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Dedekind's Axiom

  1. Jul 23, 2005 #1
    Not sure if this is the place to ask this. It concerns Dedekind's axiom. Quoting from Dantzig this says:

    "If all points of a straight line fall into two classes, such that every point of the first class lies to the left of any point of the second class, then there exists one and only one point which produces this division of all points into two classes, this severing of the straight line into two portions"

    Two questions -

    1. Is this still a fundamental axiom in some or all forms of mathematics?
    2. How is the inherent self-contradiction resolved?
     
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  3. Jul 23, 2005 #2

    AKG

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    What inherent self-contradiction? Suppose you have the real line, and you divide it into one class (-infinity, 0) and [0, infinity) then 0 is the one and only one point which produces the division.
     
  4. Jul 23, 2005 #3

    Hurkyl

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    This axiom is one of Hilbert's axioms for Euclidean geometry.

    The algebraic version of Dedekind's axiom is part of the definition of the real numbers: it's the complete part of "The real numbers form a complete ordered field".

    And I'll echo AKG: I have no idea what you mean by "inherent self-contradiction".
     
  5. Jul 23, 2005 #4

    selfAdjoint

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    There are various completeness axioms for the real numbers. Dedekind's was AFAIK the first. They all imply each other, so you could say correctly that yes, Dedekind's axiom is still a part of mathematics, either as an axiom itself or a true theorem from other axioms. And Canute, what again is that "inherent contradiction"?
     
  6. Jul 24, 2005 #5
    Thanks, that's very clear. The contradiction I refered to is that in the first part of the axiom, as given, it is stated that all points on the line fall into two classes, while the second part states that there is another point that does not belong in either class. I don't see how both these statements can be true.
     
  7. Jul 24, 2005 #6

    matt grime

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    it doesn't state that the dividing point does not lie in either subset.
     
  8. Jul 24, 2005 #7
    Doesn't it? To me it does. The axiom clearly states that there are two categories of points and that all points belong in one of these two categories. But then it immediately contradicts this rule and clearly states that there is a third category of point. I can't see how there can be both two and three categories of points. What am I misunderstanding here?

    Does it say - there are two mutually exclusive categories of points, but these categories overlap, and where they overlap there is a point that belongs in both categories? This also seems self-contradictory to me. Surely there are three categories of points according to the axiom, however one reads it? :confused:
     
    Last edited: Jul 24, 2005
  9. Jul 24, 2005 #8

    matt grime

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    Because you're misreading it. there is nothing there that states the point of division is in the left or right hand set, or that it cannot be in either. which is not surprising. it states that any divisoin of the real line is of the form

    (-inf,x) [x,inf)

    or

    (-inf,x] (x,inf)

    and that the x is unique.
     
  10. Jul 24, 2005 #9
    So does the dividing point belong in the first class or the second class?
     
  11. Jul 24, 2005 #10

    matt grime

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    That would depend upon the division as i indicated above.
     
  12. Jul 24, 2005 #11
    I'm trying understand this, but genuinely can't grasp how it is possible for the axiom to state that point x is unique having just stated that no point is unique. This problem shows up in your symbolic representation, since this shows point x as being either divided down the middle, in which case it is not a point; in both classes at the same time, in which case the classes are not mutually exclusive after all; or shows two different x's, in which case x is not unique. How can a unique individual be in two classes that are defined as mutually exclusive at the same time? I'm afraid I can't make sense of that yet.
     
  13. Jul 24, 2005 #12

    matt grime

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    what? do you understand the notation?

    (-inf,x) means the x is not included [x,inf) means the x is included. this is one way to divide into two classes

    (-inf,x] and (x,inf) means the x is in the first class and not the second. These are the two options (and the only two).
     
  14. Jul 25, 2005 #13
    Ah yes, I misunderstood that. You probably have no idea how little mathematics some people do! So x can be thought of as being in either class as long as it's in one or the other? But in this case how can it be said that x divides the classes? Say I represent the line of points as 1,2,3,4,5. Say that 1 & 2 are in the first class and 3,4 & 5 are in the second class. Does x = 2 or x =3, or is it an arbitrary choice?

