Compute any real number, given an uncountable set of real numbers? Given an uncountable set of real numbers $S$, for any real number $r$, is there a Turing machine that computes $r$ given an oracle for some countable subset of $S$?
If yes, then is there a Turing machine that computes $r$ given some finite subset of $S$?
 A: For simplicity, by "real" I mean "infinite binary sequence;" there's a slight hiccup here vis-a-vis the dyadic rationals, but since Turing machines really act on the expansions of reals and not the reals themselves this is actually more appropriate.
Note that there's an important subtlety when talking about using a countable set of reals as an oracle: you're not really using a countable set of reals, but rather a countable sequence of reals (e.g. when you say "all of the real numbers interlaced together," there are different ways to do this for the same set of reals).
Now this may seem benign, but it's definitely not:

Suppose I have a countably infinite set $A$ of reals, and $r$ is an arbitrary real. Then there is some listing $(b_i)_{i\in\mathbb{N}}$ of $A$ such that $r$ is computable from the real $\bigoplus_{i\in\mathbb{N}} b_i$ (this is just the real you get by interlacing the $b_i$s together in order - that is, its $\langle m, n\rangle$th bit is the $m$th bit of $b_n$).

An earlier version of this answer gave a vastly over-complicated argument.
Proof: First, a bit of notation. Given distinct reals $s,t\in 2^\omega$, write $s\triangleleft t$ iff $s(n)<t(n)$ where $n$ is the first point of difference between $s$ and $t$. Note that we can computably in $s$ and $t$ determine whether $s\triangleleft t$ or $t\triangleleft s$.
Now fix some enumeration $(a_i)_{i\in\omega}$ of $A$, with all the $a_i$s distinct (we can do this since $A$ is infinite). We define a new enumeration $(b_i)_{i\in\omega}$ of $A$ by permuting adjacent pairs of $a_i$s in order to code $r$ into the "pattern of first differences." Specifically, say that $i\in\omega$ is correct iff $$r(i)=0\leftrightarrow a_{2i}\triangleleft a_{2i+1}.$$
Now define reals $b_i$ for $i\in\omega$ as follows:

*

*If $i$ is correct then $b_{2i}=a_{2i}$ and $b_{2i+1}=a_{2i+1}$.


*If $i$ is not correct then $b_{2i}=a_{2i+1}$ and $b_{2i+1}=a_{2i}$.
The sequence $(b_i)_{i\in\omega}$ is then an enumeration of $A$, and for each $i$ we have $$r(i)=0\leftrightarrow b_{2i}\triangleleft b_{2i+1}.$$ So $\bigoplus_{i\in\omega}b_i\ge_Tr$.

To address the second part of your question (which I forgot about), the answer is no, and there are many ways to prove this. For example, by transfinite recursion (HINT: use the exact pair theorem) we can define a $<_T$-increasing sequence of reals $r_\eta$ (for $\eta<\omega_1$) such that no $r_\eta$ computes (say) $0'$; since the sequences is $<_T$-increasing, this means no finite set of $r_\eta$s computes $0'$ either.
Actually, though, with a bit of work we can do much better. We can produce a perfect set (closed with no isolated points - all such have size continuum) of reals no finitely many of which compute (say) $0'$. Roughly, this works as follows. We want to build a function $F$ from $2^{<\omega}$ to $2^{<\omega}$ with the following properties:

*

*$\sigma\prec\tau$ iff $F(\sigma)\prec F(\tau)$.


*For all $n\in\mathbb{N}$ - and letting $\{\sigma_j: j<2^n\}$ list the binary strings of length $n$ - we have $$\Phi_n^{g_0\oplus g_1\oplus ...\oplus g_{2^n-1}}\not=0'$$ whenever $g_j\succ F(\sigma_j)$ (for $j<2^n$).
(Of course, we must show that such a tree exists, but it's not too hard - the key point being that $0'$ isn't computable, so we can't get "stuck" at any finite stage in the construction of $F$ since otherwise we'd be able to figure out what $0'$ is from finite information; note that this is not to say that $F$ is computable (it isn't).)
Given $f\in 2^\omega$, we let $f^*$ be the real $$\bigcup_{\sigma\prec f}F(\sigma);$$ our set $S$ will then be $\{f^*: f\in 2^\omega\}$.
