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Hello, i was reading your article about non metrizability of *R. i was able to prove that the interval open topology is not metrizable by proving that the intersection of decreasing hyper-intervals contains an interval. But i do not get how you used this argument for any metric, since we do not know how our nested balls look like. Plus how do you prove that *R is not connected? thanks a lot

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"Hello, i was reading your article..."?? –  Pete L. Clark Nov 13 '10 at 21:01
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Perhaps Jessica pushed the Ask Question button while reading another post, without realizing that we wouldn't know which post she had been reading. –  Joel David Hamkins Nov 13 '10 at 22:20

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I am not sure whom you are addressing in your question, but some of your remarks relate to issues brought up at this MO question. If not, could you let us know to which post you were referring?

It is quite common to consider hyperreal structures ${}^*\mathbb{R}$ that exhibit uncountable cofinality, which means that every countable subset of the hyperreals is bounded. For example, this is true in all the hyperreal models obtained by the ultrapower method, and it is a consequence of the $\omega_1$-saturation property that is often insisted upon. (But it is possible though rarely done to construct hyperreal models with the full transfer principle, but mere countable cofinality.)

In any hyperreal model with uncountable cofinality, no point has a countable local basis in the order topology, since for any countable collection of neighborhoods, one can find an infinitesimal ball around the point contained in them all. Thus, such a space cannot be metrizable, as metric spaces always have a countable local basis at every point, a countable collection of open neighborhoods of the point, such that every open neighborhood of the point contains one of them.

In a hyperreal model ${}^*\mathbb{R}$ with countable cofinality (rare but possible), however, then the order topology is metrizable. To see this, fix an increasing unbounded $\omega$-sequence of nonstandard reals $r_n$, which will eventually include many infinitely large hyperreals, and assume $r_{n+1}\geq 2r_n$. The balls of radius $\frac{1}{r_n}$ form a countable local basis around any point. Define a metric $d$ on ${}^*\mathbb{R}$ by $d(x,y)=\frac{1}{2^n}$, if $n$ is first such that $\frac{1}{r_n}\leq |y-x|$. It generates the same topology since the ball of radius $\frac{1}{2^n}$ corresponds basically to the interval of radius $\frac{1}{r_n}$ in ${}^*\mathbb{R}$.

Finally, the hyperreals are never connected, since we may partition the space into the infinitesimals and the non-infinitesimals, and these sets are both disjoint and open. In the case that there are increasingly tiny levels of infinitesimality, a consequence of saturation, then the same argument shows that the space is totally disconnected, since for any two points I may consider the lower-level infinitesimals around $p$ and the complement of this set, which are both open.

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@Joel: It is standard (hehe) to ask for $\omega_1$-saturation when presenting ${}^*{\mathbb R}$ abstractly, so that would take care of your comment on paragraph 2. This is how Keisler used to do it, anyway, while I was learning (mid 90s); this was around the time Keisler and Fajardo were developing their theory of neometric spaces. –  Andres Caicedo Nov 13 '10 at 21:05
    
Yes, that is what I meant by "countable" saturation. But I'll edit it to $\omega_1$-saturation, to avoid any confusion. –  Joel David Hamkins Nov 13 '10 at 21:07
    
what do you mean never connected in the last paragraph? With respect to any topology or the open interval topology only? –  jessica Nov 22 '10 at 19:49
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What I mean is that no model of the hyperreals is connected in the order topology. It is not really correct to speak of "the" hyperreals, as there are many non-isomorphic models with the hyperreal features of infinitesimals and the Transfer principle, etc. But all of them are disconnected by infinitesimals $\cup$ noninfinitesimals. –  Joel David Hamkins Nov 22 '10 at 20:19

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