Infinite recognizable not decidable subsets If L is an infinite ($|L|=|\mathbb{N}| $) decidable language, prove that it contains:
a) An infinite subset that is not recognizable
B) An infinite subset that is recognizable and not decidable
For the (a) I considered all the subsets of L, i.e., $\mathbb{P}$(L)
So $|\mathbb{P} (L)| = |\mathbb{R}|$ so that means that it exists $A \subset L$ that is not recognizable
(for there are not enough Turing machines to recognize them all)
If A were finite, then it would be decidable so it would be recognizable which would be a contradiction.
Therefore A is infinite and not recognizable.
For (b) I thought about B=L\A.
Then B is recognizable because A is not?
But I'm not really sure
Let $K$ be the set of indices of Turing machines which halt on the empty input. Consider the following language:
$$X = \{ 0 w : |w| \in K \} \cup \{ 1 w : |w| \notin K \}. $$
You can check that neither $X$ nor its complement are recognizable.
Therefore your proof idea doesn't work. Here is a different idea. Let $w_1,w_2,w_3,\ldots$ be the words in $L$, enumerated according length and then lexicographically (so if $L = \Sigma^*$, the order would be $\epsilon,0,1,00,01,10,11,\ldots$). The language
$$ Y = \{ w_i : i \in K \} $$
is recognizable but not decidable.