infinity ...
in [Math]
|
Prev: math
Next: The proof of mass vector.
From: David R Tribble on 24 Oct 2005 20:08 Tony Orlow wrote: >> So, which element of the power set does not have a natural mapped to it? > Virgil said: >> {x in S:x not in f(x)} > Tony Orlow wrote: > Oh yeah, the entire set, last element and all. What element was that again? Your mapping does not include any natural that maps to the entire set *N, which it must do in order to be called a bijection, since *N is one of the members of P(*N). Show us that one single mapping, please. But none of us can figure out where you keep pulling this "last element" gibberish from. Just let it go, Tony.
From: William Hughes on 24 Oct 2005 21:00 Tony Orlow wrote: > Virgil said: <snip> > > > So, which element of the power set does not have a natural mapped to it? > > > > {x in S:x not in f(x)} > Oh yeah, the entire set, last element and all. What element was that again? No, just the entire set. The set does not have a last element. - William Hughes
From: Virgil on 24 Oct 2005 21:37 In article <MPG.1dc70e511038079d98a54e(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Virgil said: > > In article <MPG.1dc1c0b757ad81cd98a518(a)newsstand.cit.cornell.edu>, > > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > > > Virgil said: > > > > > > For what set is TO assuming that each member of its power set > > > > can be represented by a single bit in an infinite sequence of > > > > bits? > > > I never said that. > > > A member of the power set is a subset of the entire set. You are > saying that I am saying that each of these subsets can be represented > by a single bit in an infinite sequence? NO! I am saying that you say that each singleton set of P(S), set with exactly one member, must be so represented in TO's argument. > If you take the power set of > the power set, then you have sets of subsets, each of which can be > assigned a bit corresponding to its order in the enumeration of the > power set. But, that is not what you seem to be saying here now. If TO were not functionally illiterate, he could read what I actually say. > > If each subset of *N is to be represented by an infinite binary > > sequence of digits with 1 in some position representing the > > presence of a member *N and 0 representing its absence, then one > > element sets must be represented by strings with one 1 in them. > This makes sense, but doesn't jibe with what you said before, as far > as I can tell. It does, but TO can't tell. > Yes, each singleton set in P(*N) will map to a natural > whose binary representation has a single bit. This only includes N > out of 2^N subsets. But you already have a bijection from all of *N to the singleton members of P(*N), and that leaves most of P(*N) out. > > > > So that, while he may not have been aware of it, TO was saying > > precisely that. That TO is often unaware of the meaning of what he > > is saying has long been apparent. > Precisely what? You are being particularly opaque, even for Virgil. If TO is too illiterate to read what I have said, it is not my responsibility to send him back to kindergarten. > > To do that in a string, TO must not only well-order the reals, but > > order them isomorphically to the set of finite naturals. > But the finite naturals have finite numbers of bits, so what do you > mean? Each finite naturally individually requires only a finite number of bits for its expression, but there is no finite number of bits that suffices for all of them. Those who fail to understand this have a form of quantifier dyslexia. > > > So, which element of the power set does not have a natural mapped > > > to it? > > > > {x in S:x not in f(x)} > Oh yeah, the entire set, last element and all. What element was that > again? What "last element" is talking about? The set S need not even be an ordered set, as order is not essential for the existence of {x in S:x not in f(x)}, so how can an unordered set be though to have a last element? The conents of {x in S:x not in f(x)} depends entirely on the nature of the function f:S -> P(S). Any ordering of S is irrelevant to the existence, or membership, of {x in S:x not in f(x)}. The set {x in S:x not in f(x)} is defined soon as f is defined, but until f is defined for all of S, the issue of whether it can be a bijection on S is irrelevant. So that TO's complaints are out of order!
From: albstorz on 25 Oct 2005 02:52 David R Tribble wrote: > David R Tribble wrote: > >> I've got a set S = {0, 2^0, 2^2^0, 2^2^2^0, ...}, which contains > >> all the powers of 2 of the form 2^p, where p=0 or 2^q. > >> 1) If it is not an infinite set, tell me how many members it has. > >> 2) If it is an infinite set, tell me what the smallest (first) > >> infinite number is a member of it. > > > > Albrecht Storz wrote: > > A short view upon this makes me think that you are writing sensless > > symbols. S don't contain all numbers of the form 2^2^q if you think > > about the sequence 0, 2^0, 2^2^0, 2^2^2^0,... . If this is your > > intention you may have a infinite sequence 0, 1, 1, 1, ..., and a set > > {0,1}, so > > 1) 2 > > 2) ? > > Obviously, you do not understand. > > My set is: > S = {0, 2^0, 2^2^0, 2^2^2^0, 2^2^2^2^0, ...} > S = {0, 1, 2, 4, 16, 65536, ...} > I don't know what kind of math you apply here. Tribble-O-Math? I don't want discuss your very interesting system. Look at my starting posting if you want to know in what I'm interested to discuss. Start a new thread if you search for people which want talk about Tribble-O-Math. Regards AS
From: David Kastrup on 25 Oct 2005 05:05
"David R Tribble" <david(a)tribble.com> writes: > Tony Orlow wrote: >>> So, which element of the power set does not have a natural mapped to it? >> > > Virgil said: >>> {x in S:x not in f(x)} >> > > Tony Orlow wrote: >> Oh yeah, the entire set, last element and all. What element was that again? > > Your mapping does not include any natural that maps to the entire set > *N, which it must do in order to be called a bijection, since *N is one > of the members of P(*N). Show us that one single mapping, please. > > But none of us can figure out where you keep pulling this "last > element" gibberish from. "complete" = "finished" = "has an end" in Tonymatics. He'd not be able to explain the topology of a circle, except in a metamathematical way by circular reasoning, but never mind. > Just let it go, Tony. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum |