Uncountability of the Rationals?
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From: Tony Orlow on 14 Jun 2010 07:46 On Jun 12, 3:35 pm, David R Tribble <da...(a)tribble.com> wrote: > Tony Orlow wrote: > >> I would like you consider a slightly modified goal: To exactly order > >> sets according to size, and to distinguish between those that have an > >> absolute size and those that do not. For me, no countably infinite set > >> has an absolute size, but only one relative to other countably > >> infinite sets, especially the standard N, starting at 0. > > David R Tribble wrote: > >> Wouldn't it be easier to say that some chosen countably infinite > >> set has a standard "absolute" size, and then that all other such > >> sets have a size that is some fraction of that size? Thus every > >> set then has an "absolute" size, based on the standard set size. > > Tony Orlow wrote: > > That might be easier, and would lead, as suggested, to natural > > density. However, that doesn't address the relations between infinite > > sets that have other ratios or offsets besides the finite. It is much > > more fruitful to look for a unit infinity based on uncountable > > infinity, which can be much more easily combined with measure and > > topology. > > So having an "absolute" set size for countable sets is not a > good idea? You'd rather just say that non-finite countable sets > don't have an absolute size at all and leave it at that? > > Why go through the trouble of dealing with uncountable sets > but just write off the simpler domain of countable sets? > > David R Tribble wrote: > >> For convenience, you'd want to choose the "largest" countably > >> infinite set for the standard "absolute" set size. This would > >> probably be N, since it contains all of the countable naturals. > >> (Another logical choice would be Z.) Given this, you could then > >> say that every countably infinite set does indeed have an > >> "absolute" set size. > > Tony Orlow wrote: > > But, for me, they do not. If N starts at the first location with 1, > > and then the second with 2, etc, then the nth natural is n, and unless > > a set has n elements, there is no nth one. If there are omega > > naturals, then there is an omegath, which by the order=value > > definition of that set must equal omega. > > What makes you think that? > > Statements like that make it look like you don't really understand > what terms like "omega" mean. > > > However omega cannot exist within the set. Consider this. > > Are the following two statements logically equivalent? > > > AneN n<omega > > > The first says there is no ntaural that can be the size of the entire > > set of naturals. That's true. > > Yes. > > > ~EneN n>=omega > > > The second says that there is no n in N that is greater than or equal > > to this "size". > > Yes. > > > Does either imply, actually, that such a size exists? No. Both assert > > that no natural number will suffice. However, this is not to say that > > there exists ANY number which does. Logically, you must concede, > > neither statement proves that omega exists as a number. > > Yes, that's correct. Neither of your two statements implies that > there exists any number equal to card(N). > > Neither statement implies card(N) does not exist, either. Nope. They simply say that IF N has a size, it cannot be a member of N. > > > I am not > > obligated to imagine such a number, or give it any credibility. > > So you're saying that if the two statements you provided do > not imply that card(N) exists, that proves that card(N) does not > exist? That's an extremely weak argument to make. Never said that. Read carefully. > > Have you considered any of the other statements you *could* > have made but didn't, that maybe one of them might imply the > existence of card(N)? After all, your statements simply state > what does *not* exist, but nothing about what *could* exist. > > How is you argument any different than: > 1. For all x, f(x) < z; > 2. There does not exist x such that f(x) >= z; > Therefore > z does not exist. I never said the two statements I gave above prove the nonexistence of |N|, but that the statements also do not state that it exists, only that it cannot be any natural number, if it exists. So, your, "Therefore z does not exist" has no equivalent in anything I said. Are you having trouble distinguishing between "does not prove x" and "proves the opposite of x"? BTW, Card(N) can exist if you want it to. Bigulosity of countably infinite sets is always in functional relation to omega, but omega is more of a limit than a number. > > > When > > it comes to the countably infinite, we can only look at this as some > > kind of limit, and for most discriminating results, consider them as > > Big-O measures or something similar. > > But then that's sort of what omega (or Aleph_0) is, it being a kind > of (lower) limit on infinite set sizes.- Hide quoted text - > > - Show quoted text - With IFR, N is hardly the smallest caountably infinite set, consider the evens, squares of naturals, natural powers of 2, etc. Tony
