I was thinking how to say that two sets of prime numbers are close to each other, and came up with this.

**NOTATION**
<ul>
<li>$\ \Delta(A\ B)\ :=\ (A\setminus B)\cup(B\setminus A)\ \ $ is the symmetric difference of sets $\ A\ B$;</li>

<li>$\ \mathbb N := \{1\ 2\ \ldots\}\ $ is the set of natural numbers, and $\ \mu\ $ is the $\sigma$-measure in $\ 2^{\mathbb N}\ $ such that $\ \mu(\{n\}):=2^{-n}\ $ for every $\ n\in\mathbb N$;</li>

<li>$\ Span(A)\ $ is the multiplicative monoid generated by any non-empty $\ A\subseteq \mathbb N,\ $ while $\ Span(\emptyset)\ :=\ \emptyset$;

<li>$\ \mathbb P\ $ is the set of all primes, and $\ d(S\ T)\ :=\ \mu(Span(S\,\Delta\,T))\ $ is obviously a metrics in $\ 2^{\mathbb P}\ $ of diameter $\ 1$. (<i>Except that it is not true that $\ d\ $ is a metrics</i>).</li>
<li>Which metric axiom does it not satisfy - can you give an example?</li>

<li>(OPTIONAL: &nbsp; also $\ \delta(S\ T)\ :=\ \mu(Span(S)\,\Delta\, Span(T))\ $ for every $\ S\ T\subseteq\mathbb P\ $ is a metrics in $\ 2^{\mathbb P}\ $ of diameter $\ 1$).
</ul>

For example, $\ d(S\ T)\ =\ 1$ for every partition $S\ T\ $ of $\ \mathbb P$.

Is it of any interest? Has this already been defined and is it called something?

My hope is that this may find use as follows. Suppose that there is a set of objects such that for each object in the set there is a set of associated prime numbers, and suppose that each object is uniquely determined by its set of associated primes. Then the metric defined above can be used to construct a metric on this set.

For example, consider the set G of Galois extensions of Q. Each extension K/Q in G has an associated set of its unramified primes, say U(K). Define an equivalence relation ~ on G by saying that two extensions are equivalent if they have the same set of unramified primes. Consider the set of equivalence classe, H = G/~, and for an extension K/Q denote its equivalence class by K/~. Then there is a metric e on H defined by e(K/~, L/~) = d(U(K), U(L)).

Any pointers would be welcomed! Thanks, Martin