Let $X$ be a set and $\mathcal{T}$ the smallest/weakest/coarsest topology containing some family $\left\lbrace C_i\right\rbrace_{i\in I} $ of sets declared to be closed.

  • This is well-defined since the set of topologies is a lattice or just by taking finite unions of intersections of $\left\lbrace C_i\right\rbrace_{i\in I} $. So that one can now define the interior of these sets. Is that a subbase for $\mathcal{T}$?

The wikipedia article about Spacetime topology says that what physicists call Alexandrov topology is called interval topology in mathematics. And indeed it does look like it but in the article (and I also checked in books), one defines the open sets of the topology while in mathematics one defines the closed sets. The natural question is if these coincide, hence the previous question.

Maybe that is the same question as the equality of the specialization topology defined with a partial order, and that defined with a strict order?


Remarks:

  1. By morgan's law (complementary of union and intersections, holds in infinite case) $\left\{ X \backslash C_i\right\}_{i\in I}$ is a subbase of the topology
  2. If open sets were the interior of their closure then $\left\{ C^0\right\}_{i\in I}$ ( ${}^0$ denotes the interior) would be another subbase: the closure can be written as an arbitrary intersection of finite unions of $\left\{ C\right\}_{i\in I}$ and the interior of it is contained in the intersection of unions of $\left\{ C^0\right\}_{i\in I}$ (if I'm not mistaken)
  3. I stumbled over that notion of regular and semi regular space: my question can be reformulated as "Is the interval topology always semiregular?"

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.