A topological space is called Noetherian if every non-empty set of closed subsets of $X$, ordered by inclusion, has a minimal element. Clearly, every subspace of a Noetherian space is Noetherian. Now let $Y$ be a subspace of a topological space $X$, I am looking for some condition on $Y$ such that if $Y$ is a Noetherian space as a subspace of $X$, then we can deduce that $X$ is also Noetherian?

  • 1
    $\begingroup$ If by "a condition on $Y$" you mean a condition on the homeomorphism type of $Y$ (ie without mentioning $X$), clearly no such condition exists. In my opinion, this question is too vague as stated. Can you be more specific? $\endgroup$ – HJRW Apr 24 '17 at 8:28
  • 4
    $\begingroup$ If $Y$ is noetherian, and if $Y$ is everywhere dense in $X$ (its intersection with any closed subset is dense in that subset), then $X$ is noetherian. For example, the set of closed points of a spectral space is an everywhere dense subset. $\endgroup$ – js21 Apr 24 '17 at 8:51

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.