Cantor Confusion
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From: Andy Smith on 12 Jan 2007 03:07 > >> > >> >> a(i+1) > a(i) for all i<last i > >> > > You asked that a(i+1) > a(i) for all i < last i, and > indeed it is so. > Your criterion doesn't mention anything about the > last i having a > predecessor. > Let j = (last i) -1 then it requires a(j+1) > a(j) or a(last i) > a((last(i) - 1). But anyway, this is all just niggling. I will have a look at your other post. > U.S. Court of Appeals to review three issues > concerning case of Mumia Abu-Jamal. > <http://www.mumia2000.org/> yes, I looked at that. from this side of the pond the US justice system appears Kafka-esq, but I wouldn't necessarily rely on the common sense of the legal system here either. If you got rid of the death penalty, that would be a start ...
From: Andy Smith on 12 Jan 2007 03:33 > > >> The crossings are totally ordered but they are not > >> well ordered, nor is > >> the reverse ordering well ordered. > > > > How does one define "well ordered" as distinct from > > "totally ordered", please? > > A set is totally ordered if it satisfies the > trichotomy law. That is, if > x and y are members of the set, then exactly one of > the following > statements holds: > > (1) x < y, > (2) x > y, > (3) x = y. > OK, fair enough. (how could a pair of elements be otherwise if there as a well defined function f(x) of each element on which the elements are sorted ?). > A set is well ordered if every nonempty subset has a > least member. Your > set of crossings is not well ordered because the > subset of crossings for > x > 0 does not have a least element. > But presumably it is acceptable if every nonempty subset does not have a greatest member? In the context of this discussion, that is a bit circular, isn't it? I am sure that it is incompatible with Peano to have a countably infinite ordered set with finite boundaries. But the set of zero crossings exist as an ordered structure as a counter example. Of course you can pull out your (carefully self consistent) rule book & rule it off-side, but that is maybe the point? > Every ordinal is well ordered. Each natural number, > being a finite > ordinal, is likewise well ordered. The set of all > natural numbers is > identical to the ordinal w and is therefore well > ordered. Every well > ordered set is necessarily totally ordered. > > Some examples of totally ordered sets that are not > well ordered are: > > (1) Your set of crossings. > (2) The reals in [0,1]. > (3) The rationals in [0,1]. > (4) The integers. > > >> The fact that two ordered sets have the same > >> cardinality does not imply > >> that they have the same order type. You need to > >> describe more carefully > >> how you are associating those two sets, keeping in > >> mind that each bit > >> position of an integer is finite. > > > Well the point I was suggesting was that the > infinite sequence > > of crossings is an ordered set, with a start and an > end > > and even if one can't label each term with an > ascending > > integer, their locations exist on the real line as > > much as any other fairyland construction. (and, > hence, maybe, other > > infinite sequences e.g. 11...11 with a beginning > and an end might > > exist and have some meaning?) > > Certainly the crossings exist. That has never been > in dispute. > > You are free to write digit sequences such as > 11...11, but they have no > particular meaning as numbers until you explain what > you mean by them. > > > What can you say then about an ordered countably > infinite set that > > has a well defined beginning and an end - or does > the fact that it > > have a beginning and an end automatically > disqualify > > it from being a legitimate object? > > Anything that can be rigorously defined is a > legitimate mathematical > object. The set w+1 is an ordered countably infinite > set that has a well > defined beginning and an end. > But not a well defined end? Like pointing to e.g. some definition of a number on the real line, and saying "there it is". Can you give any concrete example of an actually realised set of transfinite numbers? (bearing in mind that I am an earthy physicist, I want some explicit points that I can see on a calculator, making up the transfinite set). I don't buy the set of all ratioanls in [0,1] as meeting my criteria of a well ordered infinite sequence with finite boundaries, back to 3 wishes again. The set of rationals is not acceptable because there is no last member as a predecessor to 1 (whereas, with my zero crossings, there exists a last one, its predecessor, that ones predecessor, etc ...) --- Andy
From: Dave Seaman on 12 Jan 2007 13:46 On Fri, 12 Jan 2007 13:07:56 EST, Andy Smith wrote: >> >> >> >> >> a(i+1) > a(i) for all i<last i >> >> >> >> You asked that a(i+1) > a(i) for all i < last i, and >> indeed it is so. >> Your criterion doesn't mention anything about the >> last i having a >> predecessor. >> > Let j = (last i) -1 Ordered sets don't necessarily have subtraction defined on them, and not everything needs to have a predecessor. > then it requires > a(j+1) > a(j) > or > a(last i) > a((last(i) - 1). > But anyway, this is all just niggling. > I will have a look at your other post. >> U.S. Court of Appeals to review three issues >> concerning case of Mumia Abu-Jamal. >> <http://www.mumia2000.org/> > yes, I looked at that. from this side of the pond the US > justice system appears Kafka-esq, but I wouldn't necessarily > rely on the common sense of the legal system here either. > If you got rid of the death penalty, that would be a start ... Agreed. