Consider R, or rather the equivalence class of fields isomorphic to R, endowed with an order type of the same kind. Let us consider a new object, dx, a quantity such that dx > 0, but for all a > 0 in R, dx < a. (We are considering dx, something outside of R, but without the context of a larger universe set.)(adsbygoogle = window.adsbygoogle || []).push({});

Now let add dx to R, and field-ify it. That is, we are considering the infinite field extension R[dx]. Furthermore we inherit the order of R, but extend it to include dx. The way I think about the order type is that, suppose we can sort of magnify R until we see a, b, and a, b have no more elts of R between them (there is no c with a<c<b)... we can't actually do this, but entertain this idea for a moment. Adding dx to R is like fitting another whole copy of R between a and b: we have between them a < a + k*dx < b for all k in R. But then we magnify this new copy of R until we get to a, b and add another copy of R (dx^2) and so on, and do this for all nonexistent pairs a, b satisfying there is no c st a<c<b...

My question is, what is the "dimension" of R[dx]? It is not 2.. The above thought experiment suggests that it is |Z| (aleph-1), but I don't believe that is true. It may well be |R|, but I suspect the answer is much more sinister...

Note that the standard definition of dimension in field extension doesn't apply.

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# On the infinitesimally small

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