Uncountability of the Rationals?
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From: Virgil on 15 Jun 2010 15:41 In article <f02c7a6b-0793-42db-96bf-e7f3cf9a4377(a)e5g2000yqn.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > > It seems to me that he's stated his assumption, but he doesn't get quite > > how much it assumes. > > > > � Given any real valued functions f and g, if lim (f - g) > 0, then f(x) > > � > g(x) where x is any infinite number. > > That's actually a nice succinct way to state it, however I generally > include the condition that finite-case induction applies to f-g as > well, to make clear that it is an extension of the finite form of > proof to the infinite case. Given some real function f, how does one find out the supposed values of f(x) for all those relavant x which are not real?
From: Virgil on 15 Jun 2010 15:45 In article <71251f7a-be55-4d15-9c61-5e446609453a(a)z8g2000yqz.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Okay, but the size of the set is equal to the number represented by > the von Neumann ordinal, if they start at 0. If one wants it to start > at 1, I suppose one could associate ech set with the deepest nesting > of brackets in the set, which starts with one level. No, they are not > elements of the sets, but measures thereof. The *size* of a set is its cardinality, not its ordinality, and it is only for finite sets that they must coincide.
From: Virgil on 15 Jun 2010 15:52 In article <951862ad-f7bb-4bfc-bbd1-610b0ce802ac(a)19g2000vbi.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 14, 7:46�pm, Virgil <Vir...(a)home.esc> wrote: > > In article > > <23e0c345-10ff-4172-8a1e-0c230c170...(a)f8g2000vbl.googlegroups.com>, > > �Tony Orlow <t...(a)lightlink.com> wrote: > > > > > On Jun 14, 3:23�pm, David R Tribble <da...(a)tribble.com> wrote: > > > > Brian Chandler wrote: > > > > >> I certainly don't think it would be correct to say that the > > > > >> reciprocal > > > > >> of zero is the empty set. > > > > > > Tony Orlow wrote: > > > > > In standard mathematics it is "undefined", both positively AND > > > > > negatively infinite. It doesn't exist. The set of solutions at 0 is > > > > > empty, ot perhaps a pair of hyperreals. > > > > What is undefined or does not exist is neither positively nor negatively > > infinite. 1/x does not have a value, or a solution, at x = 0. > > > > I believe there are formal systems where this value may be considered > +/- oo, as they are equal in many respects. In such systems, the functions are no longer real function but functions whose domain and range are no longer subsets of the reals. > > It is certainly not standard for a function all of whose values all > > reals to have a value which is not a real, whether hyperreal or unreal > > in some other way. > > Given that the infinite value represents a counting number greater > than any given natural The "infinite value" approached by f(x) = 1/x as x -> oo, is definitely NOT a counting number. > and that two functions have a difference > without a limit of 0 as the independent variable increases from some > natural n without bound, that difference remains in the infinite case. > There is no reason to think it disappears, and the difference is > functionally quantifiable. Unless the terms in that difference are equally quantifiable, which TO has yet to show, his analysis has huge holes in it.
From: Tony Orlow on 15 Jun 2010 16:33 On Jun 14, 7:55 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <b95c48f1-4244-4d6b-b218-5dcebc166...(a)30g2000vbf.googlegroups.com>, > Tony Orlow <t...(a)lightlink.com> wrote: > > > > > > > On Jun 14, 3:08 pm, David R Tribble <da...(a)tribble.com> wrote: > > > 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. > > > > David R Tribble wrote: > > > >> 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.) > > > > Tony Orlow wrote: > > > > 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. > > > > Sorry, but it's not clear what you mean by "the number of square > > > roots over omega is omega^2". Do you mean: > > > 1. the number of naturals in omega that are square roots is omega^2? > > > 2. the number of naturals in omega that have square roots is > > > omega^2? > > > 3. the set of the squares roots of all the naturals in omega has > > > omega^2 members? > > > 4. or something else? > > > Number 3. The set of square roots of naturals has a Bigulosity of > > omega^2. > > Since every natural has 2 square roots, why wouldn't it be 2*aleph_0?- Hide quoted text - > > - Show quoted text - First of all, by IFR, in harmony with the fact that square roots of naturals occur more often as we move along the real line than do naturals, we can conclude that over N+ there exist more naturals' square roots than naturals, omega^2 square roots of naturals. Now, if you want to include the negative square roots of naturals, then the inverse is not a function, since it includes two values for each neN. In such cases, which we haven't gotten to yet, one must divide the domain of the mapping from N to S at their extrema (mins and maxs) to get monotonically increasing or decreasing sequence segments, then add the results for each to get the total membership. Thus, the number of squeare roots should be roughly double, perhaps plus the 0th one. TOny
From: Tony Orlow on 15 Jun 2010 16:38
On Jun 14, 8:03 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <1959a4fe-5e75-4ae6-a9e0-df23d18e4...(a)y4g2000yqy.googlegroups.com>, > Tony Orlow <t...(a)lightlink.com> wrote: > > > > I'm of course assuming we're talking about ZFC and standard > > > arithmetic and so forth. In your BO ("Bigulosity Ordering"), I > > > guess not all sets have a size, so it might very well be possible > > > for statement in BO to say "if set N has a size, then ...". > > > If it's a countably infinite set then it is expressed with relation to > > standard N+. > > Then you are claiming that EVERY subset of N+ has a well defined > bigulosity? IF it has an algebraic bijection with N+ it can be compared therewith. > > If it is an uncountable number of points it begins to > > > have Lebesgue measure, and can start to be quantified with respect to > > zillions. No countable set of points has Lebesgue Measure >0. > > > > Do you have a test in BO for "size" that determines when a > > > set has a size and when it does not? > > > Countably infinite sets as normally defined have only parametric size > > in relation to N+. > > DO they all have bigulotic "sizes"? > The term is "Bigulosities", and it's always capitalized, because it "that" important. Countably infinite sets have Bigulosity relative to standard omega, N +. > > > >> Therefore, omega as a definition is self contradictory, and connot > > "particularly" exist. > > But omega as a set is merely N, so that N cannot exist either in > Bigulosity. It is more a of a lmit than a point. I've shown how, logically, such a point cannot exist, given the contradiction that its definition entails. <3 TOny |