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On page 83-85 (here, on googlebooks), An Introduction to Nonstandard Real Analysis, Albert E. Hurd, Peter A. Loeb, two steps are given to construct a monomorphism $\ast : V(X) \to V({}^{\ast} X)$. The first step is to define map $e$, for all $a \in V(X)$, $a \mapsto [\overline {a}]$. $\overline {a}$ is a sequence of $\{a_i\}_{i \in I}$ in which $a_i = a$ for all $i \in I$. $[\overline {a}]$ is the class of all $\mathcal{U}$-almost equivalent sequences of $\overline {a}$,given an ultrafilter $\mathcal{U}$ on index set $I$.

I don't understand the reason why $e$ is not qualified as the monomorphism $\ast$ . In particular, why $A$, for a subset of $\Bbb R$, $e(A) = [\overline A]$ is not a subset of ${}^{\ast}\Bbb R$? Here ${}^{\ast}\Bbb R$ is the ultrapower $\prod_{\mathcal{U}}\Bbb R$, or $\prod\Bbb {R}/=_{\mathcal{U}}$.

Here's screenshot of the relevant part: enter image description here enter image description here

The requirement of a monomorphism is on page 79 (here): enter image description here

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The way you state their definition of $e$, it is a map only defined on elements of the superstructure rather than sets. Thus, $e$ as stated is not defined for a subset $A\subset \mathbb{R}$ but rather for the element $\{A\}\in\mathcal{P}(\mathbb{R})$. In particular, at this stage it is not yet clear what the elements of $^* A$ will be when $A$ is viewed as a subset. In other words, the map $e$ does not say very much about the relation $^*\!\in$ of the enlargement.

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  • $\begingroup$ Dear sir, thank you for your answer and editing. A minor problem, do you mean $A \in \mathcal{P}{(\Bbb R)}$? $\endgroup$ – Metta World Peace Jul 17 '13 at 19:08
  • $\begingroup$ By writing $\{A\}$ I just meant to emphasize that $A$ is not viewed as a subset $A\subset \mathbb{R}$ but rather as an element $\in \mathcal{P}(\mathbb{R})$. Ignore it if you prefer. $\endgroup$ – Mikhail Katz Jul 19 '13 at 7:55

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