Turing recognizable language between languages that aren't recognizable? Let $L_{1},L_{2}$ be languages that are not Turing recognizable, and let $L$ be a language such that $L_{1} \subseteq L \subseteq L_{2}$. 

Is $L$ also not Turing recognizable? 

I am inclined to believe it is, but I really have no idea how to show this. It seems that non-recognizable languages have very little about them that could  be used to solve a problem like this. Ideas?
The answer is no, of course.

Consider, for instance, the following languages:


$L_1 = \{ \langle M \rangle \# \langle M \rangle \mid \, \text{$M$ is a TM which does not halt on input $\langle M \rangle$} \}$
$L = \{ \langle M \rangle \# \langle M \rangle \mid \, \text{$M$ is a TM} \}$
$L_2 = L \cup \{ \langle M \rangle \# w \mid \, \text{$M$ is a TM which does not halt on input $w$} \}$


Then it is easy to see neither $L_1$ nor $L_2$ are recursively enumerable (i.e., Turing-recognizable), despite $L$ being even context-sensitive. (In fact, you could even make $L$ context-free by replacing the second $\langle M \rangle$ with its reversed copy.)



The upshot is that set inclusion (on its own) tells us very little about the sets involved being recursively enumerable or not, unless the symmetric difference between them is finite. In the case of $L_1 \subseteq L \subseteq L_2$, we do know, for instance, that $|L \setminus L_1|$ or $|L_2 \setminus L|$ being finite implies that $L_2$ is also not recursively enumerable.