Let $\alpha$ be an infinite ordinal. Is there an order-preserving injection from $\langle\alpha\cdot\alpha+\omega,\in\rangle$ to $\langle\mathcal{P}(|\alpha|),\subsetneq\rangle$?
I tried to construct an argument for why such injection cannot exist for the simpler case of $\omega\cdot\omega\to\mathcal{P}(\omega)$, but as I failed doing that, I couldn't formalize a construction that works.
This is my attempt for a solution:
Lemma 1. If $A$ and $B$ are sets s.t. $|A|\le |B|$, there is an order-preserving injection $\langle \mathcal{P}(A),\subsetneq \rangle \to \langle \mathcal{P}(B),\subsetneq \rangle$.
Proof. Let $f:A\to B$ be injective. Define $h:\mathcal{P}(A)\to\mathcal{P}(B)$ by $h(X)=f[X]$ for all $X\subseteq A$. For $X,Y\in\mathcal{P}(A)$, it is clear that $f[X]\subsetneq f[Y]$ iff $X\subsetneq Y$. $~~~\blacksquare$
Lemma 2. (Without assuming choice!) For an infinite ordinal $\alpha$, we have $|\alpha|=|\alpha\cdot \alpha|=|\alpha\cdot \alpha + \omega|$.
Proof. It is known that for an infinite cardinal $\kappa$ we have $\kappa \cdot \kappa =\kappa =\kappa +\kappa$. Using the (un-recuresive) definitions of ordinal addition and multiplication as the following order types:
$$ \begin{align} \sigma +\tau &:= \operatorname{ort}(\sigma^\frown \tau)&:=\operatorname{ort}(\langle\sigma\sqcup\tau,<_{\text{lex}}\rangle), \\ \sigma \cdot \tau &:= \operatorname{ort}(\sigma\otimes\tau)&:=\operatorname{ort}(\langle\sigma\times\tau,<_{\text{lex}}\rangle), \end{align} $$
We deduce that $|\alpha\cdot\alpha|=|\alpha|\cdot|\alpha|$ and $|\alpha+\omega|=|\alpha|+|\omega|$. Therefore,
$$|\alpha|\le|\alpha+\omega|=|\alpha|+|\omega|\le|\alpha|+|\alpha|=|\alpha|,$$ so $|\alpha|=|\alpha+\omega|$. But $|\alpha\cdot\alpha|=|\alpha|\cdot|\alpha|=|\alpha|$, which finishes the proof. $~~~\blacksquare$
Proposition. For infinite ordinal $\alpha$, there exists an order-preserving injection from $\langle\alpha\cdot\alpha+\omega,\in\rangle$ to $\langle\mathcal{P}(|\alpha|),\subsetneq\rangle$.
Proof. By lemma 1, it's sufficient to find an order-preserving injection from $\alpha\cdot\alpha+\omega$ into $\langle\mathcal{P}(A),\subsetneq\rangle$ for any set $A$ with $|A|=|\alpha|$. From lemma 2, $\gamma=\alpha\cdot\alpha+\omega$ is fitting. From here it's easy - we need to define an order-preserving function $g:\gamma\to \mathcal{P}(\gamma)$, but the identity itself finishes the job, as $\subsetneq$ and $\in$ are exactly the same on an ordinal.$~~~\blacksquare$
Any notes? Is this proof OK?