Let me begin with the definition.

Suppose $u$ is a free ultrafilter on $\omega$.

Theorem. If $(r_n)$ is a bounded sequence of real numbers, then there exists a unique $l\in\mathbb R$ such that $$\{n\in\omega : |r_n-l|<\epsilon\}\in u$$ for all $\epsilon>0$. We write $l=\lim _u r_n$. $\blacksquare$

Let $$\mathbb R _\infty=\{(x_n)\in\mathbb R ^\omega:(\exists c\in\mathbb R)(\forall n\in\omega)(\frac{1}{n}|x_n|\leq c)\}.$$

Define an equivalence relation on $\mathbb R _\infty$ by $$(x_n)\sim(y_n)\iff \lim _u \frac{1}{n}|x_n-y_n|=0.$$

Let Con$_u \mathbb R=\{(x_n)/\sim :(x_n)\in \mathbb R _\infty\}$.

Theorem. Con$_u \mathbb R$ is a complete metric space with metric $$d((x_n)/\sim,(y_n)/\sim )=\lim _u \frac{1}{n}|x_n-y_n|. \blacksquare $$


1) Is there any nice geometric way to visualize Con$_u \mathbb R$? What does this space "look like"?

2) I believe Con$_u \mathbb R$ is a quotient of a subset of the ultrapower $\mathbb R ^\omega /u$. Does the quotient topology on Con$_u \mathbb R$ coincide with its metric topology?

Really any intuition you can give me about this space would be greatly appreciated!

  • $\begingroup$ In the definition of $\operatorname{Con}_u(\Bbb R)$ do you want to require, perhaps, that $(x_n)\in\Bbb R_\infty$ rather than in $\Bbb R^\omega$? $\endgroup$ – Asaf Karagila Jun 29 '14 at 21:22
  • $\begingroup$ yes, thank you I will fix it $\endgroup$ – Forever Mozart Jun 29 '14 at 21:23
  • $\begingroup$ Tom: it is obviously true that all asymptotic cones of R are isometric to R since rescaling R leads to an isometric metric space, which means you can skip the rescaling. Now appeal to local compactness of R. $\endgroup$ – Moishe Kohan Jun 29 '14 at 23:49
  • $\begingroup$ My answer shows that they are all isometric to $\mathbb{R}$. $\endgroup$ – Lee Mosher Jun 30 '14 at 0:03
  • $\begingroup$ Ok I think I did not ask the question I wanted. This space is rather trivial I guess. $\endgroup$ – Forever Mozart Jun 30 '14 at 0:06

It looks like, in fact it is isometric to, $\mathbb{R}$. Define a map $\mathbb{R}_\infty \to \mathbb{R}$ by the formula $$(x_n) \to \lim_u \frac{x_n}{n} $$ This limit exists, by the theorem you quote. This function is well-defined on $Con_u(\mathbb{R})$, it is injective, surjective, and an isometry.

As for intuition, think of watching $\mathbb{R}$ shrink away from you as time passes, multiplying the metric by $\frac{1}{n}$ at time $n$. You're allowed to peek at almost any moment of time, where "almost any" means at a set of times in the ultrafilter. What does it look like, what patterns do you see? Well, $\mathbb{R}$ multiplied by $\frac{1}{n}$ is isometric to $\mathbb{R}$. It never changes.

  • $\begingroup$ I think it is possible for two sequences to converge w.r.t. $d$ but not have the same $u$-limits $\lim _u (x_n/n)$. Does that sound right? Take for instance $(1,2,3,...),(2,4,6,...)\in \mathbb R _\infty$. The $u$-limit of the first would be $1$, the second would be $2$, yet in $d$ these sequences converge. $\endgroup$ – Forever Mozart Jun 29 '14 at 23:35
  • $\begingroup$ I'm not following you. By "two sequences", do you mean two elements of $\mathbb{R}_\infty$, or two sequences of points in the space $Con_u(\mathbb{R})$? $\endgroup$ – Lee Mosher Jun 29 '14 at 23:41
  • $\begingroup$ yes, two elements of $\mathbb R _\infty$ $\endgroup$ – Forever Mozart Jun 29 '14 at 23:41
  • $\begingroup$ Ah, I see. Yes, under the map I defined, $(1,2,3,…)$ goes to $1$ and $(2,4,6,…)$ goes to $2$. $\endgroup$ – Lee Mosher Jun 29 '14 at 23:42
  • $\begingroup$ Take a look at math.cornell.edu/~riley/Conferences/SL(n,Z)/…, for example, although it doesn't have much more detail than what I wrote. $\endgroup$ – Lee Mosher Jun 29 '14 at 23:48

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