Ultraproduct of a metric space

I do not know for sure, but here are two possible interpretations. Let $(X,d)$ be a metric space, $\omega$ a nonprincipal ultrafilter on ${\mathbb N}$. The standard definition of the ultraproduct $$ X^*:= \prod_{n\in {\mathbb N}} X/\omega $$ of $X$ as a set is the quotient of $$ \prod_{n\in {\mathbb N}} X $$ by the following equivalence relation: $(x_n)\sim (y_n)$ iff these two sequences are equal on a subset contained in $\omega$. Now, you need to define a metric on this ultraproduct. The most natural thing to do is to define a metric with values in the field of nonstandard real numbers ${}^*{\mathbb R}$: $$ d^*((x_n), (y_n))= [d(x_n, y_n)] $$ where $[t_n]$ is the nonstandard real number represented by the sequence $(t_n)$ of real numbers. For such a nonstandard metric one can define all the usual concepts like completeness, geodesic property, etc.; it is easy to see that the ultraproduct $(X^*, d^*)$ is a geodesic metric space in this sense. Suppose, however, that you want to have a traditional metric, taking values in real numbers. You need a projection from ${}^*{\mathbb R}_+$ to $[0, \infty]$ and this is what the ultralimit of sequences accomplishes: $$ lim_{\omega}: {}^*{\mathbb R}_+ \to [0, \infty]. $$ Then the only natural thing I can think of is the ultralimit of the constant sequence $(X, d, x)$ of pointed metric spaces as the ultraproduct of $(X,d)$: It picks up a certain subset of $(X^*, d^*)$ and takes its further quotient, using the map $lim_\omega$.

Then the results that you quoted show that such ultraproduct is again complete and geodesic.

Hope it helps. I will add the tags "nonstandard analysis", "model-theory" and "logic" to your question, maybe logicians on this site can provide further insight.

If you have a sequence of paths $\gamma_i$ (between $p$ and $q$) of length converging to the infimum then the equivalence class of the sequence $\langle \gamma_i:i\in\mathbb N\rangle$ will give a minimizing path in the ultrapower.