Why are there exactly $2^{\aleph_0}$ Turing degrees?

Question

I know that there are exactly $2^{\aleph_0}$ Turing degrees, but I am trying to realize the reason: I am convinced there can't be more than $2^{\aleph_0}$ of them, but why can't there be less? I know it should be trivial...

Answer 1

For each real $X$, the set of things Turing below $X$ is countable. In set theoretic notation $|\{Y : Y \leq_t X\}| = \aleph_0$. This is because there are only countably many Turing machines. Consequently, for each $X$ its (Turing) degree $\text{deg}(X) = \{ Y : Y\leq_t X \ \& \ X\leq_T Y\}$ is at most countable.

Turing equivalence $\equiv_T$ partitions the set of reals into countable equivalence classes, and you can show that if $|\bigcup_{i\in I} C_i| = 2^{\aleph_0}$ when each $C_i$ is countable, then $I$ has size at least continuum.

I don't think any 1-degree is equal to any m-degree is equal to any $t$-degree, but I could be mistaken. There are, however, continuum many 1 degrees and m-degrees too.