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

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

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

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

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

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

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