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

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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

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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

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References

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  • FOM discussion, 15 August 2007