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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.

user26857
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Demonicdrn
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    Define an equivalence relation $\sim$ by $a\sim b$ iff $a=2^nb$ for some $n\in\mathbb Z$. Then pick a representative from each equivalence class in $\mathbb R/\sim$. (This requires the axiom of choice.) The totality of all these representatives is a desired smallest set. – user1551 Sep 02 '23 at 06:15
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    Your example is incorrect, you need to include negative reals, so e.g. you could take $A = (-b, b)$. – Qiaochu Yuan Sep 02 '23 at 06:24
  • As a [very] special case: if $f$ is continuous at $0$, then $A={e^\pi}$ works. – Brian Moehring Sep 02 '23 at 06:25

1 Answers1

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The answer is not very satisfying but it is the following. We make two simplifications.

  • First of all, $A$ must contain $0$, and the value of $f(0)$ is unrelated to the value of $f$ at any other point; from here on we will ignore it.
  • Second, the values of $f$ at positive and negative reals are unrelated, and the problem decomposes into two copies of the problem for the positive reals; from here on we will restrict our attention to the positive reals.

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 $\log_2 x$.

Qiaochu Yuan
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  • I suppose you might ask from here whether there exists a transversal of measure zero. Maybe? Most of them will be non-measurable, I think. – Qiaochu Yuan Sep 02 '23 at 08:24