Dense Linear Ordering problems.
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From: David C. Ullrich on 28 Nov 2009 08:24 On Fri, 27 Nov 2009 08:53:56 -0500, David Bernier <david250(a)videotron.ca> wrote: >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. That's most of my point, but an important bit is missing. You only need to assume that ZF is consistent, because the consistency of ZFC _follows_, by the result of Godel you mention below. (It's not clear to me whether that was clear to you or not...) That's "important" in regard to "(assuming of course that ZF is consistent) if you can prove something in ZFC then you can't disprove the same thing in ZF" - a person who doesn't "accept" AC would a priori be more willing to assume the constency of ZF than of ZFC, but in fact that two are equivalent. >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 ? Yes. (Well, people can "doubt" anything...) >David Bernier > >[rest of earlier post of mine snipped] 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 4 Dec 2009 17:52 David C. Ullrich wrote: > On Fri, 27 Nov 2009 08:53:56 -0500, David Bernier > <david250(a)videotron.ca> wrote: [...] >> 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. > > That's most of my point, but an important bit is missing. > You only need to assume that ZF is consistent, because > the consistency of ZFC _follows_, by the result of > Godel you mention below. (It's not clear to me whether > that was clear to you or not...) No, it wasn't clear to me. And I still don't grasp "V = L" . David Bernier > That's "important" in regard to "(assuming of course > that ZF is consistent) if you can prove something in > ZFC then you can't disprove the same thing in ZF" - > a person who doesn't "accept" AC would a priori > be more willing to assume the constency of ZF than > of ZFC, but in fact that two are equivalent. > >> 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 ? > > Yes. > > (Well, people can "doubt" anything...) > >> David Bernier >> >> [rest of earlier post of mine snipped] > > David C. Ullrich
From: Aatu Koskensilta on 4 Dec 2009 17:54 David Bernier <david250(a)videotron.ca> writes: > And I still don't grasp "V = L" . What's unclear to you about "V = L"? -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: David Bernier on 5 Dec 2009 00:45 Aatu Koskensilta wrote: > David Bernier <david250(a)videotron.ca> writes: > >> And I still don't grasp "V = L" . > > What's unclear to you about "V = L"? > I had a look at the article: < http://en.wikipedia.org/wiki/Constructible_universe > . Suppose we have the function d: N -> N d(n) := n'th digit of pi (with d(0):= 3 ). Viewed as a relation on NxN, for each n, (n, d(n)) is in L_omega. Then say A = { (n, d(n)) , such that n in N }. Probably A lies in some L_alpha, alpha not a large ordinal. I haven't thought much about this. The difficulty for me is in finding a first-order formula with bounded quantifiers that defines A. David Bernier
From: David C. Ullrich on 5 Dec 2009 08:59
On Fri, 04 Dec 2009 17:52:38 -0500, David Bernier <david250(a)videotron.ca> wrote: >David C. Ullrich wrote: >> On Fri, 27 Nov 2009 08:53:56 -0500, David Bernier >> <david250(a)videotron.ca> wrote: >[...] > >>> 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. >> >> That's most of my point, but an important bit is missing. >> You only need to assume that ZF is consistent, because >> the consistency of ZFC _follows_, by the result of >> Godel you mention below. (It's not clear to me whether >> that was clear to you or not...) > > >No, it wasn't clear to me. And I still don't >grasp "V = L" . What I meant was it wasn't clear to me whether or not you were aware that Godel _had_ in fact proved that the consistency of ZF implies that of ZFC, not whether the proof was clear. Regarding V and L, http://en.wikipedia.org/wiki/Constructible_universe seems somewhat clear to me. >David Bernier > > > >> That's "important" in regard to "(assuming of course >> that ZF is consistent) if you can prove something in >> ZFC then you can't disprove the same thing in ZF" - >> a person who doesn't "accept" AC would a priori >> be more willing to assume the constency of ZF than >> of ZFC, but in fact that two are equivalent. >> >>> 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 ? >> >> Yes. >> >> (Well, people can "doubt" anything...) >> >>> David Bernier >>> >>> [rest of earlier post of mine snipped] >> >> David C. Ullrich 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.) |