Are there complexity classes with provably no complete problems?

Here's a really simple class that is very natural and has no complete problems: ALL, the class of all languages. The reason is that there are uncountably many problems in ALL, but only countably many Turing machines to go around (for reductions), so every problem in ALL cannot be reduced to a single problem in ALL.

Similarly, any class with advice, like P/poly, L/poly, BQP/qpoly, or even P/1 does not have complete problems (using the same argument).

Another class without complete problems w.r.t. logspace or polytime reductions (not an union of classes TIME(f) for some family of polynomially related functions f, but still relatively natural in my opinion) is

ELEMENTARY = TIME(2ⁿ) ∪ TIME(2²ⁿ) ∪ TIME(2²²ⁿ) ∪ TIME(2²²²ⁿ) ∪ ⋯

If L were ELEMENTARY-complete, then it would belong to some level of this hierarchy, and all problems above could be reduced to it. But this hierarchy is known to be proper (time hierarchy theorem), contradiction.

The zoo of complexity classes extends naturally into the realm of computability theory and beyond, to descriptive set theory, and in these higher realms there are numerous natural classes which have no members that are complete, even with respect to far more generous notions of reduction than the one you mention.

• For example, consider the class Dec of all decidable sets of natural numbers. There can be no decidable set $U$ such that every other decidable set $A$ reduces to it in time uniformly bounded by any computable function $f$ (even iterated exponentials, or the Ackerman function, etc.). In particular, it has no member that is complete in your sense. If there were such a member, then we would be able to consruct a computable enumeration $A_0$, $A_1$, $\ldots$ containing all decidable sets, which is impossible since then we could diagonalize against it: the set of $n$ such that $n\notin A_n$ would be computable, but it can't be on the list.

• The class of all arithmetically definable sets is obtained by closing the decidable sets (or much less) under projection from $\mathbb{N}^{n+1}\to \mathbb{N}^n$ and under Boolean combinations. The members of this class are exactly the sets that are defined by a first order formula over the structure $\langle\mathbb{N},+,\cdot,\lt\rangle$, and the hierarchy is stratified by the complexity of these definitions. This class has no member that is complete with respect to any computable reduction, and even with respect to any arithmetically definable reduction of bounded complexity, for in this case the hierarchy would collapse to some level $\Sigma_n$, which is known not to occur.

• A similar argument works for the hyperarithmetic hiearchy, which can have no universal member hyperarithmetic reductions of any fixed complexity.

• And similarly for the projective hiearchy on sets of reals.

The general phenomenon is that there are numerous hierarchies growing from computability theory into descriptive set theory which are all known to exhibit strictly proper growth in such a way that prevents them from having universal members.