tau')$ which is not Baire. – Giulio R Nov 11 '22 at 17:35