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I know that if a topological structure $\Omega$ on a set $X$ is linearly ordered by inclusion such that $(X,\Omega)$ contains no strictly increasing infinite chain of open sets has exactly one base, where by definition a base $B$ is a collection of nonempty open sets such that any open set in $\Omega$ is a union of sets in $B$.

My question is, 1) what about the converse: if a topological space has exactly one base, does it follow that its topological structure is linearly ordered by inclusion such that it contains no strictly increasing infinite chain of open sets?

2) Altering the definition of base and defining it to be a collection of nonempty open sets excluding the whole space, what would be the case?

This question is a bit different from What topological structures have exactly one base? so do not refer me to this link.

  • If a topological space has only one base it must be the entire topology. This means that no open subset could be written as an union of any other open subsets – Yanko Sep 20 '18 at 20:31
  • @Yanko I know, what about the question of whether it contains no increasing infinite chain of open sets or not –  Sep 20 '18 at 20:33
  • Consider $\mathbb{N}$ with the open sets being intervals starting at 0. It's not hard to see that this is a topology and clearly it has an increasing infinite chain of open sets. Also no open set can be written as the union of other open sets so it must have exactly one base. – Yanko Sep 20 '18 at 20:35
  • @Yanko your statement is wrong. I posted an answer –  Sep 20 '18 at 20:41
  • @Yanko: The collection ${[0,n] : n > 0}$ is a base. The collection ${[0,n] : n > 0} \cup {\mathbb{N}}$ is another. – Nate Eldredge Sep 20 '18 at 20:50
  • @NateEldredge You're right, indeed. – Yanko Sep 20 '18 at 20:52
  • @Yanko so your example is not related to my question –  Sep 20 '18 at 20:52
  • @NateEldredge I have defined a base not to contain the empty set –  Sep 20 '18 at 20:53
  • Sorry, I overlooked that. – Nate Eldredge Sep 20 '18 at 20:55
  • Note that it may contain arbitrary long (finite) increasing sequences. For example if you take $\mathbb{N}\cup{\infty}$ with open sets all intervals which ends with infinity – Yanko Sep 20 '18 at 20:58
  • Your example works for question 2 @Yanko –  Sep 20 '18 at 21:15

1 Answers1

1

As to question 1, the answer is positive: It cannot contain any strictly increasing infinite chain of open sets as (an amendment of) it would yeild an open set other than those appearing in the chain.

As to question 2, the answer is “not necessarily”.

Yanko
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