Are there any decision complexity classes which have no "complete" subclass?
Apart from the many answers from the Mathoverflow link commented above by Ariel, we have a very simple example as follows.

Under widely accepted conjecture that $\mathrm{P}\neq \mathrm{NP}$, we have that $\mathrm{NP}\setminus \mathrm{NP}$-$\mathrm{complete}$ does not have complete problem and therefore the "hard" part of it is well "above" it.

For the proof, use Ladner's result on $\mathrm{p}$-degree, for any two different $\mathrm{p}$-degrees $\mathrm{A}$ and $\mathrm{B}$, if $\mathrm{A}&lt;_p\mathrm{B}$, then there must exist another $\mathrm{p}$-degree $\mathrm{C}$ in between: $\mathrm{A}&lt;_p\mathrm{C}&lt;_p\mathrm{B}$. So by removing $\mathrm{NP}$-$\mathrm{complete}$ from $\mathrm{NP}$, every $\mathrm{p}$-degree below the complete degree always has another non-complete degree above it.