Non standard Algebraic Topology

Let *$\mathbb R$ a field of non-standard real numbers (or any real closed field) equipped with its natural generalized metric $d(x,y)=|x-y|$. Equip *$\mathbb R^2$ and *$\mathbb R^3$ with the $\ell^1$-(generalized)-metric.

Question: Does there exist an homeomorphism between *$\mathbb R^3$ and *$\mathbb R^2$?

Well, this is the simplest subquestion of the most general one

Question: Is there anybody developing non standard Algebraic Topology? If not, is there any particular reason?

Thanks in advance,

Valerio

-
Do I understand correctly that the topology on ${}^*\mathbb R$ is just the order topology, generated by the open intervals? Note that the set $I$ of infinitesimals is clopen. So your question is equivalent to asking whether $I^2$ and I^3\$ are homeomorphic. – Goldstern Jan 24 '12 at 20:18
Yes, it is just the order topology. – Valerio Capraro Jan 24 '12 at 20:32
See also this related question: mathoverflow.net/questions/10870/… Which topological spaces admit a nonstandard metric? – Joel David Hamkins Jan 25 '12 at 0:57
Valerio, it isn't really correct to speak of the nonstandard reals, since there are different nonstandard ordered fields each with the transfer principle (indeed, of any desired cardinality). Under the continuum hypothesis, then there is a unique saturated model of size continuum, and so one gets a little categoricity this way, but in general, one must deal with the possiblility of several different fields all looking like nonstandard reals. – Joel David Hamkins Jan 25 '12 at 1:03
OK, thank you very much! So I should say a field of nonstandard reals.. but it is now not clear to me if the answer of my question may depend on the particular chosen field of nonstandard reals. – Valerio Capraro Jan 25 '12 at 7:20

2 Answers

As mentioned in the comments, the actual topology on the non-standard extension can be quite nasty. This is illustrated for example in the first set of problems in these notes. A solution is to replace standard topological notions by definable analogues. Then things mostly work in an arbitrary o-minimal structure. This is also explained in the above notes.

More specifically on algebraic topology in the o-minimal settings, there are several papers by Berarducci and by Edmundo

-
Moshe, many thanks for the references. – Valerio Capraro Jan 24 '12 at 23:01
Good references; see also, earlier, Delfs and Knebush in the semialgebraic case. We probably want to insist though on the fact that in placing ourselves in the o-minimal setting, the homeomorphisms that we are willing to consider are tremendously restricted compared to what was Valerio's original question. On the other hand, this is a setting that we actually want to consider, i.e. where developing topological methods actually led to some useful results. – Thierry Zell Jan 26 '12 at 0:25
Not sure about tremendously; it is true formally speaking, of course. But many natural expansions of the reals have been shown to be o-minimal, e.g., the exponential and all restricted analytic functions. So a substantial amount of functions and spaces that come up in nature are included. – Moshe Jan 26 '12 at 14:10
@Moshe: I'd wager that there are more o-minimal expansions of the reals than what people originally expected when the theory took off. Nevertheless, it's still easy to get out of that framework, especially since combining two o-minimal structures can break o-minimality. – Thierry Zell Feb 12 '12 at 21:18

Qu 2. In the 1970s there were some papers dealing with the non-standard analysis and the theory of shape in the sense of Borsuk. The author was Frank Wattenberg. The reference is Fund. Math. 98 (1978), 41-60.

I do not know if more was published although I did see a preprint of another paper. I also do not know if those ideas have been followed up.

-
Tim, many thanks for the reference. – Valerio Capraro Jan 24 '12 at 23:00
Wattenberg is Andreas Blass's "mathematical father" and my "mathematical grandfather". It's great to see him mentioned! ;) – Todd Eisworth Jan 26 '12 at 0:32