Is this language decidable?

Question

I'm struggling with whether or not this is decidable:

A = {x is an element of the set of Natural Numbers | for every y greater than x, 2y is the sum of two primes}

I'm inclined to think that this is decidable given the fact that when fed into a Turing Machine, it will never reach an accept state and loop for infinity unless it rejects. However, I also do know that for a language to be decidable, there must only exist an algorithm to decide it; we don't necessarily have to know how it's done. With this, part of me think that it is decidable? Does anyone know how to prove either?

Answer 1

This language is decidable, though the proof is a bit evil.

For starters, let's think about the properties of this language. Clearly, if n is a natural number contained in the language, then every number greater than n is also in the language. Thus there are three possible forms this language can take:

1. This language contains all natural numbers, or
2. This language contains no natural numbers, or
3. This language contains all natural numbers greater than some natural number n.

Languages (1) and (2) are, respectively, {0, 1}* and the empty language, both of which are decidable (so there are TMs that always halt that accept those languages). Every language of form (3) is also decidable, because for any n we can easily write a TM with n hardcoded into it that simply checks whether the input is at least n. Consequently, no matter which case is true (either 1, 2, or 3), there exists some TM that always halts whose language is the language you've provided, so your language is decidable.

But that said, this proof is nonconstructive. We can show that the language has to be decidable, but we can't actually find the TM that always halts that accepts it! In fact, no one knows which TM it is, because Goldbach's Conjecture (whether every even number greater than two is the sum of two primes) is an open problem in mathematics.

Hope this helps!