Is NaN a valid key value for associative containers?

Question

Consider the ordered and unordered associative containers in C++ keyed on double.

Is NaN a valid key type?

With ordered containers,

Answer 1

This is because

std::less(NaN, NaN) == false

Like you said, the weak total ordering (required for std::map<>) is ok, the equality (or equivalence, extra requirement for any hash-based container) isn't ok to satisfy the key requirements for the hash (unordered) map

Irreflexivity   f(x, x) must be false. 
 Antisymmetry   f(x, y) implies !f(y, x) 
 Transitivity   f(x, y) and f(y, z) imply f(x, z). 
 Transitivity of equivalence     

         Equivalence (as defined above) is transitive: if x 
         is equivalent to y and y is equivalent to z, then x is 
         equivalent to z. (This     implies that equivalence does 
         in fact satisfy the mathematical definition of an equivalence 
         relation.)

Seeing that for std::map, equivalence is when !less(a,b) && !less(b,a), I'd say all constraints are met.