tag:blogger.com,1999:blog-11295132.post114866460799717971..comments2014-11-13T21:22:20.572-08:00Comments on A Neighborhood of Infinity: Defining the RealsDan Piponihttps://plus.google.com/107913314994758123748noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-11295132.post-74254472234408851792007-04-29T16:34:00.000-07:002007-04-29T16:34:00.000-07:00I found it very interesting that the construction ...I found it very interesting that the construction of the reals can be done in any topos, and then Cauchy's and Dedekind's way give different results (that's in "Elementary Categories, Elementary Toposes" you're consuming).<BR/>I guess this construction can also be done in any topos and gives Cauchy reals, there is an obvious sequence of rationals lurking behind these grid-points.<BR/>So, while this method of not mentioning rationals is nice and doesn't feel unnatural, it might be provable that this construction is essentially(?) the same as Cauchy's.Christian_Snoreply@blogger.comtag:blogger.com,1999:blog-11295132.post-1149985921760325462006-06-10T17:32:00.000-07:002006-06-10T17:32:00.000-07:00There is a FOM post at http://www.cs.nyu.edu/piper...There is a FOM post at http://www.cs.nyu.edu/pipermail/fom/2003-August/007128.html<BR/>about generalizing this construction.<BR/><BR/>>I'd like to show that \E(R) is always a ring<BR/><BR/>If you start with an abelian group and replace "bounded" by "finite" in the definition of almost homomorphisms and the equivalence relation, the analogous construction always gives a ring. The IsarMathLib project (http://www.nongnu.org/isarmathlib/) has a formal verification of this fact.Slawekhttp://www.blogger.com/profile/01357411631017735279noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-1149025390144445812006-05-30T14:43:00.000-07:002006-05-30T14:43:00.000-07:00Theo,Interesting. My (admittedly not very reliable...Theo,<BR/><BR/>Interesting. My (admittedly not very reliable) intuition is that generalising this construction isn't going to go very far.<BR/><BR/><EM><BR/>I'd like to show that \E(R) is always a ring<BR/></EM><BR/>Did you get anywhere?<BR/><BR/><EM><BR/>we didn't kill any genuine homos<BR/></EM><BR/>You wouldn't want to read that sentence out of context :-)sigfpehttp://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-1148884379429184542006-05-28T23:32:00.000-07:002006-05-28T23:32:00.000-07:00Or maybe the point is that I need a generalized-to...Or maybe the point is that I need a generalized-to-arbitrary-ring idea of "measure".Theohttp://www.blogger.com/profile/10449695704601441213noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-1148882562686919312006-05-28T23:02:00.000-07:002006-05-28T23:02:00.000-07:00If you don't mind, I'd like to think out loud a bi...If you don't mind, I'd like to think out loud a bit about other directions of generalization, possibly already pursued elsewhere.<BR/><BR/>This paper considers the space of "almost homomorphisms" of the \Z-module (i.e. ab gp) \Z; given an arbitrary ring R, it seems that we might be interested in the almost homomorphisms of R _as an R module_.<BR/><BR/>Well, why? Because vital to this argument is that \Z is naturally isomorphic to the space of homomorphisms of \Z: any homo f:\Z\to\Z is uniquely determined by f(1), which can take any value. In a general ab gp, this is not the case --- the space of \Z-endomorphisms of \Z^2 is not even commutative (it is the space of 2x2 \Z-matrices). But it is true that the space of R-endos of R is naturally isomorphic to R, again by identifying f with its value at 1. And, moreover, composition is naturally identified with multiplication.<BR/><BR/>So, great. But what would an R-almost-endo look like? Because we certainly want R-almost-endos. If we take the space of \Z-almost-endos of, say, R=\Z[i], then we cannot expect to get something small and commuting, whereas we expect the relationship between \Z and \R to be reflected by the relationship of \Z[i] to \C. The point of an R-homo is that not only does f(p+q)=f(p)+f(q), but f(mp)=mf(p). An R-almost-homo (which I think I'll start writing as an R-almo, short for "almomorphism", and it will be clear from context if I'm talking about almost endos (R-almo of A) or almost homos (R-almo from A to B)) has a bounded error in the first property, but I think that we cannot expect simply a bounded error in the second. Certainly the best bounds in the \Z case (as described in the paper) for f(mp) is that f(mp)-mf(p) < (m-1)C, where C is the bound on f(p+q)-f(p)-f(q), and m is a nonnegative integer. And, indeed, if we expected f(mp)-mf(p) to be bounded, then f(p) could be only boundedly far from pf(1), and modding by bounded functions reduces us back to \Z, rather than getting a construction of the reals.<BR/><BR/>No, we need something more flexible for f(mp). But what? Bounds don't work --- I want to extend to rings with measures, not rings with orders. Perhaps the best we can do is to say that, for any given m, the difference d^m_f (p) = f(mp)-mf(p) is bounded as p ranges, where the bound depends on m? It would be nice to make the bound linear in m --- no, almost linear. Perhaps, then, we want to say this: the volume of the range of d^m_f is bounded by some \Z-almo C(m) from R to \Z. Yes, perhaps that's it.<BR/><BR/>Oh, another thing concerns me. Vital to the embedding of \Z into \R in this construction was that, when modding out by bounded almos, we didn't kill any genuine homos. Can we be sure of that here? Well, say R is infinite (otherwise, naively, all almos are bounded, unless we use a different measure, but let's cross that bridge later). Then if f_c:p\mapsto cp is has finite range, then pigeonhole guarantees infinitely many p s.t. cp=0. We fix our problems if we demand that R is a domain --- if it isn't, the probably we have no right to expect R to embed into some continuum limit of R, because this continuum limit really is a continuum limit of the field of fractions of R. (You say you want the space of slopes of almost-lines --- an important subclass of almost-lines are those with rational slopes.)<BR/><BR/>Anyway, looking at \Z[i], certainly any almo of \Z extends to an almo of \Z[i] by f(p+iq)=f(p)+if(q). So \E(\Z[i]) certainly contains \R[i]=\C. (Naturally, \E(R) is the R-module of R-almos of R modulo boundeds.) Showing that that's all there is will take more time than I have tonight --- I'd like to show that \E(R) is always a ring (so an R-algebra, and, if R is a domain, a ring extension of R), and then perhaps that \E(R) is in fact a field, but doing so will require getting multiplication to work.<BR/><BR/><BR/>Going further, I'd be curious to know what happens with other measures. For instance, there are very few interesting measures on finite rings --- I could imagine endowing 0 with finite measure and all other points with measure \infty, in which case almos are required to be genuine homos. But other directions are also possible.<BR/><BR/>Hmm.Theohttp://www.blogger.com/profile/10449695704601441213noreply@blogger.com