Logarithms Assuming the axiom of choice and, given an infinite cardinal κ and a finite cardinal μ greater than 1, there may or may not be a cardinal λ satisfying . However, if such a cardinal exists, it is infinite and less than κ, and any finite cardinality ν greater than 1 will also satisfy .
Logarithms Assumi,ng the axiom of choice and, given an infinite cardinal κ and a finite cardinal μ greater than 1, there may or may not be a cardinal λ satisfying . However if such a cardinal exists, it is infinite and less than κ, and any finite cardinality ν greater than 1 will also satisfy .