How to approach to solve this question and the likes of it?
Let $L$ be the set of strings $\langle M\rangle$ such that $M$ accepts all strings of even length and does not accept any strings of odd length.
a) Is $L$ Turing-recognizable? Prove your answer.
b) Is the complement of $L$ Turing-recognizable? Again prove your answer.
Please have a look at my approach, is it correct? :
Since the string representation of set of all the Turing Machines (say S) is Infinite countable and Recursively Enumerable, there fore there exists a Turing Machine that accepts S. Now, if we choose the Turing Machines M1, M2, ... from S, in the given order, such that all the chosen machines accept even length string and reject odd length strings, then the set we get (Say T) will also be Infinite Countable and recursively Enumerable and will have a Turing Machine which accepts it. That is why it is language T is Turing Recognizable. Similarly complement of T is Turing Recognizable.
@Rick if I use the following code :
RL(<M>) = for n = 0, 1, ... for s = s0, ..., sn \\in the standard order on all strings run M on s for one move if M(s) = accept AND |s| is odd reject // accept as string in L complement if not rejected for any string then accept as a string in L
This code will select all the $M$ which accept all strings of even length and do not accept any string of odd length. Then now since we have a membership algorithm for $L$ (as well as for $\overline L$), then can we say that the $L$ and $\overline L$ are recursive?