Let ${\displaystyle (F,+,\cdot ,<)}$  be an arbitrary ordered field, and ${\displaystyle M}$  a nonempty set; a function ${\displaystyle d:M\times M\to F^{+}\cup \{0\}}$  is called a metric on ${\displaystyle M,}$  if the following conditions hold:

1. ${\displaystyle d(x,y)=0}$  if and only if ${\displaystyle x=y}$ ;
2. ${\displaystyle d(x,y)=d(y,x)}$  (symmetry);
3. ${\displaystyle d(x,y)+d(y,z)\geq d(x,z)}$  (triangle inequality).

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

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

However, under axiom of choice, every general metric is monotonically normal, for, given ${\displaystyle x\in G,}$  where ${\displaystyle G}$  is open, there is an open ball ${\displaystyle B(x,\delta )}$  such that ${\displaystyle x\in B(x,\delta )\subseteq G.}$  Take ${\displaystyle \mu (x,G)=B\left(x,\delta /2\right).}$  Verify the conditions for Monotone Normality.

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

proof.

Case I: ${\displaystyle F}$  is an Archimedean field.

Now, if ${\displaystyle x}$  in ${\displaystyle G,G}$  open, we may take ${\displaystyle \mu (x,G):=B(x,1/2n(x,G)),}$  where ${\displaystyle n(x,G):=\min\{n\in \mathbb {N} :B(x,1/n)\subseteq G\},}$  and the trick is done without choice.

Case II: ${\displaystyle F}$  is a non-Archimedean field.

For given ${\displaystyle x\in G}$  where ${\displaystyle G}$  is open, consider the set ${\displaystyle A(x,G):=\{a\in F:{\text{ for all }}n\in \mathbb {N} ,B(x,n\cdot a)\subseteq G\}.}$ 

The set ${\displaystyle A(x,G)}$  is non-empty. For, as ${\displaystyle G}$  is open, there is an open ball ${\displaystyle B(x,k)}$  within ${\displaystyle G.}$  Now, as ${\displaystyle F}$  is non-Archimdedean, ${\displaystyle \mathbb {N} _{F}}$  is not bounded above, hence there is some ${\displaystyle \xi \in F}$  such that for all ${\displaystyle n\in \mathbb {N} ,}$  ${\displaystyle n\cdot 1\leq \xi .}$  Putting ${\displaystyle a=k\cdot (2\xi )^{-1},}$  we see that ${\displaystyle a}$  is in ${\displaystyle A(x,G).}$ 

Now define ${\displaystyle \mu (x,G)=\bigcup \{B(x,a):a\in A(x,G)\}.}$  We would show that with respect to this mu operator, the space is monotonically normal. Note that ${\displaystyle \mu (x,G)\subseteq G.}$ 

If ${\displaystyle y}$  is not in ${\displaystyle G}$  (open set containing ${\displaystyle x}$ ) and ${\displaystyle x}$  is not in ${\displaystyle H}$  (open set containing ${\displaystyle y}$ ), then we'd show that ${\displaystyle \mu (x,G)\cap \mu (y,H)}$  is empty. If not, say ${\displaystyle z}$  is in the intersection. Then

From the above, we get that ${\displaystyle d(x,y)\leq d(x,z)+d(z,y)<2\cdot \max\{a,b\},}$  which is impossible since this would imply that either ${\displaystyle y}$  belongs to ${\displaystyle \mu (x,G)\subseteq G}$  or ${\displaystyle x}$  belongs to ${\displaystyle \mu (y,H)\subseteq H.}$  This completes the proof.

• Ordered topological vector space
• Pseudometric space – Generalization of metric spaces in mathematics
• Uniform space – Topological space with a notion of uniform properties

• FOM discussion, 15 August 2007