I'm reading Goldblatt's Lectures on the Hyperreals, and he provides the following proof that $^*\mathcal{P}(A)\subseteq \mathcal{P}(^*A)$:
- Given sets $A,\mathcal{P}(A)\in\mathbb{U}$, the statement $\forall x\in\mathcal{P}(A):(\forall y\in x)(y\in A)$ is true if $\mathcal{P}(A)$ is the power set of $A$.
- Therefore $\forall x\in {^*}\mathcal{P}(A):(\forall y\in x)(y\in {^*}A)$ holds by transfer. So every element of $^*\mathcal{P}(A)$ is a subset of $^*A$.
If I understand correctly, transfer applies here because $\forall x \in X: Q(x)$ is equivalent to $\forall x : x\in X \rightarrow Q(x)$, so the quantification is first order.
But I don't understand why we can't reverse this proof. I think $x\subseteq X $ is equivalent to $\forall y (y\in x \rightarrow y\in X)$ so even though the we're talking about subsets, it's still first order. Then given $\mathcal{P}(A)$ and $A$, the formula $\forall x (x\subseteq A \rightarrow x\in \mathcal{P}(A))$ is true.
What fails when this is transferred? Because $\forall x (x\subseteq {^*}A \rightarrow x\in {^*}\mathcal{P}(A))$ would mean that $\mathcal{P}(^*A)\subseteq ^*\mathcal{P}(A)$ which is false.
I'm not sure where my mistake is. I think I have a misunderstanding of how transfer works, and which statements are first order. It seems like the transferred $\forall y (y\in x \rightarrow y\in A)$ is going to miss some subsets of $^*A$, but I don't know why.
The point is that transfer only applies to formulas in the $\in$-language. Your formula, as you acknowledge, is not a formula in the $\in$-language. That's why transfer doesn't apply. The simplest example is of course the subset $\mathbb N\subseteq {}^\ast\mathbb N$. It is not an internal subset and therefore not an element of ${}^\ast\mathcal P$.