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Find a function $f: \mathbb{R} \setminus \{0\} \rightarrow ??$, which satisfies

$$\ker(f) = ~ \sim$$

for the equivalence relation

$$x \sim y \iff \exists ~ p,q \in \mathbb{Q}: \frac{x}{y} = p+q \sqrt7 \text{ for }x,y \in \mathbb{R} \setminus \{0\},$$

where $\ker(f)$ is defined as $\ker(f):= \{(x,y) \in A \times A: f(x) = f(y) \}$

jjagmath
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almac
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    Has this $f$ been given a concrete domain and codomain in the problem statement? Because if $\ker(f) = {\sim}$ is to make rigorous sense, $f$ actually has to have domain $(\Bbb R\setminus {0})\times (\Bbb R\setminus {0})$, and it's more like you need $f(a, b) = 0$. – Arthur Feb 09 '23 at 20:43
  • The domain is $\mathbb{R} \setminus {0}$, and the codomain was left open with ???. – almac Feb 09 '23 at 20:46
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    Well, then. Considering the fact that ${\sim}\subseteq (\Bbb R\setminus {0})\times (\Bbb R\setminus {0})$, I don't know how one ought to interpret $\ker(f)={\sim}$ in that case. – Arthur Feb 09 '23 at 20:47
  • I also have that issue. I wrote down, how I understood it. I believe, $f(a) = f(b)$ shall be true if $a$ and $b$ are within the same equivalence class (And as far as I can tell, all classes consist of infinite elements). – almac Feb 09 '23 at 20:51
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    @Arthur For any function $f \colon A \to B$, its kernel is $$\ker(f) = {(x,y) \in A \times A : f(x)=f(y)}.$$ – azif00 Feb 09 '23 at 20:53
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    @azif00 That is not a definition of kernel that I have ever seen. It makes sense in this context, and maybe that's what they want you to use, but I have never seen it used before. Kernels have only ever been subsets of the domain in my experience. – Arthur Feb 09 '23 at 20:54
  • To be clear, we want a function such that $f(x) = f(y)$ if and only if $x \sim y$, yes? – eyeballfrog Feb 13 '23 at 19:18
  • @Arthur: That's the definition of kernel that works for any function in any category. It is only in very special contexts (such as groups, rings, modules, and vector spaces) that it can be "coded" as a subset/substructure. See here, for example for how the notion generalizes. – Arturo Magidin Feb 13 '23 at 19:35
  • Well, if you're allowed to choose the codomain, there is the function $f:\mathbb R\setminus{0}\rightarrow 2^\mathbb{R}$ defined as $f(x) = {z : x \sim z}$. Clearly satisfies $f(x) = f(y)$ if and only if $x \sim y$. – eyeballfrog Feb 13 '23 at 19:40
  • Note that the classes don't consist of "infinite elements"; since elements are real numbers, they are all finite. What you probably meant is that every equivalence class is infinite (that is, has infinitely many elements). – Arturo Magidin Feb 13 '23 at 19:44
  • @ArturoMagidin Well, at least it works in concrete categories. And it makes sense. It was just new to me is all. – Arthur Feb 13 '23 at 19:46
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    @Arthur: If you talk "functions", then the category is concrete. Otherwise, you talk about "arrows" or "morphisms". But fair enough. – Arturo Magidin Feb 13 '23 at 19:46
  • Do you search for an explicit function $f$? I think it can not be measurable, so it can not be explicit. Indeed, given a countable subgroup $H$ of $\mathbb{R}$, no Borel subset of $\mathbb{R}$ can contain exactly one element in each equivalence class of $\mathbb{R}/H$. This applies to $H := {\ln(p+\sqrt{7}) : p,q \in \mathbb{Q}}$. – Christophe Leuridan Feb 13 '23 at 20:25

2 Answers2

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One can say $x \sim y$ iff $\frac{x}{y}\in\mathbb{Q(\sqrt{7})}(:=\{a +b\sqrt{7}~|~a,b\in\mathbb{Q} \})$.
So if we want $ker(f)=~\sim ~$, we need: $f(x)=f(y)$ iff $\frac{x}{y}\in\mathbb{Q(\sqrt{7})}$.

Intuition:
If $f(1)=a$, we get $f(x)=a$ iff $x=\frac{x}{1}\in\mathbb{Q(\sqrt{7})}$.
So we get all elements from $\mathbb{Q}(\sqrt{7})$ go to $a$.
In the same manner, we get for some $r\in\mathbb{R}$:
if $f(r)=k$ than $f(x)=k$ iff $\frac{x}{r}\in\mathbb{Q(\sqrt{7})}\iff x\in r\cdot\mathbb{Q(\sqrt{7})}$
By this logic, we see that each equivalence class is of the type:
$r\cdot\mathbb{Q(\sqrt{7})}$.

We can indeed see that:$$r\cdot\mathbb{Q(\sqrt{7})} = k\cdot\mathbb{Q(\sqrt{7})}\iff r(a +b\sqrt{7})=k(c +d\sqrt{7})\iff\frac{r}{k}=\frac{a +b\sqrt{7}}{c +d\sqrt{7}}\in\mathbb{Q}(\sqrt{7})$$

And the same in the opposite way if $\frac{r}{k}\in\mathbb{Q}(\sqrt{7})$ we can get $r\cdot\mathbb{Q(\sqrt{7})} = k\cdot\mathbb{Q(\sqrt{7})}$.

Define $A:=\{r\cdot\mathbb{Q(\sqrt{7})}~|~r\in\mathbb{R}\backslash\{0\}\}$.

We can define $f$ as follows: $$f:\mathbb{R}\backslash \{0\}\rightarrow A\\x\mapsto x\cdot\mathbb{Q}(\sqrt{7})$$

We get $f(x)=f(y) \iff x\cdot\mathbb{Q(\sqrt{7})} = y\cdot\mathbb{Q(\sqrt{7})}\iff\frac{x}{y}\in\mathbb{Q}(\sqrt{7})\iff x \sim y$

Dinor
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Recall that equivalence relations on a set $X$ correspond to partitions of $X$:

  1. Given an equivalence relation $\sim$, the corresponding partition of $X$ is the set of equivalence classes of $X$ modulo $\sim$.

  2. Given a partition $\mathcal{P}$ of $X$, the corresponding equivalence relation is the relation $x\sim y$ if and only if there exists $A\in\mathcal{P}$ such that $x,y\in P$.

In particular, given an equivalence relation $\sim$ on a set $X$, we can let $X/\sim$ be the seet of equivalence classes on $X$. Then we always have a map $$\pi\colon X\to X/\sim$$ given by $\pi(x) = [x]_{\sim}$, where $[x]_{\sim}$ is the equivalence class of $x$ under $\sim$. Note that by definition we will have $$\pi(x)=\pi(y)\iff [x]_{\sim}=[y]_{\sim}\iff x\sim y.$$ This is the "canonical projection" onto the set of equivalence classes.

In addition, one often defines the kernel of a function $f\colon A\to B$ to be the set $$\ker f= \{ (x,y)\in X\mid f(x)=f(y)\}.$$ This is the set that defines the equivalence relation "have the same image under $f$".

Given these facts, it should be straighforward to find a set and function that satisfy the desired properties.

Arturo Magidin
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