1
$\begingroup$

A topological space $X$ is called Noetherian if closed subsets satisfy the descending chain condition, equivalently, the open subsets satisfy the ascending chain condition.

Let $A$ and $B$ be two topological space such that $B$ is Noetherian. If $f:A\rightarrow B$ is a continuous function, is there any condition $T$ on $f$ such that any of the following to be true:

1)if $f$ satisfies $T$, then $A$ is Noetherian.

2) if $A$ if Noetherian, then $f$ satisfies $T$.

3) $f$ satisfies $T$ if and only is $A$ if Noetherian.

Improve this question
$\endgroup$
1
  • $\begingroup$ A condition for (1) would be that $f$ is both open and injective. If $U_1\subseteq U_2\subseteq\ldots$ is an ascending chain of open subsets of $A$, then $f$ being open assures that $f[U_1]\subseteq f[U_2]\subseteq\ldots$ is an ascending chain of open subsets of $B$, which is Noetherian, whence the chain should stabilize: there is some $n$ such that $f[U_i]=f[U_n]$ for each $i\geq n$. Injectivity of $f$ assures now that $U_i=U_n$ for each $i\geq n$, i.e., the original ascending chain stabilizes. So $A$ is Noetherian. $\endgroup$
    – Bert Lindenhovius
    Commented Jun 3, 2017 at 0:10

0

Reset to default

You must log in to answer this question.