The Definition of Points
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From: Virgil on 18 Apr 2007 17:34 In article <46266bb2(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Mike Kelly wrote: > > On 18 Apr, 06:17, Tony Orlow <t...(a)lightlink.com> wrote: > >> Mike Kelly wrote: > >> > >> < snippery > > >> > >>> You've lost me again. A bad analogy is like a diagonal frog. > >>> -- > >>> mike. > >> Transfinite cardinality makes very nice equivalence classes based almost > >> solely on 'e', > > > > Why "almost solely"? > > > > The successor relation is a relation not related to 'e' except in a > particular model of the naturals upo which the rest is built. Can TO define or construct anything like successorship in any set theory which doe not depend, even indirectly, on 'e'? > > >> but in my opinion doesn't produce believable results. > > > > Which "results" of cardinality are not believable? You don't think the > > evens and the naturals and the rationals are mutually bijectible? Or > > you do think the reals and the naturals are bijectible? Or did you > > mean you have a problem not with what cardinality actually is but with > > what you hallucinate cardinality to be? > > > > As long as transfinite "equivalence" classes are referred to as such, > and not said to denote "equal" size or numerosity, then I have no problem. If we choose to define "size" and "equinumerosity' in terms of injections and bijections, then TO's problems are purely his own, and need not affect anyone else. > > >> I'm working on a better theory, bit by bit. Of which TO so far has zip. > > > > You have never managed to articulate a problem with what cardinality > > actually says (all it says is which sets are bijectbile). You've never > > managed to explain the motivation behind your "better theory". I don't > > even know whether your ideas are supposed to be formulated in ZF or in > > something else, possibly a new foundation of your own devising(or, > > perhaps, never formalised at all?). FWIW I don't think it would be > > very hard to define, say, "density in the naturals" in ZF. I just > > don't see the motivation behind doing so. > > > > The idea is to create a method that works for comparing countable and > uncountable sets alike, and properly defining the relationship between > the two. That implies, contrary to fact, that has already been done is improper. > I suppose it's more the way it's spoken about. Leave out "size" and > "equinumerous" and talk about "equivalence classes" and not "equality", > and I have no problem with the consistency of standard theory. Since we define size and equinumerosity in terms of injections and bijections, TO will just have to learn to deal with his problems on his own.
From: Mike Kelly on 18 Apr 2007 17:42 On 18 Apr, 20:13, Tony Orlow <t...(a)lightlink.com> wrote: > MoeBlee wrote: > > On Apr 17, 10:02 am, Tony Orlow <t...(a)lightlink.com> wrote: > > >> I am saying it's obvious that any countable set has a well ordering. It > >> is not obvious for uncountable sets. > > > It's not only obvious, but it's trivial to prove that every countable > > set has a well ordering. > > > So, just to be sure you understand, what is at stake with an axiom of > > countable choice is not at all that every countable set has a well > > ordering (since we don't need any choice principle to prove that) but > > rather that for any countable set there is a function that chooses > > exactly one member from every nonempty subset of that countable set > > (and actually, we need concern ourselves only with denumerable sets, > > since we don't need a choice principle to prove that for every finite > > set there is a function that chooses exactly one member from every > > nonempty subset of the finite set). > > > MoeBlee > > It's not trivial to prove any such thing for an uncountable set. That > requires full Choice, does it not? In fact, it was Virgil, I believe, > who said that was the MOTIVATION behind Zermelo's formulation of Choice, > so that ALL sets could be considered well orderable. It is my understanding (I may be mistaken) that Zermelo noticed when formalising his proof that he was assuming one could choose an element from an arbitrary set, and this was not justified by just ZF. So he proposed Axiom of Choice to make his proof work. It was also noticed that a number of previous theorems used AoC implicitly. So.. in a sense AoC was "motivated" by wanting a proof for the wellordering theorem (to allow transfinite induction on all sets). But really Zermelo's was more of an acknowledgement that mathematicians had been using AoC implicitly already and that it was needed for proving wellordering. > However, while I see that a choice function produces a well ordering for any countable set, Wellordering for countable sets doesn't need choice at all. Think about it a little.. > I don't see that it does for an uncountable set. Well... have you looked at the proof? It's not trivial, really.. -- mike.
