In mathematics, the concept of a generalised metric is a generalisation of that of a metric, in which the distance is not a real number but taken from an arbitrary ordered field.

In general, when we define metric space the distance function is taken to be a real-valued function. The real numbers form an ordered field which is Archimedean and order complete. These metric spaces have some nice properties like: in a metric space compactness, sequential compactness and countable compactness are equivalent etc. These properties may not, however, hold so easily if the distance function is taken in an arbitrary ordered field, instead of in

Preliminary definition

Let be an arbitrary ordered field, and a nonempty set; a function is called a metric on if the following conditions hold:

  1. if and only if ;
  2. (symmetry);
  3. (triangle inequality).

It is not difficult to verify that the open balls form a basis for a suitable topology, the latter called the metric topology on with the metric in

In view of the fact that in its order topology is monotonically normal, we would expect to be at least regular.

Further properties

However, under axiom of choice, every general metric is monotonically normal, for, given where is open, there is an open ball such that Take Verify the conditions for Monotone Normality.

The matter of wonder is that, even without choice, general metrics are monotonically normal.

proof.

Case I: is an Archimedean field.

Now, if in open, we may take where and the trick is done without choice.

Case II: is a non-Archimedean field.

For given where is open, consider the set

The set is non-empty. For, as is open, there is an open ball within Now, as is non-Archimdedean, is not bounded above, hence there is some such that for all Putting we see that is in

Now define We would show that with respect to this mu operator, the space is monotonically normal. Note that

If is not in (open set containing ) and is not in (open set containing ), then we'd show that is empty. If not, say is in the intersection. Then

From the above, we get that which is impossible since this would imply that either belongs to or belongs to This completes the proof.

See also

  • FOM discussion
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