Let $f: \mathbb{R} \to \mathbb{R}$ satisfy $f(2x)=f(x)$ and let $A \subset \mathbb{R}$ such that if I then specify $f(a)$, $\forall a \in A$ then $f(x)$ will be defined $\forall x \in \mathbb{R}$. What is the "smallest" set $A$ that satisfies this property?
I know that any set of the form $A = (-b, b), \forall b \in \mathbb{R}_+$ satisfies this property, and that all of those sets have no smallest member, since $\forall b \in \mathbb{R}, \exists c < b$. The question then reduces to asking if there are any sets A which have this property that are not of the aforementioned form. Thank you in advance.
The answer is not very satisfying but it is the following. We make two simplifications.
Now consider the equivalence relation on $\mathbb{R}^{+} = \{ x \in \mathbb{R} : x > 0 \}$ generated by $x \sim 2x$; explicitly this equivalence relation is given by $x \sim y$ iff $\frac{x}{y} = 2^k$ for some $k \in \mathbb{Z}$. By definition, if we know the value of $f(x)$ then we know the value of $f(y)$ for all $y \sim x$; moreover this is best possible. So the minimal (with respect to inclusion) subsets $A \subset \mathbb{R}^{+}$ which determine the values of $f$ are exactly those subsets which contain exactly one element of each equivalence class. Such subsets always exist by the axiom of choice but generally can't be constructed explicitly. These are sometimes called transversals.
However, one set of transversals which can be constructed explicitly is
$$A_k = \left[ 2^k, 2^{k+1} \right), k \in \mathbb{Z}.$$
None of them contain each other (in fact they partition $\mathbb{R}^{+}$) and their measures get arbitrarily small. To see that these are in fact transversals think about $\log_2 x$.