    It seems to me that if x is in the first class then the division between the classes occurs to the right of x, and if x is in the second class then the division is to the left of x. In neither case does x represent the point of division between the classes. Would it be right to say that the axiom is a formal convention and that this ambiguity is therefore not important? Or am I still seeing an ambiguity where there isn't one?
     
  15. Jul 25, 2005 #14

    Zurtex

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    You have chosen an entirely different number line to what Dedekind's Axiom is defined on. Dedekind's Axiom is defined on the real number line and it is a way of helping construct the Real number line which isn't as easy as it sounds.

    Now here we can compare integers to the real number line, say you have the division in the integers: (-inf, 0] and [1, inf) or we could write this as (-inf, 1) and [1, inf) or we could even write this (-inf, 0] and (0, inf). As you can see, there is no single point that is the division.

    However if we look a real number division: (-inf, 1) and [1, inf), as we see here the point 1 is the unique point where there is a division, it's quite important to note here that unlike in the integers there is no other way of writing this.
     
    Last edited: Jul 25, 2005
  16. Jul 25, 2005 #15
    Does it make a difference what sort of line it is? I assumed it is a line of points, seeing as this is what the axiom says it is. Is it not this? If it is a line of points then how we name them is entirely arbitrary. They can be represented a 1,2,3,4,5... or a,b,c,d,e, ... or anything at all. Surely they're just the names we use to distinguish the different points so we can talk about them?

    I'm not worried about formalisms here. Of course within mathematics the axiom is not problematic. My question is meta-mathematical, which I forgot to say earlier, and asks whether the axiom can be applied to, say, a line of apples.
     
  17. Jul 25, 2005 #16

    AKG

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    Think geometrically. Which of the following is a line:

    . . . . . . . . . . .

    or

    ______________

    The first is just a bunch of dots and we'd say that they're in line or they're in a line, but the dots themselves don't form a line. The second thing is a line.
     
  18. Jul 25, 2005 #17

    matt grime

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    Dedekind's axiom exactly distinguishes the real line from, say, the rational numbers, it shows that all convergent sequences have a limit, that is its whole raison d'etre! It doesn't apply to the rationals, or the integers, it is a completeness axiom! For example, the classes: S whcih is the set of positive raitonal numbers whose square is greater than two, and its complement in the rationals are a divisoin of the rationals into two distinct classes alll the elements of one lying to the left of all the elements of the other yet there is no rational number that splits the rationals into those two classes.
     
  19. Jul 25, 2005 #18
    Well, this is just my problem. The axiom states of itself that it applies to a line made of points, or a line of points. Yet a line of points does not seem able to behave in the way the axiom states. If the axiom defines a continuous line by stating that it's a series of points then I will be confused, but assume that the contradiction involved is down to the difficulty of representing the continuous mathematically. Is that it?

    That seems ok to me, since you do not state anywhere that there is a unique rational dividing the two classes. What I can't handle is the word 'all' at the start of the axiom. If all means all then where does the unique dividing point come from?

    In oversimplified terms is the axiom saying that between any two points there is another point? That's my naive reading of it. Or is it not a stand alone axiom, and so perhaps my problem stems from not knowing the others that go with it?

    But maybe I need to understand what an axiom of completeness is. I know this is painfully elementary but could you briefly explain? Sorry to be a bore but I want to understand this. After what's been said I'm not even sure now whether the axiom defines a series of points or a continuum, even though I started out taking it at face value, as the former.
     
