It is well known that we have many different definitions of noetherianity for rings. Namely, given a ring $R$, the following are equivalent:
1) every ideal of $R$ is finitely generated.
2) $R$ satisfies a.c.c. (ascending chain condition) on ideals.
3) every non-empty set of ideals of $R$ has a maximal element.
I would like to state something similar for topological spaces. First a definiton: we call a closed subset of a topological space irreducible if it cannot be decomposed in the union of two proper closed subsets. Now i want to prove:
Let $X$ be a topological space. Then the following are equivalent:
1') every closed subset $Y$ of $X$ has a decomposition $$Y=Y_1\cup\ldots\cup Y_r$$ where $Y_j$ is irreducible and $Y_j\nsubseteq Y_m$ for $j\neq m$.
2') $X$ satisfies d.c.c. (descending chain condition) on closed subsets.
3') every non-empty set of closed subsets of $X$ has a minimal element.
I have a proof for $2')\rightarrow 3')$ and for $3')\rightarrow 1')$, but not for $1')\rightarrow 2')$, maybe because it is false? Can you help me?
EDIT: condition $Y_j\nsubseteq Y_m$ for $j\neq m$ in 1') is needed only for unicity of decomposition, but i'm non intersted in, so we can skip it. I would like 1') to be "similar" to 1) in the sense that 1) tells every ideal is finitely generated, and 1') that every closed subset has finitely many irreducible components....