Learning rigorous set theory for the first time as a freshman in UNI here. I am working through some problems regarding equivalence relations, where one of such was to prove an isomorphism $\mathbb{R}^2/_\sim\cong [0,\infty)$, where $f:\mathbb{R}^2 \rightarrow \mathbb{R}$, $f(x,y) := \sqrt{x^2 + y^2}$ and $v \sim v' \Leftrightarrow f(v)=f(v')$.

Now the idea for the proof is pretty straight forward, as $f$ is essentially the Euclidean norm on $\mathbb{R}^2$, so each equivalence class is a circle with a certain radius, except for the class $[(0,0)]_\sim$, which is obviously just the origin. Therefore one would take a representative from an equivalence class $\xi \in \mathbb{R}^2/_\sim$, say $v \in \xi$ and then use $f$ to map $v$ to $[0,\infty)$, call this whole map $F:\mathbb{R}^2/_\sim \rightarrow [0,\infty)$, where $F([v]_\sim):=f(v)$.

Obviously, $F$ is surjective, since for any $r \in [0,\infty)$ we construct the pair $w_r:=(r,0)$, therefore $F([w_r]_\sim) = r$, as $v \in [w_r]_\sim \iff f(v)=f(w_r)=r$. Moreover, $F$ is surely injective by the definition of the relation $\sim$, therefore $F$ is a bijection, hence an isomorphism too, which completes the proof.

The question that bugged me lies at taking the representative part. I asked myself, can I always specify a unique representative for any class in the quotient with reason(i.e. can I construct a map that will take a representative with what I have), or do I need AC to "provide" me with such a map? My answer was that yes I can, following the logic bellow:

  1. For every class $\xi \in \mathbb{R}^2/_\sim$ there exists $r \in [0,\infty)$ so that $(r,0) \in \xi$.

Proof: Let $\xi \in \mathbb{R}^2/_\sim$. By definition there exists $v \in \mathbb{R}$ so that $\xi = [v]_\sim$, so $\xi$ is not empty. We notice that for any pair $u:=(a,b) \in \mathbb{R}^2$, $f(u) = \sqrt{a^2 + b^2} = \sqrt{\sqrt{(a^2 + b^2)^2} + 0^2} = f(f(u),0)$, which means that $u\sim(f(u),0)$! Therefore, as $\xi$ is not empty, there exists $r\in[0,\infty)$ so that $(r,0)\in \xi$ (We would otherwise get that $\xi$ is empty, a contradiction).

  1. Such an $r \in [0,\infty)$ is unique.

Proof: Assume $r,r' \in [0,\infty)$ and that $(r',0) \in \xi:=[(r,0)]_\sim$. Then $f(r',0)=f(r,0)$, or $\sqrt{r'^2 + 0^2} = \sqrt{r^2 + 0^2}$, thus $r=r'$, for both are nonegative adn that completes the proof.

  1. Because of the above proven unique existence of such an $r\in[0,\infty)$, we can define a map $S:\mathbb{R}^2/_\sim \rightarrow \mathbb{R}^2$, where $S: \xi \mapsto (r,0), where r\in[0,\infty) and (r,0)\in \xi$. Therefore this map $S$ chooses a representative for a class from the quotient $\mathbb{R}^2/_\sim$.

Is this a sound and rigorous approach? Can it be done better? Is there a possibility that I am missing something here? I really want to understand the significance of the axiom of choice, so I try to really understand where I need to apply it and where not.

Hopefully this is a well fromed and good question. I am a novice in LaTeX and MD still, so take the outline/format with a grain of salt. Thank you for your help in advance!


The axiom of choice tells you that for every indexed set of sets there exists a choice function.

But if you have a particular indexed set of sets such as your set $\{f^{-1}(r) \mid r \in [0,\infty)\}$, and if you can yourself with your bare hands construct a choice function by writing down a concrete formula such as $\xi(r)=(r,0)$, then you have no need of the axiom of choice.

You don't need a tool factory to build a lever.

Nonetheless, there are abstract situations where you just don't have any concrete way of producing a choice function. I don't know what your level is, but one of the first places that one encounters this situation is proving the existence of a basis of an arbitrary vector space. For finite dimensional vector spaces, in a rigorous linear algebra course one learns how to employ the Gram-Schmidt process to construct a basis, based on a finite sequence of single choices --- single vectors --- which does not rely on the axiom of choice. But for infinite dimensional vector spaces, there is no recourse but to apply the axiom of choice.

  • $\begingroup$ I altered my notation to try to match more with the notation of your question. $\endgroup$ – Lee Mosher Jan 25 at 15:25
  • $\begingroup$ Thank you for the answer. I focused on the equivalent of AC that there is always a set of representatives for a quotient, which was more confusing. I would still need to do some of the argument above to prove I can partition the quotient with the index set of $[0,\infty)$ though right(indexes uniquely coreesponding with the classes)? We did prove the existance of the basis for any vector basis using the Zorn's lemma, but yes we did prove it for finite d. v. spaces by extending a linearly independent subset using a finite spanning list by concating the two and removing the dependent vectors. $\endgroup$ – Berserker1984 Jan 26 at 16:17
  • $\begingroup$ Once you have the choice function $\xi : [0,\infty) \to \mathbb R^2$, you can then simply do function operations. For example (and I think this answers your last question), the map $S$ which takes each element of $\mathbb R^2$ to the representative of its equivalence class is simply the composition $S = \xi \circ f$: $$S : \mathbb R^2 \xrightarrow{f} [0,\infty) \xrightarrow{\xi} \mathbb R^2$$ $\endgroup$ – Lee Mosher Jan 26 at 16:44

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