TITLE: How does the Constructibility Degree of a real compare with its Turing Degree?
QUESTION [6 upvotes]: Specifically, is it the case that (for $a,b\in\omega^\omega$) $a$ $\leq_T$ $b$ implies $a$ $\leq_c$ $b$?
I suspect it might be trivial, but not knowing much Recursion Theory, it's hard to see how it could.
Thank you in advance.

REPLY [5 votes]: As Bjorn pointed out, $a\leq_Tb$ implies $a\leq_cb$.  But it should also be mentioned that constructible degrees are much coarser then Turing degrees:  Suppose $\aleph_1^L$ is the real $\aleph_1$.  Then every constructible degree (set of $\leq_c$-equivalent reals) is uncountable (take your real $a\in\omega^\omega$ and consider all coordinate wise sums of $a$ and a constructible real.  These sums are all of the same constructible degree as $a$).  Every Turing degree is countable since there are only countably many Turing machines.
This observation explains Carl Mummert's statement "in $L$ there are extremely large Turing degrees".