Cantor Confusion
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From: Andy Smith on 13 Jan 2007 06:06 Thanks for the education all. Anyway, re the set of zero crossings of sin(pi/x). Ok, the set including subset of zero crossings in x>0, x<=1 while ordered is not "well-ordered" according to set theory because it contains no least member. But the subset in the closed interval x>=-1,x<=0 is, and we can define a set as that subset? Each element of that well defined set has an attribute x, its real crossing point, and is monotonically ordered on that attribute such that x(m)>x(n) for m>n. This set has the cardinality of N (can be placed in a 1:1 correspondence with the natural numbers)? It has the property that it is countably infinite; you cannot count from the first to the last element. However there exists a last member of the set. So, if one wanted to, one could define the natural numbers in such a way that there was (or was not) a greatest member of the natural numbers, as a matter of axiomatic choice, without affecting other properties of the natural numbers?
From: David Marcus on 13 Jan 2007 17:29 Andy Smith wrote: > Anyway, re the set of zero crossings of sin(pi/x). > > Ok, the set including subset of zero crossings in x>0, x<=1 > while ordered is not "well-ordered" according to set theory > because it contains no least member. Not quite. Definition: An ordered set X is "well ordered" if every non- empty subset has a smallest element. > But the subset in the closed interval x>=-1,x<=0 is, and > we can define a set as that subset? You can pretty much define any set you like. There are some restrictions, but you aren't likely to run into them here. However, the set you defined is not well ordered. > Each element of that well defined set has an attribute x, its real > crossing point, You are distinguishing things that are the same. A crossing point is a real number. > and is monotonically ordered on that > attribute such that x(m)>x(n) for m>n. What are m and n? > This set has the cardinality of N (can be placed in a 1:1 > correspondence with the natural numbers)? Yes. > It has the property that it is countably infinite; Yes, although redundant. > you > cannot count from the first to the last element. Don't know what this means. > However > there exists a last member of the set. The set has a largest element where "largest" is referring to the ordering as a subset of the real numbers. This ordering has nothing to do with the bijection with N. > So, if one wanted to, one could define the natural numbers > in such a way that there was (or was not) a greatest member > of the natural numbers, as a matter of axiomatic choice, > without affecting other properties of the natural numbers? One can define the "natural numbers" to be pink butterflies, if you wish. But, I don't know what you mean by "without affecting other properties of the natural numbers". Seems to me that any set with a largest element isn't going to look much like the natural numbers at all. -- David Marcus
From: Andy Smith on 13 Jan 2007 08:52 David Marcus wrote: ... > > One can define the "natural numbers" to be pink > butterflies, if you > wish. But, I don't know what you mean by "without > affecting other > properties of the natural numbers". Seems to me that > any set with a > largest element isn't going to look much like the > natural numbers at > all. > > I can't argue with you on your terms, don't have the grounding. And I am doubtless just wasting your time. In layman's language (mine) I just wanted to express the thoughts: 1) the basic property of the natural numbers, starting from 0, is that they increase one after the other without limit. That is basically what Peano says, plus an axiom of induction to keep everything unique (as I understand it). 2) I was previously of the opinion that you couldn't have a countable infinite monotonically increasing sequence that had a greatest member. 3) But you can, e.g. the infinite succession of zero crossings in [-1,0] of sin(pi/x) which has a last member at 0.00.. 4) So there is no illogicality in positing that the natural numbers could also increase without limit, but still assert that there is a maximum (infinite) number? Of course that is not kosher re Peano, because Peano defines a natural number as one that has a successor. But one might as well say that all numbers have a predecessor, except 0. So it isn't that unreasonable to regard both first and last numbers as having special properties (one without predecessor, one without successor) ? The idea of a greatest (infinite) number is distinct as I understand it from "omega". Numbers are viewed as sets of elements, omega is the set of all of {},{{}} etc. i.e. omega is in a different category to a hypothetical infinite greatest number.
