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
, iff the following conditions hold:
;
, commutativity;
, 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: F is an Archimedean field.
Now, if x in
open, we may take
, where
, and the trick is done without choice.
Case II: F is a non-Archimedean field.
For given
where G is open, consider the set
.
The set A(x, G) is non-empty. For, as G is open, there is an open ball B(x, k) within G. Now, as F is non-Archimdedean,
is not bounded above, hence there is some
with
. Putting
, we see that
is in A(x, G).
Now define
. We would show that with respect to this mu operator, the space is monotonically normal. Note that
.
If y is not in G(open set containing x) and x is not in H(open set containing y), then we'd show that
is empty. If not, say z is in the intersection. Then
.
From the above, we get that
, which is impossible since this would imply that either y belongs to
or x belongs to
.
So we are done!
Discussion and links
- Carlos R. Borges, A study of monotonically normal spaces, Proceedings of the American Mathematical Society, Vol. 38, No. 1. (Mar., 1973), pp. 211–214. [1]
- FOM discussion link