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Are all ordered fields dense?

  1. Mar 9, 2007 #1
    Following a random discussion with a friend (which started on the topic of 0.9r=1 :uhh: ), the question of denseness in fields came up.

    By my understanding, to define denseness the field needs to be ordered otherwise you can't say A<B<C. I get the vague feeling all ordered fields are dense, but that's by assuming the two field operations are always addition and multiplication, since they allow you to always generate 0.5(A+B) from A and B and by ordering A<B => A<0.5(A+B)<B.

    Usually when I attempt such general lines of logic I go horrifically wrong, so are there ordered fields which aren't dense? The list of ordered field Wiki gives doesn't have any which immediately jump out as non-dense, but some I'm not exactly familiar with.

    Thanks :smile:
     
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  3. Mar 9, 2007 #2

    morphism

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    I've never seen fields described as being dense; usually this concept is attached to topological spaces.

    I take it you're defining density for an ordered field if between any two of its elements there is a third. What about the rational subfield of the formal Laurent series over a field?
     
  4. Mar 9, 2007 #3

    AKG

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    I think you're right.

    0.5(A+B) < B iff 0.5A + 0.5B < B iff 0.5A < B - 0.5B iff 0.5A < 0.5B iff A < B

    and indeed A < B. Note most of the above follows directly from the axioms of an ordered field. The only thing that isn't direct is the implicit assumption 0 < 2. But this is practically direct. Note 0 < 2 iff 0 < 1. Suppose 0 > 1. Then 0 < -1, so 0 < (-1)(-1) = 1, contradiction. Hence 0 < 2 indeed. Actually, there's another tacit assumption, the one that 0.5 even exists. 0.5 exists iff 2 exists. 2 exists unless the field is {0,1}. But it's easy to check that no order makes {0,1} into an ordered field.
     
  5. Mar 10, 2007 #4

    JasonRox

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    If you're thinking of the dense as in between any two numbers there is another number, then no not all ordered fields are dense.

    Take the field Z_p, where p is prime. With the usual definitions of addition and multiplication, but with modulo p.
     
  6. Mar 10, 2007 #5

    matt grime

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    That field is not ordered, Jason. It is implicit that the order has something to do with the algebraic stucture (like at least respecting it), so that a<b implies a+c<b+c. It is easy to derive contradictions if we assume any order on Z/pZ (I think you meant that - you actually wrote the p-adic integers, which aren't a field).

    Denseness is a property of topological spaces inside other topological spaces.

    It does not make sense to say the rationals, as a field, are dense. It only makes sense to say the rationals are dense inside the reals. And the fieldness or otherwise has nothing to do with that. (Q^n is dense in R^n, but these are not fields).
     
  7. Mar 10, 2007 #6

    Gib Z

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    Yay! I spotted that too! The primes can not be a field because to be a field
    it must have an Existence of Inverse: Every non-zero number must have a multiplicative inverse such that their product is 1, and an additive inverse sum that their sum is 0. The primes do not include negative numbers, nor non-integers, so the primes can't be a field! Yay I got something!
     
  8. Mar 10, 2007 #7

    matt grime

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    I'm afraid you made a mistake, but only one of interpreting what some one else said. Certainly the primes with the obvious multiplication and addition are not a field - they are not even closed under any of the operations, do not contain 0 or 1. But no one has said that they were.
     
  9. Mar 10, 2007 #8

    Gib Z

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    :'( I'm going to go cry now...
     
  10. Mar 10, 2007 #9

    matt grime

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    But you correctly reasoned that the primes are not a subfield of the rationals. It was just that no one said they were. Jason talked of 'addition mod p'. That means fix a prime and consider the numbers 0,1,2,..,p-1 with the operations of addition and multiplication taken 'remainder after division by p'. At least that is what I think he meant. The notation Z_p is ambiguous. Frequently and normally and it means the p-adic integers, and Z_[p] means the p-locals. Z_(p) might mean the field Z/pZ as well. All confusing. The standard notation for the field with p elements is F_p. (All these should be thought of as blackboard bold fonts, eg [itex]\mathbb{F}_p[/itex]
     
    Last edited: Mar 10, 2007
  11. Mar 10, 2007 #10

    JasonRox

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    I just meant the set {1,2,3....,p-1} along with the operations I mentionned.

    And I'm totally aware that denseness doesn't make any sense, but I was just going along with whatever definition they went with. I mentionned that in my post.

    You mentionned that there are contracdiction that occur on this set when we think of it as an ordered set. What are they?

    The order relation is satisfied.
     
  12. Mar 10, 2007 #11

    matt grime

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    The point of an ordered field is that the order has something to do with the algebraic structure. It isn't just a field with any old order - since every set is well orderable, you've asserting every field is ordered. But that isn't true, if only for the meta-reason that if it were we'd never bother mentioning ordered fields at all.

    The order must respect the algebra. In particular, if x>0 and y>z, then xy>xz, and for any x, y>z implies x+y>x+z. Just try to use the natural order on F_p, and it is trivial to show that these rules are broken. (just add 1 to two numbers x>y, until you wrap around, and you'll get a contradiction).
     
  13. Mar 10, 2007 #12

    JasonRox

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    I agree with you matt. I won't deny that. I won't debate too much as I don't know too much.

    Anyways, although it is implicit that the definition of ordered is with respect to its algebra. I won't deny that. I wasn't really introduced to that idea before, so I was thinking the simple order. But now that you mention it, it does imply all fields are well orderable. I almost forgot about that important theorem. This is a great way to never forget about it now. :biggrin:
     
  14. Mar 10, 2007 #13
    Thanks for the explaination guys. Upon reading the replies I remembered that the typical statement is "A is dense in B" rather than just "A is dense", which suggests something a bit more subtle than I was thinking of.
     
  15. Mar 10, 2007 #14

    JasonRox

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    Even asking whether A is dense in B can be meaningless.

    For example, create a topology on the set {1}. (I know you may not know how to do so, but imagine you did.) The set {1} itself is dense in {1}.

    Or, if you learn some topology, you'll learn about the indiscrete topology (in the first chapter). Any set in the indiscrete topology, except the empty set, is dense in the underlying set. For example, the indiscrete topology on the set of real numbers, the set {0} is dense in R (indiscrete topology).

    Therefore, talking about whether a set is dense within another is meaningless if you don't know what the sets are or what the topology is.
     
  16. Mar 11, 2007 #15

    AKG

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    The statement "X is dense in Y" is not more typical than "X is dense". They are two totally different sentences. The first one is normally used when talking about topology, which is not the case of the question asked in the original post. The second one is used when talking about order theory, which is what the thread is about. See here.
     
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