Dense Linear Ordering problems.
in [Math]
|
Prev: Is this a valid statement?
Next: lagrange identity
From: David Bernier on 6 Dec 2009 11:22 David C. Ullrich wrote: > On Sat, 05 Dec 2009 10:44:24 -0500, David Bernier > <david250(a)videotron.ca> wrote: > >> David C. Ullrich wrote: >>> On Sat, 05 Dec 2009 00:45:15 -0500, David Bernier >>> <david250(a)videotron.ca> wrote: >>> >>>> 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. >>> Well surely you don't want to try to write down an explicit >>> formula for this in the language of set theory - that would be >>> very long and incomprehensible. >>> >>> There _are_ first-order formulas that "say" the following: >>> >>> N(n): n is a natural number >>> Pair(x,y,z): z = (x,y) >>> Pi(n,m): n, m are natural numbers and m is the n-th digit of pi >>> >>> (seems to me the fact that pi is irrational meams that a >>> formula Pi exists that's simpler than, say, G(n,m), >>> saying that m is the n-th digit of Euler's constant gamma - >>> we have an actual algorithm for computing the n-th digit >>> of pi, while.if it appear that the 20-th digit of gamma is 3 >>> and then there's a large number of 9's it could be that they're >>> actually all 9's from that point, so we never find out that >>> the 20-th digit is really 4. So G(n,m) would involve >>> something like "there exists a sequence of digits such >>> that...", while Pi(n,m), it seems to me, only involves >>> quantifying over natual numbers. This has some relevance >>> to whether A is in L_{w+1} or not...) >>> >>> Then it seems to me that A is the set of z in L_w such that >>> >>> En Em (pair(n,m,z) & pi(n,m)) >>> >>> holds in L_w, hence A is in L_{w+1}. >> [...] >> >> Yes. One section in Wikipedia says: >> >> "All arithmetical subsets of w and relations on w belong to L{w+1} >> (because the arithmetic definition gives one in L{w+1})". >> >> If we look at L_{w+1}, then there are countably many formulas phi >> with countably many choices of parameters z_1, ... z_n in L_w . >> >> So I think L_{w+1} is countably infinite if we accept >> >> ZFC to provide bijections. > > Yes. In fact there's no need for AC here. An explicit enumeration > of L_w and an explicit enumeration of the first0order formulas > give an explicit enumeration of L_{w+1}/ That's a nice argument. In ZF, obviously we can still form the power set of N ( another name for omega), to get P(omega). Under the V=L condition/hypothesis, for some ordinal alpha, P(omega) is an element of L_{alpha}. The Wikipedia article states that each L_alpha is transitive, i.e. if A is in L_alpha, and x is an element of A, then x is an element of L_alpha. I think I understand that now. There's a least alpha such that P(omega) e L_alpha. Either alpha is a limit ordinal or not. It's easy to see that the minimal alpha can't be a limit ordinal, so it must be a successor alpha = beta+1. Then P(omega) is in L_{beta+1} and there is a formula defining P(omega) which allows P(omega) to be in L_{beta+1}. Then, by that formula, P(omega) has its elements in L_beta. So, under V=L, P(omega) is in some L_alpha and since L_alpha is transitive, P(omega) is a subset of L_alpha. I don't have any question for now, but I'm contemplating the least ordinal alpha such that P(omega) is an element of L_alpha, under V=L. David Bernier |