is not countable: contradiction by diagonalization
is not countable: contradiction by diagonalization
Idea: For any alphabet, there are uncountable infinite many languages, and countable infinite many Turing Machines to recognize them
The set of all possible languages over binary strings has the same cardinality as , so it's uncountable.
Claim: the number of Turing Machines is countable.
Strategy: for each list all well-formed TMs that have less than characters in description: every TM can be encoded as a finite string of characters.
Let be the tape alphabet and there are at most many TMs. List them all and increment . That list all TMs.
Encoding of TM, <> as "writing the 7-tuple as binary string", describing how TM works.
Prove by diagonalization.
Undecidable Language
is recognizable, but not decidable. There's no way to detect a loop.
Suppose by contradiction is decidable by decider that decides .
Construct a new TM (kinda like diagonalization) when the input to is its own description.
On input a TM, runs on <> and output the opposite: