Every decidable lanugage $L$ has an infinite decidable subset $S \subset L$ such that $L \setminus S$ is infinite


Given an infinite decidable language $ L$ , then if $ S \subset L$ such that $ L \setminus S$ is finite, then $ S$ must be decidable. This is true since given a decider of $ L$ we contruct a decider for $ S$ :

Simulate the decider of $ L$ on the input, if it accepts, go over $ L \setminus S$ and check if it is there, if it is, reject. If it isn’t accept. If the decider of $ L$ rejects – reject.

Another point is if $ S \subset L$ is finite then $ S$ also must be decidable, this is immediate that every finite language is decidable.

Now we have the last case where $ S$ is infinite and $ L \setminus S$ is infinite. We know that there must be some subsets $ S$ corresponding to this case that are undecidable. This is since there are $ \aleph$ such $ S$ but only $ \aleph_0$ deciders. Denote $ D(L) = \{ S \subset L : |S|= |L \setminus S|=\infty \wedge S \text{ is decidable} \}$

Is it true that for all infinite decidable languages $ L$ we have $ D(L) \neq \phi$ ?

If this is true then as a conclusion we will have for all infinite decidable languages $ L$ a sequence of decidable languages $ L_n$ such that $ L_0=L$ and $ L_{n+1} \subset L_n$ and $ |L_n \setminus L_{n+1}| = \infty$

We will also have a limit-set $ L_\infty = \{ e \in L : \forall n \in \mathbb{N} \text{ } e \in L_n \}$ and can dicuss if it is empty/finite/infinite and decicable or not.

This seems like a nice way to study decidable languages, and curious to know if this direction is indeed interesting and whether there are articles published regarding these questions

Thanks for any help