From: Virgil on 18 Apr 2007 17:46 In article <46266dd9(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > MoeBlee wrote: > > On Apr 17, 10:02 am, Tony Orlow <t...(a)lightlink.com> wrote: > > > >> I am saying it's obvious that any countable set has a well ordering. It > >> is not obvious for uncountable sets. > > > > It's not only obvious, but it's trivial to prove that every countable > > set has a well ordering. > > > > So, just to be sure you understand, what is at stake with an axiom of > > countable choice is not at all that every countable set has a well > > ordering (since we don't need any choice principle to prove that) but > > rather that for any countable set there is a function that chooses > > exactly one member from every nonempty subset of that countable set > > (and actually, we need concern ourselves only with denumerable sets, > > since we don't need a choice principle to prove that for every finite > > set there is a function that chooses exactly one member from every > > nonempty subset of the finite set). > > > > MoeBlee > > > > It's not trivial to prove any such thing for an uncountable set. That > requires full Choice, does it not? In fact, it was Virgil, I believe, > who said that was the MOTIVATION behind Zermelo's formulation of Choice, > so that ALL sets could be considered well orderable. However, while I > see that a choice function produces a well ordering for any countable > set, I don't see that it does for an uncountable set. TO's selective blindness strikes again. A function which assigns to every nonempty subset of a given set a particular element in the set, which is what the AOC states exists,IS a well ordering of that set that set. One merely identifies as the first element of each set the value of that function. > > Maybe I am confused about something. To is confused about almost everything. > Can a well ordering include an > uncountable number of limit elements? Why not? All any well ordering of any set requires is that every non-empty subset of that set have a first member under that ordering. > I just don't see that a well order > can be accomplished in principle on an uncountable set, whether one > declares an axiom to that effect of not. There is a critical difference in mathematics between proving something exists and instantiating it. Given the AOC, one can trivially prove that for every set there exists a well ordering. Actually constructing an explicit well ordering may be a whole different ball of wax.
From: stephen on 18 Apr 2007 17:51 In sci.math Tony Orlow <tony(a)lightlink.com> wrote: > stephen(a)nomail.com wrote: >> In sci.math Tony Orlow <tony(a)lightlink.com> wrote: >>> Mike Kelly wrote: >>>> The point that it DOESN'T MATTER whther you take cardinality to mean >>>> "size". It's ludicrous to respond to that point with "but I don't take >>>> cardinality to mean 'size'"! >>>> >>>> -- >>>> mike. >>>> >> >>> You may laugh as you like, but numbers represent measure, and measure is >>> built on "size" or "count". >> >> What "measure", "size" or "count" does the imaginary number i represent? Is i a number? >> The word "number" is used to describe things that do not represent any sort of "size". >> >> Stephen > Start with zero: E 0 > Define the naturals: Ex -> Ex+1 > Define the integers: Ex -> Ex-1 > Define imaginary integers: Ex -> sqrt(x) > i=sqrt(0-(0+1)), so it's built from 0 and 1, using three operators. It's > compounded from the naturals. That does not answer the question of what "measure", "size" or "count" i represents. And it is wrong on other levels as well. You just pulled "sqrt" out of the air. You did not define it. Claiming that it is a primitive operator seems a bit like cheating. And if I understand your odd notation, the sqrt(2) is an imaginary integer according to you? And sqrt(4) is also an imaginary integer? You also have to be careful about about claiming that i=sqrt(-1). It is much safer to say that i*i=-1. If you do not see the difference, maybe you should explore the implications of i=sqrt(-1). So what is wrong with Start with zero: E 0 Define the naturals: Ex -> Ex+1 Define omega: Ax I did that using only one operator. > A nice picture of i is the length of the leg of a triangle with a > hypotenuse of 1 and a leg of sqrt(2), if that makes any sense. It's kind > of like the difference between a duck. :) That does not make any sense. There is no point in giving a nonsensical answer, unless you are aiming to emulate Lester. Stephen
From: MoeBlee on 18 Apr 2007 18:17
P.S. This looks like a good article stressing some of the history: http://citeseer.ist.psu.edu/cache/papers/cs/21901/http:zSzzSzwww.math.ucla.eduzSz~aslzSzbslzSz0303zSz0303-001.pdf/kanamori97mathematical.pdf MoeBlee |