Generalised metric

Preliminary definition

Let ( F , + , ⋅ , < ) be an arbitrary ordered field, and M a nonempty set; a function d : M × M → F + ∪ { 0 } is called a metric on M , iff the following conditions hold:

1. d ( x , y ) = 0 ⇔ x = y ;
2. d ( x , y ) = d ( y , x ) , commutativity;
3. d ( x , y ) + d ( y , z ) ≥ d ( x , z ) , triangle inequality.

It is not difficult to verify that the open balls B ( x , δ ) := { y ∈ M : d ( x , y ) < δ } form a basis for a suitable topology, the latter called the metric topology on M , with the metric in F .

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

Further properties

However, under axiom of choice, every general metric is monotonically normal, for, given x ∈ G , where G is open, there is an open ball B ( x , δ ) such that x ∈ B ( x , δ ) ⊆ G . Take μ ( x , G ) = B ( x , δ / 2 ) . 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 G , G open, we may take μ ( x , G ) := B ( x , 1 / 2 n ( x , G ) ) , where n ( x , G ) := min { n ∈ N : B ( x , 1 / n ) ⊆ G } , and the trick is done without choice.

Case II: F is a non-Archimedean field.

For given x ∈ G where G is open, consider the set A ( x , G ) := { a ∈ F : ∀ n ∈ N , B ( x , n ⋅ a ) ⊆ G } .

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, N F is not bounded above, hence there is some ξ ∈ F with ∀ n ∈ N : n ⋅ 1 ≤ ξ . Putting a = k ⋅ ( 2 ξ ) − 1 , we see that a is in A(x, G).

Now define μ ( x , G ) = ⋃ { B ( x , a ) : a ∈ A ( x , G ) } . We would show that with respect to this mu operator, the space is monotonically normal. Note that μ ( x , G ) ⊆ G .

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 μ ( x , G ) ∩ μ ( y , H ) is empty. If not, say z is in the intersection. Then

From the above, we get that d ( x , y ) ≤ d ( x , z ) + d ( z , y ) < 2 ⋅ max { a , b } , which is impossible since this would imply that either y belongs to μ ( x , G ) ⊆ G or x belongs to μ ( y , H ) ⊆ H .

So we are done!

Discussion and links