SCI.MATH POLL - uncountable infinity
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From: George Greene on 4 Jun 2010 23:55 On Jun 4, 2:37 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > In particular, the > results of a poll which asks "Is CH true?" are > more likely to be accepted No, they are not. Such a poll is incoherent. That question simply cannot have a true Xor false answer. Polling as though it did is flaunting ignorance. What is actually the case is that CH is true in some models of ZFC AND FALSE IN OTHERS. True AND False. AT THE SAME TIME. Though not in the same place. That is just all there is to it. Polls are simply not relevant. There is a fact of the matter. People DO NOT GET TO HAVE opinions. Either they know or they don't.
From: Barb Knox on 5 Jun 2010 01:57 In article <86od4nFja0U1(a)mid.individual.net>, "|-|ercules" <radgray123(a)yahoo.com> wrote: [SNIP] Ah, the Townsville looney returns. Mate, you just gotta stop licking those cane toads. -- --------------------------- | BBB b \ Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum videtur. | BBB aa a r bbb | -----------------------------
From: herbzet on 5 Jun 2010 08:58 herbzet wrote: > |-|ercules wrote: > > > I'll wait and see if someone else takes the bait. > > > > >> The proof of higher infinities than 1,2,3...oo infinity relies on > > >> the fact that there is no box that contains all and only all the > > >> label numbers of the boxes that don't contain their own label number. > > > > TRUE OR FALSE > > Um, false, so far as I know. > > We have > > 1) |N| < |P(N)| > 2) |P(N)| <= |R| > -------------- > .: |N| < |R| > > but neither of Cantor's proofs that |N| < |R| involves either of > premises (1) or (2), as far as I can recall. Why do you ask? -- hz
From: |-|ercules on 5 Jun 2010 09:03 "herbzet" <herbzet(a)gmail.com> wrote > > > herbzet wrote: >> |-|ercules wrote: >> >> > I'll wait and see if someone else takes the bait. >> > >> > >> The proof of higher infinities than 1,2,3...oo infinity relies on >> > >> the fact that there is no box that contains all and only all the >> > >> label numbers of the boxes that don't contain their own label number. >> > >> > TRUE OR FALSE >> >> Um, false, so far as I know. >> >> We have >> >> 1) |N| < |P(N)| >> 2) |P(N)| <= |R| >> -------------- >> .: |N| < |R| >> >> but neither of Cantor's proofs that |N| < |R| involves either of >> premises (1) or (2), as far as I can recall. > > Why do you ask? > Because the most widely used proof of uncountable infinity is the contradiction of a bijection from N to P(N), which is analagous to the missing box question. Does the proof of higher infinities than 1,2,3...oo infinity rely on this verbage being true? http://en.wikipedia.org/wiki/Cantor's_theorem Suppose that N is bijective with its power set P(N). Let us see a sample of what P(N) looks like: P(N) = { {}, {1,2}, {1,2,3}, {4}...} P(N) contains infinite subsets of N, e.g. the set of all even numbers {2, 4, 6,...}, as well as the empty set. Now that we have a handle on what the elements of P(N) look like, let us attempt to pair off each element of N with each element of P(N) to show that these infinite sets are bijective. In other words, we will attempt to pair off each element of N with an element from the infinite set P(N), so that no element from either infinite set remains unpaired. Such an attempt to pair elements would look like this: 1 <-> {} 2 <-> {1,2} 3 <-> {1,2,3} (different example than text refers to, see wiki) Given such a pairing, some natural numbers are paired with subsets that contain the very same number. For instance, in our example the number 2 is paired with the subset {1, 2, 3}, which contains 2 as a member. Let us call such numbers selfish. Other natural numbers are paired with subsets that do not contain them. For instance, in our example the number 1 is paired with the subset {4, 5}, which does not contain the number 1. Call these numbers non-selfish. Likewise, 3 and 4 are non-selfish. Using this idea, let us build a special set of natural numbers. This set will provide the contradiction we seek. Let D be the set of all non-selfish natural numbers. By definition, the power set P(N) contains all sets of natural numbers, and so it contains this set D as an element. Therefore, D must be paired off with some natural number, say d. However, this causes a problem. If d is in D, then d is selfish because it is in the corresponding set. If d is selfish, then d cannot be a member of D, since D was defined to contain only non-selfish numbers. But if d is not a member of D, then d is non-selfish and must be contained in D, again by the definition of D. This is a contradiction because the natural number d must be either in D or not in D, but neither of the two cases is possible. Therefore, there is no natural number which can be paired with D, and we have contradicted our original supposition, that there is a bijection between N and P(N). Herc
From: herbzet on 5 Jun 2010 09:06
|-|ercules wrote: > herbzet wrote: > > herbzet wrote: > >> |-|ercules wrote: > >> > >> > I'll wait and see if someone else takes the bait. > >> > > >> > >> The proof of higher infinities than 1,2,3...oo infinity relies on > >> > >> the fact that there is no box that contains all and only all the > >> > >> label numbers of the boxes that don't contain their own label number. > >> > > >> > TRUE OR FALSE > >> > >> Um, false, so far as I know. > >> > >> We have > >> > >> 1) |N| < |P(N)| > >> 2) |P(N)| <= |R| > >> -------------- > >> .: |N| < |R| > >> > >> but neither of Cantor's proofs that |N| < |R| involves either of > >> premises (1) or (2), as far as I can recall. > > > > Why do you ask? > > > > Because the most widely used proof of uncountable infinity is the > contradiction of a bijection from N to P(N), which is analagous to > the missing box question. Perhaps so, but why do you ask? -- hz |