computational philosophy of mathematics

[gusl]

I only believe in mathematical objects with finite information (i.e. with a finite expression). Things can expressed in any way you wish, intensionally, extensionally, whatever. So, while the natural numbers exist, not all subsets of it do.

While the set of real numbers exists (R can be expressed as the power set of N), not all of the traditional real numbers exist (only the computable ones do). So, in my definition, the cardinality of R is aleph_0 (i.e. the same as N). In fact, no set can have greater cardinality, for that would imply it had non-existing elements (since only computable things exist).

Can we design a set theory this way? How much of traditional set theory can be translated? Does we lose any good mathematics this way?

Whereas people seem to view uncomputability as a fundamental property of a problem, I tend to view it as another form of self-reference paradoxes. All "well-defined" problems are computable. Again, this is not a theorem or mathematical insight, but a re-definition, just like my "R only has countably many existing numbers" is a definition. But the point isn't just to change names and keep everything the same... it's to change the intuition that goes with the names.

Comments

[gustavolacerda.livejournal.com]

That's even easier then!
When does the diagonalization argument go through then?

[easwaran.livejournal.com]

Diagonalization requires that every entry in the table be well-defined, so we see that there is no recursive enumeration of the recursive functions, though there is of the partial recursive functions. Though I believe there is no enumeration without repetition of the partial recursive functions.

[gustavolacerda.livejournal.com]

Diagonalization requires that every entry in the table be well-defined,

so we see that there is no recursive enumeration of the recursive functions,

Hmmm... ok: if there were a computable enumeration of the computable real numbers, then we could use the diagonalization algorithm to add a new computable real number (since all the inputs are computable) (of course, this also means it has finite information, which I think is equivalent to being computable). Therefore, there is no computable enumeration of the computable real numbers.
Is this your proof?

[easwaran.livejournal.com]

Exactly.