Uncountability of the Rationals?
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From: Tony Orlow on 5 Jun 2010 14:06 On Jun 5, 12:37 pm, David R Tribble <da...(a)tribble.com> wrote: > Tony Orlow wrote: > > Perhaps simple bijection as a proof of equinumerosity is superficial. > > That's also a possibility. :) > > If you have two sets, and every single member of both are paired > together, in what logical sense can you say that they are not really > (only "superficially") equinumerous? Hi David - I can say so in the sense that one may be a proper subset of the other, or is less dense in the natural quantitative order, or that bijection is only part of a problem which is only properly solved by taking into account the mapping function between the two sets. Take your choice. Smiles, Tony
From: Tony Orlow on 5 Jun 2010 14:09 On Jun 5, 12:44 pm, David R Tribble <da...(a)tribble.com> wrote: > Virgil wrote: > >> Note that the reals with their standard order satisfy Tony's definition > >> of "sequence" though there is not even any explicit well-ordering of them. > > Tony Orlow wrote: > > There are always the H-Riffics. Remember "Well Ordering the Reals"? > > Yeah. Remember how several of us demonstrated that the H-riffics > is only a countably infinite set, and omits vast subsets of the reals > (e.g., all the multiples of powers of integers k, where k is not 2)? Sure when using 2 as a base, the numbers you mention are uncountably distant from the beginning of the uncountable sequence. But then, I am using "uncountable sequence" in a rather nonstandard way. :) Tony
From: Tony Orlow on 5 Jun 2010 14:09 On Jun 5, 12:48 pm, David R Tribble <da...(a)tribble.com> wrote: > Virgil wrote: > >> Both the rationals and the reals, with their usual orders, satisfy YOUR > >> definition of sequences, and while the rationals, with a suitable but > >> different ordering may be a sequence, there is no ordering on the reals > >> which is known to make them into a sequence, at least for any generally > >> accepted definition of "sequence". > > Tony Orlow wrote: > > Surely you remember the T-Riffics? > > Yeah. Surely you remember how you could never come up with > a self-consistent notation for them? Or a self-consistent definition > for incrementing from one T-riffic to the next? Or several other > missing critical pieces of your theory? That doesn't ring a bell. :) TOny
From: Tony Orlow on 5 Jun 2010 14:10 On Jun 5, 12:48 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <ed1b8d08-87d2-427b-bb10-9fcc46cd8...(a)d8g2000yqf.googlegroups.com>, > Tony Orlow <t...(a)lightlink.com> wrote: > > > Perhaps simple bijection as a proof of equinumerosity is superficial. > > That's also a possibility. :) > > It is certainly adequate as such a proof of equal cardinality. Duly conceded. :) Tony
From: Tony Orlow on 5 Jun 2010 14:11
On Jun 5, 12:53 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <9083dd54-a2f1-46af-ab41-421b3c253...(a)k39g2000yqb.googlegroups.com>, > Tony Orlow <t...(a)lightlink.com> wrote: > > > > > > > On Jun 4, 4:24 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > On Jun 4, 3:20 pm, David R Tribble <da...(a)tribble.com> wrote: > > > > > Tony Orlow wrote: > > > > > One might think there were something like aleph_0^2 rationals, but > > > > > that's not standard theory. > > > > > Actually, there are Aleph_0^2 rationals. And Aleph_0^2 = Aleph_0. > > > > Orlow can't be bothered to learn such basics. > > > > MoeBlee > > > Piffle to you both. I already stated that very fact very early in this > > thread. Don't start crying "quantifier dyslexia". You know better. > > > Tony > > To declare, as TO does above, that the cardinality of the rationals > being equal to aleph_0^2 is NOT part of the standard theory, is just > plain wrong!- Hide quoted text - > > - Show quoted text - I didn't say that was its cardinality, and if I had, it wouldn't matter because aleph_0^2=aleph_0 in standard theory. :) Tony |