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: Dave Seaman on 12 Jan 2007 14:07 On Fri, 12 Jan 2007 13:33:44 EST, Andy Smith wrote: >> >> The crossings are totally ordered but they are not >> >> well ordered, nor is >> >> the reverse ordering well ordered. >> > How does one define "well ordered" as distinct from >> > "totally ordered", please? >> A set is totally ordered if it satisfies the >> trichotomy law. That is, if >> x and y are members of the set, then exactly one of >> the following >> statements holds: >> (1) x < y, >> (2) x > y, >> (3) x = y. > OK, fair enough. (how could a pair of elements be otherwise > if there as a well defined function f(x) of each > element > on which the elements are sorted ?). And what if there isn't? For example, consider the set of all subsets of N, ordered by inclusion. This is a partial order. It's not a total order, because it doesn't satisfy the trichotomy law. For example, if E = set of even numbers and P = set of primes, then neither is a subset of the other and they are not equal. >> A set is well ordered if every nonempty subset has a >> least member. Your >> set of crossings is not well ordered because the >> subset of crossings for >> x > 0 does not have a least element. > But presumably it is acceptable if every nonempty > subset does not have a greatest member? In the context > of this discussion, that is a bit circular, isn't it? The definition of a well ordering does not say anything about greatest members. How can failure to mention something possibly make a definition circular? > I am sure that it is incompatible with Peano to have > a countably infinite ordered set with finite boundaries. Then you are wrong, because the rationals in [0,1] are such a set. > But the set of zero crossings exist as an ordered structure > as a counter example. Of course you can pull out your (carefully > self consistent) rule book & rule it off-side, but that is > maybe the point? When have I ever said there was anything wrong with your set of crossings? You are simply not listening. If I say a red ball is not blue, am I finding fault with the ball? >> Every ordinal is well ordered. Each natural number, >> being a finite >> ordinal, is likewise well ordered. The set of all >> natural numbers is >> identical to the ordinal w and is therefore well >> ordered. Every well >> ordered set is necessarily totally ordered. >> Some examples of totally ordered sets that are not >> well ordered are: >> (1) Your set of crossings. >> (2) The reals in [0,1]. >> (3) The rationals in [0,1]. >> (4) The integers. >> >> The fact that two ordered sets have the same >> >> cardinality does not imply >> >> that they have the same order type. You need to >> >> describe more carefully >> >> how you are associating those two sets, keeping in >> >> mind that each bit >> >> position of an integer is finite. >> > Well the point I was suggesting was that the >> infinite sequence >> > of crossings is an ordered set, with a start and an >> end >> > and even if one can't label each term with an >> ascending >> > integer, their locations exist on the real line as >> > much as any other fairyland construction. (and, >> hence, maybe, other >> > infinite sequences e.g. 11...11 with a beginning >> and an end might >> > exist and have some meaning?) >> Certainly the crossings exist. That has never been >> in dispute. >> You are free to write digit sequences such as >> 11...11, but they have no >> particular meaning as numbers until you explain what >> you mean by them. >> > What can you say then about an ordered countably >> infinite set that >> > has a well defined beginning and an end - or does >> the fact that it >> > have a beginning and an end automatically >> disqualify >> > it from being a legitimate object? >> Anything that can be rigorously defined is a >> legitimate mathematical >> object. The set w+1 is an ordered countably infinite >> set that has a well >> defined beginning and an end. > But not a well defined end? When I say something has an end, that means it has a well defined end. What other kind of end could it have? The last element of w+1 is w. >Like pointing to e.g. > some definition of a number on the real line, and saying > "there it is". Can you give any concrete example of > an actually realised set of transfinite numbers? (bearing > in mind that I am an earthy physicist, I want some > explicit points that I can see on a calculator, making up > the transfinite set). Can you give any concrete example of an actually realised number 3? I don't mean an example of three objects, but an instance of the actual number 3. > I don't buy the set of all ratioanls in [0,1] as meeting > my criteria of a well ordered infinite sequence with > finite boundaries, back to 3 wishes again. The set of > rationals is not acceptable because there is no last member > as a predecessor to 1 (whereas, with my zero crossings, All that proves is that you failed to ask for what you really wanted. We have been talking in circles because you have never specified just what it was that you wanted to know. Perhaps this will help: we say a set X is *finite* if X can be well ordered in such a way that each nonempty subset of X has a greatest element. A different way of saying it is that a set X is finite if it can be placed in bijection with some natural number. It's fairly easy to show by induction that these two notions of finiteness are equivalent. > there exists a last one, its predecessor, that ones predecessor, etc ...) -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: Virgil on 12 Jan 2007 14:53
In article <32631958.1168598767888.JavaMail.jakarta(a)nitrogen.mathforum.org>, Andy Smith <andy(a)phoenixsystems.co.uk> wrote: > > > So you can have an infinite set with a first and > > last member? > > > > Sure. Consider the following sets of real numbers: > > > > {x| 0 <= x <= 1} > > {x| 0 < x <= 1} > > {x| 0 < x < 1} > > > > Perhaps I should have asked, can you have a "countably > infinite" ordered set with a first and last member? The reciprocals of non-zero integers, for example. -1 is first and 1 is last in the usual rational number ordering. Or the set of rationals between -1 and 1 inclusive. |