What is known about the ultra-inverse limit?

Question

Given a nonprincipal ultrafilter $\mu$ on $\mathbb{N}$ and a sequence of groups $G_i$, one can define its ultraproduct as:

$$ ^*\prod_{i\in \mathbb{N}}G_i:=\{(x_i)_{i \in \mathbb{N}}| x_i\in G_i\}/\sim$$, where $(x_i)_i\sim (y_i)_i$, iff $x_i=y_i$ $\mu$-almost everywhere.

Suppose you are given also group homomorphisms $f_{i+1}:G_{i+1}\rightarrow G_i$, then one could also consider something like an ultra- inverse limit:

$$\{[x_i]_i|f_i(x_i)=x_{i-1} \quad \mu-\mbox{almost everywhere}\}$$

Has this been studied before? Is there a good source, where I could learn about this?

I've never seen this before, but it looks like a fascinating idea. Are you taking this as a subgroup of the Cartesian product, or a subgroup of the ultraproduct? One guess I would make is that it won't be possible to put a nice topology on this construction, in the way that e.g. profinite groups have a nice topology. — Colin Reid, Oct 21, 2010 at 12:39
I'll bet he means a subgroup of the ultraproduct; the square brackets in the definition surely indicate an equivalence class? — Harald Hanche-Olsen, Oct 21, 2010 at 13:41

Colin McQuillan · Accepted Answer · 2010-10-21 12:34:31Z

7

Either the even numbers or the odd numbers are $\mu$-large, so the condition degenerates to ultraproduct.

answered Oct 21, 2010 at 12:34

Colin McQuillan

8827 silver badges9 bronze badges

$\begingroup$ OK this explains why I didn't find anything about it $\endgroup$
– HenrikRüping
Oct 21, 2010 at 13:43
1

$\begingroup$ But does the question still make sense for more general ultra-inverse limits (on large index sets, etc.)? $\endgroup$
– Andrés Villaveces
Mar 16, 2012 at 4:37

Add a comment |

1 Answer 1