From: Tony Orlow on 14 Jun 2010 07:49 On Jun 12, 3:47 pm, FredJeffries <fredjeffr...(a)gmail.com> wrote: > On Jun 12, 12:42 pm, David R Tribble <da...(a)tribble.com> wrote: > > > Jesse F. Hughes wrote: > > >> You [Tony] sometimes say that you > > >> don't think omega is the smallest infinite ordinal, but I'm not sure > > >> what you mean when you say that. > > > Tony Orlow wrote: > > > I don't use the word "ordinal" in my arguments (sure, go find one > > > mention from 1996 or whatever). I say there are a wide spectrum of > > > countable and uncountable infiniites, given the right techniques. > > > What Tony means is that size(E) < size(N) for the sets N and > > E = {0,2,4,6,...}. Specifically, he means that size(E) = size(N)/2, > > where "size(S)" is his personal definition of "set size". > > No, he told me that Size(E) = Size(N)/2 + 1http://groups.google.com/group/sci.math/msg/df77005159c91c90 > because of 0. > > At least, I THINK that's what he told me. > > > > > > > He also allows for things like sqrt(size(N)) and log(size(R)). > > These "values" of his are "different infinities", all based on his > > idea of a "unit infinity".- Hide quoted text - > > - Show quoted text - Actually, you're right. N should always start with 1 for IFR. I made an error of 1 by including 0. If we include 0, then we may say E=N/2+1 and O=N/2, because of that extra even element. Apologies. Tony
From: Tony Orlow on 14 Jun 2010 07:52 On Jun 12, 4:02 pm, David R Tribble <da...(a)tribble.com> wrote: > Tony Orlow wrote: > > ELiminating proper subset equinumerosity is worthwhile. > > That kind of statement hints at a deep misunderstanding of > the way mathematics works. > > You seem to be suggesting that by inventing an alternative way > of measuring set sizes (a more "intuitive" way), that the existence > of bijections between infinite sets and their proper infinite subsets > will disappear or stop working. You mean like your ability to read and retain? Remember that IFR is EXACTLY dependent of the establishment of a bijection. The definition of an infinite set as one wherein a bijection may be established between it and a proper subset works for both the countably and uncountably infinite. No one said bijections would disappear, unlike that infinite number of balls in the vase when you change their labels. > > Unless you invent a system of "sets" that flat out does not allow > relations like injections and surjections to be formulated at all, > you can't do this. Either that or come up with a system having > completely different meanings for "subset" and "mapping". I don't know what you think you're blathering about. It's kind of annoying to deal with staunch objections to things one never even implied. Tony
From: Tony Orlow on 14 Jun 2010 07:53 On Jun 12, 7:27 pm, David R Tribble <da...(a)tribble.com> wrote: > Jesse F. Hughes wrote: > >> You [Tony] sometimes say that you > >> don't think omega is the smallest infinite ordinal, but I'm not sure > >> what you mean when you say that. > > Tony Orlow wrote: > >> I don't use the word "ordinal" in my arguments (sure, go find one > >> mention from 1996 or whatever). I say there are a wide spectrum of > >> countable and uncountable infiniites, given the right techniques. > > David R Tribble wrote: > >> What Tony means is that size(E) < size(N) for the sets N and > >> E = {0,2,4,6,...}. Specifically, he means that size(E) = size(N)/2, > >> where "size(S)" is his personal definition of "set size". > > Fred Jeffries wrote: > > No, he told me that Size(E) = Size(N)/2 + 1 > >http://groups.google.com/group/sci.math/msg/df77005159c91c90 > > because of 0. > > Actually, I think he said that size(E) = size(N)/2 +1 *and* > that size(E) = size(N)/2 + 1/2.- Hide quoted text - > > - Show quoted text - Depends on whether you start N with 1 or 0, and the 1/2 drops off with the floor() function.
From: Tony Orlow on 14 Jun 2010 08:06
On Jun 13, 2:52 pm, David R Tribble <da...(a)tribble.com> wrote: > David R Tribble wrote: > >> For convenience, you'd want to choose the "largest" countably > >> infinite set for the standard "absolute" set size. This would > >> probably be N, since it contains all of the countable naturals. > >> (Another logical choice would be Z.) Given this, you could then > >> say that every countably infinite set does indeed have an > >> "absolute" set size. > > Tony Orlow wrote: > > Actually, in Bigulosity, there are much larger countably infinite > > sets. The square roots of the naturals have a higher "density" than > > the naturals. > > How is that? The two sets should be exactly the same size > (your size(), not just cardinality). Every natural has a square root, > and every natural square root has a corresponding natural. > (I assume you're only dealing with the positive roots.) Holy cannoli! There exists a bijection, f:N->S and g:S->N, so indeed they have the same cardinality. They do not share the same bigulosity, since f and g are not the identity function. The number of square roots over omega is omega^2. Granted, in standard theory those are equivalent, but in Bigulosity, omega^2>omega, as x^2>x for x>2 and omega>2. QED > > > In terms of natural density, they probably have a > > measure of oo? > > The set of positive natural square roots is not a subset > of the naturals, so it does not have a natural density. That is probably correct. The squares have a natural density of 0. If natural density is actually restricted to naturals, then there is no natural density, but qactually natural density allows for elements outside of N, such as the multiples of 1/2 with a density of 2. So, your objection is based ona false presumption about a theory that's not my invention. My guess is that this set would have an infinite natural density if any. > > > Not sure. For me, there are omega^2 number of square > > roots of naturals within the range of N. > > Are you sure about this? Shouldn't there be omega square > roots, or 2*omega if you're also counting negative roots? > Are you sure you're not talking about the square roots of R? No, the inverse of square root is square. Apply IFR in the finite case, and you'll see why this is the correct answer. Tony |