Can limit of hard languages be easy? Can the following all hold simultaneously?


$L_s$ is contained in $L_{s+1}$ for all positive integers $s$.
$L = \bigcup_s L_s$ is the language of all finite words over $\{0,1\}$.
There is some complexity class $C$ and a notion of reduction appropriate for $C$ such that for each $s$, $L_s$ is hard for $C$.
I think we can just start with some base language $L$, then take $L_0 = L$ and $L_{s+1} = L_s \cup \{0,1\}^{s+1}$.

That is, each $L_s$ is the union of $L$ with all strings of length up to $s$. Each $L_s$ is at least as hard as $L$ but is no harder (in an asymptotic sense), assuming we can count to $s$.

I also thought about the opposite "limit", so each $L_{s+1}$ is contained in $L_s$, and $L = \cap_s L_s$ is easy while each $L_s$ is hard. But I think we could just start with a hard (but countable) language $L_0$ and just remove one word at each step; the intersection should be empty (every word is eventually removed).