I don't think that would solve this issue - a perfect hash function is guaranteed to not have any collisions for any element in some predefined set. What Bloom'

I don't think that would solve this issue - a perfect hash function is guaranteed to not have any collisions for any element in some predefined set. What Bloom's proof requires is that all of the k hash functions should not have any collision for any input that is inserted into the Bloom filter, which is not covered by just having each function alone be perfect. That aside, Bloom does not make any assumptions about the chosen hash functions being perfect or not.