Let $G$ be a group (or even a set for our purposes here) and consider functions from $G$ to $\mathbb{R}$. Now after choosing a non-principal ultrafilter of $\mathbb{N}$, we can construct ultrapowers $^*G$ and $^*\mathbb{R}$. The usual metric on $\mathbb{R}$ extends to an ultrametric on $^*\mathbb{R}$

Suppose we have the space $C=\{ f:^*G \to ^*\mathbb{R}\}$. Now consider subspaces $$C_1 = \{ f \in C: \|f\| \text{ is bounded}\}$$ $$C_0 = \{f \in C: \|f\| \text{ is infinitesimal} \}$$ Here $\|f\|$ is the infinity norm on the function, or the supremum of $f$ over all inputs, and is now an element of $^*\mathbb{R}$.

Note that $C_0 \subset C_1$. Consider the quotient $$C_1/C_0$$ which is the space of equivalence classes of bounded functions modulo infinitesimals.

I have a couple of questions:

Can $C_1/C_0$ simply be treated as the space of functions $$\{g: ^*G \to \mathbb{R}\}$$ since $^*\mathbb{R}_{bounded}/^*\mathbb{R}_{infinitesimal} \equiv \mathbb{R}$?

Can the correspondence be given as follows: for $f \in C_1$, define $\tilde{f} \in C_1/C_0$ by $$\tilde{f}(g)=st(f(g))$$, where $st$ is the standard part function which takes a finite hyperreal and returns its closest real number?

Now suppose we further restrict ourselves to functions that are internal, or in other words, themselves arise as ultraproducts of functions from $G$ to $\mathbb{R}$. Now we define $\tilde{C}$, $\tilde{C}_1$ and $\tilde{C}_2$ the same way $$\tilde{C}=\{ f:^*G \to ^*\mathbb{R} \text{ is internal}\}$$ $$\tilde{C}_1 = \{ f \in \tilde{C}: \|f\| \text{ is bounded}\}$$ $$\tilde{C}_0 = \{f \in \tilde{C}: \|f\| \text{ is infinitesimal} \}$$ Again, consider $\tilde{C}_1/\tilde{C}_0$.

Assuming that the elements of $\tilde{C}_1/\tilde{C}_0$ can also be thought of as functions from $^*G$ to $\mathbb{R}$, is there any way in which the fact that they are internal reveals itself here?

  • $\begingroup$ Eh, I don't think the supermum exists. It certainly doesn't transfer, being a second order property. $\endgroup$ – PyRulez Feb 13 at 14:43

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