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matt grime
matt grime is offline
#39
Jul30-05, 08:25 AM
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let's take the standard example we have been using again.

consider the reals, take them and split them into two parts L and R one to the left of the other as it says in the axiom. In the reals, which satisfy the axiom, we are saying that L and R are one of L=(-inf, x) R=[x,inf) or L=(-inf,x] (x,inf) and x is the unique point creating the division (note there are two ways to split it up). In the rationals the axiom does not hold example: R is the set of positive rationals whose square is bigger than 2, and L its complement. If there were a rational creating this division then it would have to be square root of 2 which we know is not rational, ie there is no rational that is smallest and satisfies the property x^2>2

We can think of the axiom possibly applying to any linearly ordered set (the line of points in the axiom) whence it may be satisfied or not, let;s say that the class of linearly ordered sets that satisfy dedekinds axiom are called dedekind complete. so it may apply to any set of linearly ordered points. just as the axiom 'a binary operation * has a two sided identity' may or may not be true of any given operation.

in this case the reals are 'dedekind complete' (as are the integers) but the rationals are not dedekind complete.

The axiom is equivalent to: given any splitting L,R of the set of points S as in the axiom then there is a unique x in S which is sup(L) and inf(R) and necessarily lies in one of the L and R (and exactly one) since S is the disjoint union of L and R.

you may apply it to the set of points corresponding to time and it appears your author thinks that they do not satisfy the axiom in his view, or the assumption that they do is problematic, ie the set of points of time are not dedekind complete. well, that's fine, I suppose, but it isn' a problem with the axiom really any more than it is a 'problem' that the rationals don't satisfy the axiom. this is a 'problem' with your (his) view of time.