Dense Linear Ordering problems.
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From: David Bernier on 26 Nov 2009 16:34 Virgil wrote: > In article<m9atg5h0jvcun23n7oj9dq286ut82i9ti2(a)4ax.com>, > David C. Ullrich<dullrich(a)sprynet.com> wrote: > >> On Thu, 26 Nov 2009 01:27:20 -0800 (PST), Butch Malahide >> <fred.galvin(a)gmail.com> wrote: >> >>> On Nov 25, 10:05 pm, Bill Taylor<w.tay...(a)math.canterbury.ac.nz> >>> wrote: >>>> >>>> Consider subsets of the reals such that both the subset and >>>> its complement are continuum many in every interval. >>>> >>>> Are any two such subsets order-isomorphic? [Spelling typo corrected.] >>> >>> No, but the counterexample won't impress you, because it uses AC. >> >> A counterexample that depends on AC _should_ be of interest >> even to people who don't "accept" AC! If AC implies that >> there is no gazebo then it follows, whether one accepts >> AC or not, that the existence of a gazebo cannot be proved >> in ZF. > > Nice! And nicely sneaky!! I'm not sure what the gazebo-property is here... Under ZF + AC something was shown to exist. So one reading is that what I'll call the gazebo* property for a set X is this: gazebo*(X): X = (A, B) , A union B is Reals, A and B are disjoint, A \intersect [a, b] has cardinality 2^(aleph_0) , any a< b Reals, B \intersect [a, b] has cardinality 2^(aleph_0) , any a< b Reals, and [ Exists f: A -> B a bijection, and f order-preserving under canonical '<' .] Then, in ZFC, there is an X with the gazebo* property, or ? ... David Bernier >>> Since you're not going to accept the example anyway, I'll just sketch >>> the argument. >>> >>> Recall that a subset of R is "totally imperfect" if it contains no >>> perfect set (and therefore no uncountable Borel set of any type.) Use >>> AC to obtain a so-called Bernstein decomposition of R, i.e., a totally >>> imperfect set A whose complement B = R\A is also totally imperfect. >>> (Needless to say, they are nonmeasurable sets.) Of course each of the >>> sets A and B contains continuum many points in every interval. As far >>> as I know, A and B may be order-isomorphic. Now, let C be the Cantor >>> set, and consider the set X = A union C and its complement Y = B\C. >>> The modified sets X and Y still contain continuum many points in each >>> interval. I claim that X and Y are not isomorphic. To see this, >>> observe that X contains a subset which is order-isomorphic to R (in >>> fact C contains such a subset), while Y, being totally imperfect, >>> contains no subset order-isomorphic to R. >> >> David C. Ullrich >> >> "Understanding Godel isn't about following his formal proof. >> That would make a mockery of everything Godel was up to." >> (John Jones, "My talk about Godel to the post-grads." >> in sci.logic.)
From: Bill Taylor on 26 Nov 2009 21:22 > > A counterexample that depends on AC _should_ be of interest > > even to people who don't "accept" AC! If AC implies that > > there is no gazebo then it follows, whether one accepts > > AC or not, that the existence of a gazebo cannot be proved > > in ZF. This is OC the contrapositive case of the usual thing. The earlier proof shows that an alleged example is provably a true example, therefore no proof of non-existence can exist in ZF. A direct example would be a proof that gazebos exist, using ZF + indpt-hypothesis. Thus making it is useless to look for one. > Nice! And nicely sneaky!! Yes, I have long been advocating prople continue to use AC (and CH, and Suslin etc - anything now proved independent of ZF), in order to circumscribe the limits of what ZF can do. It's nice, though I wouldn't necessarily call it sneaky. An example of the direct form, would be that ~CH proves that there exists a non-continuum uncountable subset of Z; but as the hypothesis is independent, it is useless to look for and hope to find any such actual set! (Strictly speaking, it is only useless to come up with a set AND a proof that it is unctble-cntuum, but the philosophical or practical difference is minuscule.) -- Wriggling William
From: David C. Ullrich on 27 Nov 2009 06:56 On Thu, 26 Nov 2009 16:34:58 -0500, David Bernier <david250(a)videotron.ca> wrote: >Virgil wrote: >> In article<m9atg5h0jvcun23n7oj9dq286ut82i9ti2(a)4ax.com>, >> David C. Ullrich<dullrich(a)sprynet.com> wrote: >> >>> On Thu, 26 Nov 2009 01:27:20 -0800 (PST), Butch Malahide >>> <fred.galvin(a)gmail.com> wrote: >>> >>>> On Nov 25, 10:05 pm, Bill Taylor<w.tay...(a)math.canterbury.ac.nz> >>>> wrote: >>>>> >>>>> Consider subsets of the reals such that both the subset and >>>>> its complement are continuum many in every interval. >>>>> >>>>> Are any two such subsets order-isomorphic? [Spelling typo corrected.] >>>> >>>> No, but the counterexample won't impress you, because it uses AC. >>> >>> A counterexample that depends on AC _should_ be of interest >>> even to people who don't "accept" AC! If AC implies that >>> there is no gazebo then it follows, whether one accepts >>> AC or not, that the existence of a gazebo cannot be proved >>> in ZF. >> >> Nice! And nicely sneaky!! > >I'm not sure what the gazebo-property is here... >Under ZF + AC something was shown to exist. I'm not sure whether what's below is a comment or a question, nor what the question is. But note in any case I wasn't paying any attention to the specific question dealt with in this thread - I was commenting on the statement "No, but the counterexample won't impress you, because it uses AC. Since you're not going to accept the example anyway, I'll just sketch the argument." pointing out that examples "constructed" using AC are nonetheless far from irrelevant to a hypothetical being