The following questions were proposed to me by professor Zvonko Iljazović.
Let . We define the metric be the metric defined as
for all . Let .
Let and . Suppose is a sequence in such that the sequences and are computable sequences ? Does the sequence need to be computable in the sense that its components are computable sequences ?
Let where is an incomputable real. Then however, is not a computable sequence.
Now the second question. Let , and . Let be a sequence such that , and are computable sequences . Does the sequence need to be computable?
The answer is yes. The proof is available in our upcoming publication "Dense Computability Structures".
Proposition:
The metric space has at least two maximal computability structures in which points and are computable.
Proof:
Let be a computable sequence in such that . Let be a computability structure induced by . Since is separable, it is also maximal.
Let where is an incomputable real. Then the finite sequence has the property hence by definition is an effective finite sequence.
Therefore,
is a computability structure and there exists a maximal computability structure such that and are computable points in .
On the other hand, the point is not computable in since that would contradict the fact that is a non-computable real. This is equivalent to the fact that . Therefore, . From this we conclude that and are two maximal computability structures on in which points and are computable.