Skip to main content
Notice removed Draw attention by Kripke Platek
Bounty Ended with Joel David Hamkins's answer chosen by Kripke Platek
edited tags
Link
Alex Ravsky
  • 5.4k
  • 1
  • 17
  • 31
Notice added Draw attention by Kripke Platek
Bounty Started worth 50 reputation by Kripke Platek
Source Link

Infinite multiplicity set of continuous functions

Definitions:

Fix a function $f:\mathbb{R}\to\mathbb{R}$ such that $f$ obtains each value only finite (possibly $0$) number of times. We say $E \subset \mathbb{N}$ is the "multiplicity set" of $f$ iff
$1$. For every $n \in E$, there exists at least one value that $f$ takes exactly $n$ times.
$2$. If $f$ takes on a value exactly $n$ times, then $n \in E$

In other words, $n \in E$ iff $n$ is the multiplicity of some value under $f$.

A subset $E $ of $\mathbb{N}$ is said to be "constructable" iff there exists a continuous function $f$ on $\mathbb{R}$ such that $E$ is the multiplicity set of $f$.

The problem:

If $E$ is a computable subset of $\mathbb{N}$, then is there a way to effectively decide whether or not $E$ is constructable or not?

I asked a similar question about this almost 2 years ago but only for finite subsets of $\mathbb{N}$. MSE user "user21820" posted an answer for the problem which solves the finite case. They also made a conjecture related to the infinite case. So far no progress has been made after that.