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.