This question has been bothering me for some while now and I still can't figure it out:

Prove that: (1) \aleph(X)\preceq \mathcal{P}(\mathcal{P}(\mathcal{P}(X))) without the choice-axiom.

The way \aleph(X) is defined in my courses is the following:
Let W(X) = \left\{\left\langle A,R\right\rangle: A\in \mathcal{P}(X),R\in \mathcal{P}(X\times X) \right\} (where R is a wellorder on A).

Then \aleph{(X)}=\left\{\left\langle x,\alpha\right\rangle<br />
:x\in W(X)\right\} ( \alpha is a ordinal and the ordertype of x)

I define:  \mathcal{I}_A^R  = \left\{\left\{x_0\right\},\left\{x_0,x_1\right\},\  cdots\right\} to be the set of all initial segments of A with respect to the wellorder R, wich is well-ordered by \subset

Now is f:\aleph(X)\to \mathcal{P}(\mathcal{P}(X)) defined by \left\langle x,\alpha \right\rangle = \left\langle  \left\langle A,R \right\rangle,\alpha \right\rangle\mapsto \mathcal{I}_A^R<br />
not an injection?

In this case the (1) would become trivial, so I expect it's not. But Why?
It seems evidently true that \mathcal{I}_A^{R_1} = \mathcal{I}_B^{R_2} \Leftrightarrow R_1= R_2 and A=B