# Does $\mathcal{P}(\alpha)$ have maximal well-founded subsets?

Let $\alpha$ denote an ordinal. Then clearly, the powerset $\mathcal{P}(\alpha)$ has well-founded subchains that are maximal with respect to this condition (of being well-founded subchains), since $\alpha$ is itself such a subchain. Nonetheless, $\alpha$ is not maximal with respect to the condition of being well-founded, since adjoining any element of $\mathcal{P}(\alpha) \setminus \alpha$ to $\alpha$ yields a larger subset that is still well-founded.

Assume $\alpha$ is an infinite ordinal; without loss of generality, assume it is the initial ordinal of some cardinal.

Question. Does $\mathcal{P}(\alpha)$ have well-founded subsets that are maximal with respect to this property (of being well-founded)?

I feel that this is impossible. Suppose for a contradiction that $\mathcal{X}$ is a maximal well-founded subset of $\mathcal{P}(\alpha)$. Then $\mathcal{X}$ must contain every finite subset of $\alpha$. Hence $\mathcal{X}$ must contain an infinite element, call it $A$. Assume for simplicity that $\alpha$ is countable, and hence that $A$ is countable. Let $f : \omega \rightarrow A$ denote a bijection.

Okay, where can we go from here?

You're right, there can be no maximal well founded subset of $\mathcal{P}(\alpha)$ if $\alpha$ is infinite.
Like you said, if we assume $\mathcal{X}$ is one, then it must contain an infinite subset $A$ of $\alpha$.
$\mathcal{X} \cup \{A - \{\min(A)\}\}$ is well founded:
If there were a strictly decreasing sequence $u$ in $\mathcal{X} \cup \{A - \{\min(A)\}\}$, since $u(n)$ couldn't be in $\mathcal{X}$ for every integer $n$, there would be a unique integer $k$ such that $u(k) = A - \{\min(A)\}$. But then you could skip $k$ and get a stricly deacrising sequence in $\mathcal{X}$.
By applying this, one can define a strictly decreasing sequence in $\mathcal{X}$: set $u(0) = A$ and $\forall n \in \mathbb{N},u(n+1) = u(n) - \{\min(u(n))\}$.