  20. Jul 25, 2005 #19

    matt grime

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    an axiom is jsut an axiom, for instance the axiom

    G contains a unique element e such that e*x=x*e=x for all x

    is an axiom from group theory.

    it itself has no truth value and is only true of false when applied to some G and *.

    here dedekind's axiom is not true in and of itself or true for all "lines". it only makes sense to say it is true or false when applied to something. and it is true for the real line and false fro the rationals. the reals are complete (all cauchy sequences have a limit) and the rationals aren't, ie it is possible to find a sequence of rationals increasing and bounded above that do not converge to a rational.


    the axiom states that if we define two classes of elements of the reals, one allto the left of the other that this is the same as picking a unique point of division.

    thus the example i gave abuot the positive rationals whose square is bigger than 2 and its complement divide the line into two parts, but there is no rational that is the point of divisoin.

    another way is this: if we define two classes dividing the line up then we an take the sup of the left hand one and the inf of the right hand one and these agree, and that this element is in either the left or right hand sets.
     
    Last edited: Jul 25, 2005
  21. Jul 26, 2005 #20
    all is all, and if is if.
     
  22. Jul 26, 2005 #21
    So is the dividing point both the sup of the LH class and the inf of the RH class, or is it just one of them? If so which one? But there's no need to answer that, the axiom still makes no sense to me but I've probably pushed your patience far enough. At the moment I rather agree with bao ho, but I'll go away and do some reading around it. Sorry to be so dense but thanks for your help.
     
    Last edited: Jul 26, 2005
  23. Jul 26, 2005 #22

    AKG

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    It's both the sup of the LH class and the inf of the RH class. I can't understand how this axiom is troubling you. For one, did you even see what matt wrote about axioms? They're axioms. It's not a statement of a law of nature. If I define a word "spen" as a pen that has a mismatched cap, will you say, "that definition makes no sense to me?" Dedekind's axiom helps to define the reals, so the reals are by definition, some set of numbers that is, put briefly, complete. So I hope you don't disagree that the Reals are complete, since that wouldn't make sense. Do you just not understand what it means to be complete? What is it that's so confusing?

    Suppose you have the integers. If you cut the integer line at the point 2.5, then you'll split the integers into two classes, the class A = {..., -3, -2, -1, 0, 1, 2} and the class B = {3, 4, 5, ...}. Here, every point of class A is to the left of class B. However, if you divided the line at 2.75, you'd have the same two classes, but with a different dividing point. So the integers are not complete. In the case of the reals, is it evident to you that any splitting of the Reals into two classes such that all of one class is less than all of the other must split in one of the following two ways:

    (-infinity, x) and [x, infinity)

    or

    (-infinity, x] and (x, infinity)

    Is it also not obvious to you that in these cases, there is one and only one dividing point, and that point is x?
     
  24. Jul 26, 2005 #23

    Zurtex

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    They're both the same uniquie dividing point!

    bao_ho's post makes no sense other than refering to the definition of the words!
     
  25. Jul 26, 2005 #24

    matt grime

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    one thing about AKG's post is that to "split the integers to the left and right of 2.75" assumes thart such as thing as 2.75 exists, but it may not, well it does, but we dont' know that, or more accurately the completeness axiom doesn't say what does or deosn't exist. It may be that *nothing* satisfies the completeness axiom. As we can show there is something that does satisfy the axiom.


    ok to be honest i *wanted* to use this kind of example but thought of the several ontological issues. actually the integers are a bad example to use since in any reasonable sense they are complete since a cauchy sequence is eventually constant and hence converges. but the integers aren't a field.

    let's assume that we konw the reals exist and the rationals are inside them


    define a splitting of Q via L as the rationals less than pi and R as the rationals greater than pi.

    sup(L) and inf(L) are not in Q, there is no rational that is the cut point. the rationals are not complete.

    what is hard about this?
     
  26. Jul 26, 2005 #25

    AKG

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    I thought of this too, but I don't see how it's a bad example. As Canute is having trouble grasping this simple axiom, I figured that this example would be more instructive even if it isn't technically that good. Is he going to know what a Cauchy sequence, or even a field is?

    Also, if it's bad to split the integers by 2.75, then it's bad to split the rationals by pi. I think if you wanted, you could introduce surreal numbers and split the reals into classes where the splitting point is not in the reals.
     
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