From: David Marcus on 13 Jan 2007 19:54 Andy Smith wrote: > David Marcus wrote: > > One can define the "natural numbers" to be pink > > butterflies, if you > > wish. But, I don't know what you mean by "without > > affecting other > > properties of the natural numbers". Seems to me that > > any set with a > > largest element isn't going to look much like the > > natural numbers at > > all. > > I can't argue with you on your terms, don't have the grounding. > And I am doubtless just wasting your time. It is my time to waste! Actually, the discussion is rather interesting. Clearly, there is something about the way we do mathematics that is confusing you, but I can't say I can quite figure out what it is. > In layman's language (mine) I just wanted to express the thoughts: > > 1) the basic property of the natural numbers, starting > from 0, is that they increase one after the other without limit. > That is basically what Peano says, plus an axiom of induction > to keep everything unique (as I understand it). I wouldn't say that. The natural numbers are a specific thing. The Peano axioms characterize this thing in the sense that if you have two things that satisfy the Peano axioms, then the two are basically the same--the only difference is that you've changed the names of the elements in one of the things to the names in the other. > 2) I was previously of the opinion that you couldn't > have a countable infinite monotonically increasing > sequence that had a greatest member. Even layman's language must be precise or you will only confuse yourself. You can have a countably infinite ordered set that has a largest element. You can't have a sequence with a last element. > 3) But you can, e.g. the infinite succession of zero crossings > in [-1,0] of sin(pi/x) which has a last member at 0.00.. What do you mean by "succession"? And, I don't think that is a good example, since sin(pi/0) isn't zero (it isn't defined). > 4) So there is no illogicality in positing that the natural > numbers could also increase without limit, but still > assert that there is a maximum (infinite) number? It is illogical in the sense that what you are positing is false. The natural numbers are unbounded, every natural number is finite, and there is no maximum natural number. > Of course that is not kosher re Peano, because Peano defines a natural number > as one that has a successor. The Peano axioms do not define the natural numbers. The Peano axioms characterize the natural numbers. The problem isn't with Peano. The problem is that the natural numbers don't have the properties you are "positing" for them. Notice that the phrase "natural numbers" refers to a specific thing. Just as "Andy Smith" does. I could posit that Andy Smith is twenty feet tall, but presumably that is false. > But one might as well say that all numbers have a predecessor, except 0. All natural numbers except zero do have a predecessor. > So > it isn't that unreasonable to regard both first and last > numbers as having special properties (one without predecessor, > one without successor) ? Except there is no last natural number. That's like saying we could regard your second head as having a special property. > The idea of a greatest (infinite) number Do you mean the idea you have tried to explain? The natural numbers do not have a largest number. That's basic to our concept of the natural numbers as what you get if you keep adding one. > is distinct as > I understand it from "omega". Numbers are viewed as > sets of elements, omega is the set of all of {},{{}} etc. In set theory, we define the ordinals. It turns out that the finite ordinals satisfy the Peano axioms. This means that we can consider the "natural numbers" to be the set of finite ordinals. Mathematicians like to do this because the Peano axioms become theorems (the fewer axioms the better). This doesn't mean the natural numbers really are the finite ordinals (whatever "really are" means). As for omega, it is the smallest infinite ordinal. Ordinals are the order types of sets that are well ordered. Consider 0 1 2 3 ... w where w stands for omega, the natural numbers and omega are listed in order, and we stick all the missing natural numbers where the dots are. Then w is greater than every natural number and the set {0,1,2,...,w} is well ordered (something you might like to convince yourself of). > i.e. omega is in a different category to a hypothetical > infinite greatest number. Not sure what you mean by "hypothetical". Do you mean non-existent? -- David Marcus
From: Andy Smith on 13 Jan 2007 11:41
David Marcus wrote: > > > 2) I was previously of the opinion that you > couldn't > > have a countable infinite monotonically increasing > > sequence that had a greatest member. > > Even layman's language must be precise or you will > only confuse > yourself. You can have a countably infinite ordered > set that has a > largest element. You can't have a sequence with a > last element. > > > 3) But you can, e.g. the infinite succession of > zero crossings > > in [-1,0] of sin(pi/x) which has a last member at > 0.00.. > > What do you mean by "succession"? And, I don't think > that is a good > example, since sin(pi/0) isn't zero (it isn't > defined). > Isn't sin(pi/0) defined by the antisymmetry of the function? Surely it must cross the x-axis at x=0 ? And actually, that is rather the point about this example that is different from others. You can have a countably infinite sequence converging on some limit point e.g. {1/2,3/4,7/8, ...} but you could not then say add the value {2} as the last member of the sequence and claim that it was infinite and with a maximum member. But isn't the zeo crossing at x=0 of sin(pi/x) subtly different? even though the function is oscillating infinately fast at x=0, it is continuous, and the zero crossing at x=0 is attached by the function itself (like a piece of string) to the preceeding zero crossings, even if you can't identify the number of the last crossing before x = 0.0...? So the sequence of zero crossings up to and including the point x=0 exist, and are ordered? I'll think about your other comments on Monday, its 2.30 a.m here. Thanks ---- Andy Incidentally, on this web Usenet interface, is there some snappy way to scroll to the bottom of the thread? Takes ages on my machine with this huge thread ... |