1
$\begingroup$

I'm studying the topology in $\mathbb{R}$ of open and closed intervals, and

I have some doubts regarding why I can have a topology of open intervals (in which I can have the generalized union but only the intersection of a finite number of sets) but not a topology of closed intervals (in which I can have the generalized intersection but only the union of a finite number of sets)

I think this as a natural fact considering that an interval is open if the complementary set is closed and vice versa.

Can someone help me to understand why? it is a convenction or is there a reason?

$\endgroup$
  • $\begingroup$ It is not really clear what you are asking. One requirement is that arbitrary unions are also open. Any set can be written as the arbitrary union of closed sets (points) so that would mean all sets would have to be closed which would not be very useful. $\endgroup$ – copper.hat Jun 13 '18 at 14:40
  • $\begingroup$ it is a convention that you call the family of all open sets "topology". But if you prefer to describe the closed sets instead, that is fine (as long as you indicate you are talking about the closed sets). Then the topology would be the family of the corresponding open sets (the complements of the closed sets that you have defined). So you could have the topology of closed intervals, in a way, it is just the terminology, that always means the topology is the family of all open sets. $\endgroup$ – Mirko Jun 13 '18 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.