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From: David Kastrup on 25 Oct 2005 10:35 albstorz(a)gmx.de writes: > Daryl McCullough wrote: >> albstorz(a)gmx.de says... >> >> >David R Tribble wrote: >> >> >> 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? >> >> It's quite ordinary math, except for the fact that >> perhaps you are unfamiliar with the use of ^ to mean >> exponentiation? >> >> 2^0 = 1 >> 2^1 = 2 >> 2^2 = 2*2 = 4 >> 2^3 = 2*2*2 = 8 >> 2^4 = 2*2*2*2 = 16 >> etc. >> >> David's sequence is 0, 2^0, 2^(2^0), 2^(2^(2^0)), etc > > I had read 0, 2^0, (2^2)^0, (2^2^2)^0, etc > since he has written: > "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." > > Maybe I had misunderstood. Maybe you should look in a math dictionary of your choice. Presumably because (a^b)^c is the same as a^(bc), anyway, the power operator is defined as right-associative. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: albstorz on 25 Oct 2005 11:23 Randy Poe wrote: > albst...(a)gmx.de wrote: > > > David's sequence is 0, 2^0, 2^(2^0), 2^(2^(2^0)), etc > > > > I had read 0, 2^0, (2^2)^0, (2^2^2)^0, etc > > since he has written: > > "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." > > Which means the elements are 0, and 2 raised to the power of q > where q is another element. If something should means something, it should be written like this. q isn't defined when it is used, so everyone could think about it what he wants. > > Thus, since 2^0 is an element, 2^(2^0) is an element, and so > is 2^(2^(2^0))), etc. > > > > > Maybe I had misunderstood. > > Yes. I don't see how to interpret p = (2^2)^0 as 2^q where > q is another element of the set. Show me the definition which defines that q is another element of the set. > > It's the same idea as the Peano definition of the naturals, > defining elements as 0, or as the successor to other > elements. But here the successor operation is > succ(q) = 2^q, instead of succ(q) = q + 1. There is no new or exciting aspect. That is all well known. > > Peano doesn't specify what the successor operation is. This > is a perfectly good Peano set. You could, for instance, do > induction on this set. > > - Randy You must be teachers. You are so boring. Regards AS
From: David R Tribble on 25 Oct 2005 12:19 David R Tribble: >> 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, ...} Albrecht Stortz: > 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. It's obvious that you don't understand simple arithmetic. 2^x is 2 raised to the x power. 2^2^x is 2^(2^x), using ordinary arithmetic. I'm giving you set S so that you can tell us whether it is an infinite set or not. You said: >> either there are infinite natural numbers or there is no infinite set. So I'm giving you set S, which obviously does not contain any infinite numbers. So by your rule, the set is finite, right?
From: David R Tribble on 25 Oct 2005 12:29 David R Tribble wrote: >> It is reasonable to say that the number of elements in the set is >> greater than any member in the set. Therefore, the size cannot be >> a member of the set, and it cannot be a natural number. > Albrecht Storz wrote: > It's not reasonable in the case of the natural numbers. The size of a > set could not be greater than any member in the case if the members > count themself. This is so obvious. This may be obvious to you, but it is not to anyone who understands infinite sets. David R Tribble wrote: >> Your contradiction is wrong. The contradiction comes from saying >> that the size of the set is also a member of the set. > Albrecht Storz wrote: > Since it is constructed like this it is like this: the size of the set > is a member of the set. > If you just deny the obvious facts, there is no further communication > possible. You are claiming that one of the members of the infinite set is equal to the size of the set. This is not obvious to us because it contradicts proven set theory. It is your responsibility to prove that your claim is true, and that standard set theory is wrong. Saying that your claim is "obvious" is not a proof.
From: Tony Orlow on 25 Oct 2005 15:44
David R Tribble said: > Virgil said: > >> 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. > > > > Tony Orlow wrote: > > This makes sense, but doesn't jibe with what you said before, as far as I can > > 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. > > > > Your scheme listed these mappings: > f(0) = {} > f(1) = {0} > f(2) = {1} > f(3) = {0,1} > f(4) = {2} > f(5) = {0,2} > f(6) = {1,2} > f(7) = {0,1,2} > f(8) = {3} > ... > > This list defines only the mappings of finite naturals in *N to > the finite subsets in P(*N). Your list does not show any infinite > naturals, nor does it show any infinite subsets, nor does it show any > subsets containing infinite naturals. I was asked about that and gave examples for the sets of odds and evens, which map to 2N/3 and N/3. There are other examples. It should suffice to say the same patter continues without end. > > Your list shows an infinite number of finite naturals mapping to an > equally infinite set of finite subsets. That's easy. There is no infinite set of finite naturals, as Albrecht has tried valiantly to illustrate. > But it's not > a bijection between *N and P(*N). It's not a bijection between > *N and P(N), or even between N and P(N). It is a bijection between > N and some of the members of P(N). Obviously that's not good enough. That's not what it is. I have declared it to be between *N and P(*N). It includes such numbers in *N as 0:01.....0101 for the evens and 0:1010...1010 for the odds. You have no reason to think this set is limited to finite values. Sorry. > > Even if we were to grant you that the list somehow includes infinite > naturals in *N mapping to subsets, you would still be missing a lot > of subsets in P(*N). Like which, for instance? Keep in mind that my bits never end. > > So fill in the blanks, and show us those other mappings. Otherwise > your "bijection" is incomplete and is not, in fact, a bijection > between *N and P(*N). WHich other mappings do you want? What would satisfy you? > > -- Smiles, Tony |