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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?

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    $\begingroup$ 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
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    $\begingroup$ 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

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