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(Both finite and infinite) product of topological spaces are often expressed through open sets.

Can it be expressed instead in terms of Kuratowski closure axioms (directly in terms of Kuratowski axioms, not through the isomorphism between the structure defined by Kuratowski closure axioms and and sets of open or closed sets)?

porton
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  • I would imagine the answer is yes, the wikipedia article even shows how the Kuratowski closure axioms define a topology in terms of its closed sets. Any theorem about open sets can be stated instead in terms of closed sets (due to the duality between open and closed sets). – Justin Benfield Oct 25 '16 at 13:54
  • I'm sure someone who can formulate the closure operator categorically can answer this succinctly! – rschwieb Oct 25 '16 at 14:35
  • It turns out a difficult problem (even if it has any solution). I tried to modify my definition of subatomic product (see my research book http://www.mathematics21.org/algebraic-general-topology.html), but this way fails – porton Oct 25 '16 at 15:14

1 Answers1

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About a month ago I was one of organizers of a conference dedicated to the 120-th anniversary of Kuratowski. There were many talks devoted to his contribution in mathematics, but no one related to the question you asked. I am afraid that a direct application of the closure operator to product topology definition is problematic, because this topology is defined by its base of open sets and (as far as I know) there is no direct counterpart of a base for closed set, only dual, that is via complements.

PS. You also may search about your question in the topology book by Kuratowski.

Alex Ravsky
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  • Is there any way, direct or indirect, to define a product of Čech closure spaces? – bof Oct 25 '16 at 20:22
  • @bof Saying the truth, I’m not acquainted with Čech closure spaces, so I googled “product Čech closure spaces”. It seems that it is defined. For instance, maybe a definition at p.3 of the paper “Proper and admissible topologies in closure spaces” by Mila Mrševič is the required one. – Alex Ravsky Oct 28 '16 at 17:27
  • That is not quite true. You can define a subbasis of closed sets in complete duality to the definition of a subbasis of open sets. But that doesn't help much with the closure operator, I don't think. – tomasz Apr 23 '18 at 19:40