I was asking this question Another way for partition of perfect set two days ago and there was a nice discussion over there but now I need to change the question by adding more conditions.
- Question: Prove in ZFC that for any perfect $P\subset\Bbb R$ there exists a family $\{P_{\alpha}\subset P\colon \alpha<\mathfrak c\}$, none of them are a Cantor set , of pairwise disjoint perfect subsets such that $$P=\bigcup_{\alpha<\mathfrak c} P_{\alpha}.$$ I know it is possible under CH, but quite sure in ZFC
Any help will be appreciated greatly
@bof gave an answer in one of his nice comments but I would like to make it clear by using the well-known fact:
Fact$1$. Every family $\mathcal U$ of pairwise disjoint of open subsets of $\Bbb R$ is at most countable.
By way of contradiction, let $P\subset\Bbb R$ be perfect set and $\{P_\alpha\subset P\colon \alpha<\mathfrak c\}$ be a family of pairwise disjoint perfect sets that are NOT Cantor set. This means that $P_{\alpha}^{\circ}\neq\emptyset$ for every $\alpha<\mathfrak c$, where $\circ$ denotes the interior set. Then, we have $$\{V_\alpha\subset P\colon \alpha<\mathfrak c\},$$ a family of pairwise disjoint open sets which is a contradiction with the Fact $1$. This shows use we can not have a family of continuum many of pairwise disjoint perfect subsets that are not Cantor set.