Proof of Countable Reals

Claim: The real numbers are countable
Proof using bijection

1. Create a bijection between the natural numbers > 1 and the set of all multisets of prime numbers using the definition of prime factorization and index each correspondence using the natural numbers enumerated by n.
   N:n>1⇒ {M(p)}
   1: 2 ⇒ {2}
   2: 3 ⇒ {3}
   3: 4 ⇒ {2,2}
   4: …
2. Using a bijection between the natural numbers and the prime numbers in sequence of least to greatest, create a bijection between the set of all multisets of primes and the set of all multisets of positive integers
   N:n>1⇒ {M(p)) ⇒ {M(n)}
   1: 2 ⇒ {2} ⇒ {0}
   2: 3 ⇒ {3} ⇒ {1}
   3: 4 ⇒ {2,2} ⇒ {0,0}
   4: ...
3. Represent each subset of the set of multisets of positive integers as the concatenation of each of its elements. If we place a decimal point in the middle of these permutations (the average of the number of digits floored) then we’ve created a bijection between all natural numbers and the parent of the set of all real numbers. To clarify, this bijection includes all possible permutations of the real numbers’ symbols, including infinitely many that are not numbers or that are duplicates.
   N:n>1 ⇒ {M(p)} ⇒ {(M(n)} ⇒ Concatenation
   1: 2 ⇒ {2} ⇒ {0} ⇒ 0 (0.)
   2: 3 ⇒ {3} ⇒ {1} ⇒ 1 (1.)
   3: 4 ⇒ {2,2} ⇒ {0,0} ⇒ 00 (0.0)
   4: …

Conclusion: Creating a bijection between the natural numbers and a set of multisets that contains all possible real numbers proves this set of multisets and by extension the real numbers it holds must be countable.

 

I only just realized I forgot to include the negative reals. I'm sure it's possible bijecting with M(p), but I'm working on a more efficient method instead.