Originally posted by Palynka
I knew it would be straightforward for you. 🙂
Give me a method for assigning a real number to an infinite, but countable set. What happens if X is uncountable?
Edit - Adam, did his proof made it clear what I meant by that?
Somebody had to give in to your demands, because I'm very intrigued to see what happens next 🙂.
There is rather more to this than I thought before -- we will have to POP and LOAD*! I'm going to try a topological solution**.
We consider X/I, totally ordered by P, as a topological space with the order topology, i.e. the topology generated by sets of the form {x: xPy and zPx}.
Suppose we have a function u:X/I --> R with the desired property, namely that xPy if and only if u(x)<u(y). Then let Q denote the image of X/I in R. Note that if u(x) = u(y), then x~Py and y~Px, so x and y represent the same I-class, i.e. u is injective. Thus u:X/I-->Q is a bijection.
Q inherits a topology from R as a subspace, namely the topology generated by the intersections of open intervals, and the order-preservingness of u guarantees that u and its inverse (on Q) are continuous with respect to this topology.
Thus a function of the type Palynka asked for is exactly an embedding (wrt the order topologies) of X/I into R. Now Palynka's question is: Under what conditions is such an embedding possible and how is it done explicitly when X/I is countable?
For the case when X/I is countable, pick your favourite real number r > 1. Pick some arbitrary enumeration of X/I = {x(i) : i in N}. For any x in X/I, let A(x) = {i in N : x(i)Px}. Let u(x) be the sum over A(x) of r^(-i). Then if xPy, A(x) is contained in A(y), so that u(x)<u(y). On the other hand, u(x) < u(y) means that the sum determining u(y) is taken over a set of natural numbers than that for u(x), since all terms are positive, i.e. A(y) contains A(x), so xPy. This is a bijection since X/I is totally ordered by P, and it has the Palynka Ordering Property (POP).
The question of when such an embedding exists is a harder. Since we can do it for countable ordered sets, the natural next step is to do it for sets with countable subsets which are dense in the order topology. I feel like the converse is rather nastier, i.e. there are embeddable X/I which are not separable.
In any case, the idea is to take a countable subset D of X/I and construct u: D-->R as above, and then find v: X/I-->R such that v restricts to u on D, and then show that this v has the POP. Given x in X/I, choose a sequence of x(n)s in D converging to x (by density). Then since u is an embedding, the sequence {u(x(n))} has a limit v in R (this can be had by pulling back the metric on R and observing that the x(n)s are a Cauchy sequence, and so must be their images, and applying completeness of R, among other ways). We have to check that this gives the same value when we choose a different sequence y(n) converging to x, but this is immediate since u is an open map and any neighbourhood of x contains all but finitely many of the x(n) and y(n). Thus we define v(x) to be the pointwise limit of u(x(n)) for any sequence x(n) --> x. v restricts to u on D.
Now take xPy in X/I. Now we actually have to assume that X/I is connected, which gives some w with xPwPy, since otherwise the stuff larger than y and smaller than x would be a separation of X/I by open sets. By sticking an element of D between x and w and between w and z, and applying the definition of v, we get that v(x)<v(y).
Thus:
u exists when X/I is connected and has a countable dense subset, with respect to the order topology.
I don't think the converse is true; As I said, I think there is something a little weaker than separability that does the trick, and I am going to call it Linear Order Approximately Dense, or LOAD, and I'll try to figure out what it might be tomorrow evening.
*This is somewhere between popping and locking and locking and loading.
**When the only tool you have is a hammer etc.
28 Jan '09 11:01
Its A Question Of Science
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