Generalised metric

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

$${\displaystyle \exists a\in A(x,G)\colon d(x,z)<a;\;\;\exists b\in A(y,H)\colon d(z,y)<b.}$$

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.

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• Uniform space – Topological space with a notion of uniform properties

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