who does not "accept" AC. >So one reading is that what I'll call the gazebo* property for a set X >is this: > >gazebo*(X): >X = (A, B) , A union B is Reals, A and B are disjoint, >A \intersect [a, b] has cardinality 2^(aleph_0) , any a< b Reals, >B \intersect [a, b] has cardinality 2^(aleph_0) , any a< b Reals, >and >[ Exists f: A -> B a bijection, and f order-preserving > under canonical '<' .] > >Then, in ZFC, there is an X with the gazebo* property, or ? ... > >David Bernier > > >>>> Since you're not going to accept the example anyway, I'll just sketch >>>> the argument. >>>> >>>> Recall that a subset of R is "totally imperfect" if it contains no >>>> perfect set (and therefore no uncountable Borel set of any type.) Use >>>> AC to obtain a so-called Bernstein decomposition of R, i.e., a totally >>>> imperfect set A whose complement B = R\A is also totally imperfect. >>>> (Needless to say, they are nonmeasurable sets.) Of course each of the >>>> sets A and B contains continuum many points in every interval. As far >>>> as I know, A and B may be order-isomorphic. Now, let C be the Cantor >>>> set, and consider the set X = A union C and its complement Y = B\C. >>>> The modified sets X and Y still contain continuum many points in each >>>> interval. I claim that X and Y are not isomorphic. To see this, >>>> observe that X contains a subset which is order-isomorphic to R (in >>>> fact C contains such a subset), while Y, being totally imperfect, >>>> contains no subset order-isomorphic to R. >>> >>> David C. Ullrich >>> >>> "Understanding Godel isn't about following his formal proof. >>> That would make a mockery of everything Godel was up to." >>> (John Jones, "My talk about Godel to the post-grads." >>> in sci.logic.) David C. Ullrich "Understanding Godel isn't about following his formal proof. That would make a mockery of everything Godel was up to." (John Jones, "My talk about Godel to the post-grads." in sci.logic.)
From: David Bernier on 27 Nov 2009 08:53 David C. Ullrich wrote: > On Thu, 26 Nov 2009 16:34:58 -0500, David Bernier > <david250(a)videotron.ca> wrote: > >> Virgil wrote: >>> In article<m9atg5h0jvcun23n7oj9dq286ut82i9ti2(a)4ax.com>, >>> David C. Ullrich<dullrich(a)sprynet.com> wrote: >>> >>>> On Thu, 26 Nov 2009 01:27:20 -0800 (PST), Butch Malahide >>>> <fred.galvin(a)gmail.com> wrote: >>>> >>>>> On Nov 25, 10:05 pm, Bill Taylor<w.tay...(a)math.canterbury.ac.nz> >>>>> wrote: >>>>>> >>>>>> Consider subsets of the reals such that both the subset and >>>>>> its complement are continuum many in every interval. >>>>>> >>>>>> Are any two such subsets order-isomorphic? [Spelling typo corrected.] >>>>> >>>>> No, but the counterexample won't impress you, because it uses AC. >>>> >>>> A counterexample that depends on AC _should_ be of interest >>>> even to people who don't "accept" AC! If AC implies that >>>> there is no gazebo then it follows, whether one accepts >>>> AC or not, that the existence of a gazebo cannot be proved >>>> in ZF. >>> >>> Nice! And nicely sneaky!! >> >> I'm not sure what the gazebo-property is here... >> Under ZF + AC something was shown to exist. > > I'm not sure whether what's below is a comment or a > question, nor what the question is. I think I understand your point now. So I've snipped out the part you refer to above. > But note in any case I wasn't paying any attention to > the specific question dealt with in this thread - I was > commenting on the statement > > "No, but the counterexample won't impress you, because it uses AC. > Since you're not going to accept the example anyway, I'll just sketch > the argument." > > pointing out that examples "constructed" using AC are > nonetheless far from irrelevant to a hypothetical > being who does not "accept" AC. [...] If there is a proof in ZFC of a sentence T, then there is no proof in ZF of the negation of T. Proof by contradiction: Suppose there is a proof in ZF of the negation of T. Then there is a proof in ZFC of the negation of T. By hypothesis, there is a proof in ZFC of T. So there is a proof in ZFC of (T and not(T)). But ZFC is consistent, so it can't be that ZFC has a proof of (T and not(T)). So we conclude that "there is a proof in ZF of the negation of T" is false. So there is _no_ proof in ZF of the negation of T. QED In my proof above, I used as an assumption that ZFC is consistent. I read somewhere: "In 1963, Paul Cohen showed that AC couldn't be proved with the ZF axioms." And also: "In 1940, Goedel proved that AC couldn't be disproved with the ZF axioms." Recently, I've been wondering in which formal systems the Independence proofs of Goedel and Cohen can be carried out ... If a doubter accepts ZF but no more, would the doubter accept as proved that AC is independent of ZF ? David Bernier [rest of earlier post of mine snipped]
From: Frederick Williams on 27 Nov 2009 09:23
David Bernier wrote: > > ... > > In my proof above, I used as an assumption that > ZFC is consistent. I read somewhere: > "In 1963, Paul Cohen showed that AC couldn't > be proved with the ZF axioms." > > And also: > "In 1940, Goedel proved that AC couldn't be > disproved with the ZF axioms." > > Recently, I've been wondering in which formal > systems the Independence proofs of Goedel and > Cohen can be carried out ... > > If a doubter accepts ZF but no more, would the > doubter accept as proved that AC is > independent of ZF ? Cohen's proof was carried out within ZFC, G\"odel's was carried out within NBG. -- Which of the seven heavens / Was responsible her smile / Wouldn't be sure but attested / That, whoever it was, a god / Worth kneeling-to for a while / Had tabernacled and rested. |