Let $X$ be a noetherian topological space.
Prove the following statements:
(a) If $F \subset X$ is closed, then there exist $n \in \mathbb N$ and irreducible closed subsets $F_1,\ldots, F_n \subset X$ such that $F=F_1 \cup \cdots \cup F_n$
(b) If $n,m \in \mathbb N$ and $F_1,\ldots, F_n,{F_1}',\ldots,{F_m}' \subset X$ are irreducible closed subsets such that
- $F_i \not \subset F_j$ if $i,j \in \{1,\ldots,n\}$ and $i \neq j$
- ${F_i}' \not \subset {F_j}'$ if $i,j \in \{1,\ldots,m\}$ and $ I \neq j$
- $F_1 \cup \cdots \cup F_n={F_1}' \cup \cdots \cup {F_m}'$,
then $n=m$ and there is a permutation $\sigma \in S_n$ such that ${F_i}'=F_{\sigma(i)}$
I don't know how to show these two statements so I would appreciate any hints to have an idea of what should I do.
Hint: For (a), suppose that $F= G_1\cup G_2$ for $G_1$ and $G_2$ proper closed sets. If $G_1$ and $G_2$ are irreducible, we are done. If not, WLOG assume $G_1$ is reducible. Write $G_1=H_1 \cup H_2$ for $H_1$ and $H_2$ proper closed subsets. Continue in this way, expressing a reducible closed set from the previous step as a union of proper closed sets, if a reducible set exists. Do you see how the noetherian hypothesis implies that this process terminates? Try to formalize this argument.
For (b), the key is to use irreducibility to show that for each $i$, $F_i \subset F'_j$ for some $j$. The same argument shows $F'_j \subset F_k$ for some $k$, so
$$F_i \subset F_j' \subset F_k.$$
Use your hypotheses to conclude $F_i=F_k$, and so $F_i = F_j'$. You can then argue that $n=m$.