1
$\begingroup$

In my mind the axiom of choice is about being able to choose one element from each subset of some universal set $X$. I dont see the connection to the Axiom of choice in the part of the proof below.

Let $u(x): \mathbb{R}^2 \rightarrow \mathbb{R}$ with $u(x,2) > u(x,1) \; \forall \; x \in \mathbb{R}$. By density of $\mathbb{Q}$ in $\mathbb{R}$, $ \exists \, q(x) \in \mathbb{Q} : u(x,2) > q(x) > u(x,1).$

Having done that for every $x \in \mathbb{R}^+$, we obtain a function $q: \mathbb{R}^+ \rightarrow \mathbb{Q}$ by the Axiom of choice.

$\endgroup$
2
  • $\begingroup$ What is your question? $\endgroup$ Oct 21, 2017 at 8:25
  • $\begingroup$ How is the choosing of one element from each subset related to "we obtain a function $q: \mathbb{R}^+ \rightarrow \mathbb{Q}$" ? $\endgroup$
    – stollenm
    Oct 21, 2017 at 8:27

4 Answers 4

3
$\begingroup$

To directly address the question

How is the choosing of one element from each subset related to "we obtain a function $q: \mathbb{R}^+ \rightarrow \mathbb{Q}$"?

without going into whether you actually need Choice for this:

Say you for each $x\in \Bbb R^+$ have a prescribed, non-empty $U_x\subseteq \Bbb Q$ (specifically, the interval $(u(x, 1), u(x, 2))\cap \Bbb Q$ of rational numbers). Then what the Axiom of Choice says is exactly that there exists a function $q:\Bbb R^+\to \Bbb Q$ that for each $x\in \Bbb R^+$ gives an element $q(x)\in U_x$.

$\endgroup$
3
  • $\begingroup$ Thank you. I still don't entirely get it: When we choose $q(x) \in U_x$, don't we then just have a function with the domain being only the rationals $q: \mathbb{Q} \rightarrow \mathbb{Q}$? $\endgroup$
    – stollenm
    Oct 21, 2017 at 8:45
  • $\begingroup$ But $x$ isn't rational. It's real and positive. What we've said is "We want a function from $\Bbb R^+$ to $\Bbb Q$, but for each $x\in \Bbb R^+$ we have a designated subset of possible function values that we allow". What the Axiom of Choice does is exactly telling you that there actually exists such a function (although the AoC works with any other sets as well, not just the reals and the rationals). $\endgroup$
    – Arthur
    Oct 21, 2017 at 8:48
  • $\begingroup$ Ahhh. I get it now. Thank you very much! $\endgroup$
    – stollenm
    Oct 21, 2017 at 8:49
3
$\begingroup$

While we do make infinitely (in fact, continuum-many) choices here, we do not need the Axiom of Choice in order to show that a such function $q$ exists. This is because we can explicitly describe an enumeration of $\Bbb Q$ and then simply always pick the "first" rational in the allowed range.

$\endgroup$
1
$\begingroup$

For every $x$, the set $Q_x=\{q\in\Bbb Q\mid u(x,2)>q>u(x,1)\}$ is non-empty. Therefore using the axiom of choice we can choose $q_x\in Q_x$ for all $x\in\Bbb R^+$ and thus we have a function $q(x)=q_x$.

However, the rational numbers are countable, so we don't need the axiom of choice. We can take the rational with the least numerator and the least denominator of that numerator (when considering a rational as a fraction $\frac pq$ with $p$ and $q$ integers without a common divisor other than $1$, and $q>0$).

$\endgroup$
0
$\begingroup$

We have a collection of non empty sets

$$\big((u(x, 1), u(x,2))\cap {\mathbb Q}\big)_{x\in {\mathbb R}}$$

The axiom of choice says that there is a collection of elements $(q_x)_{x\in {\mathbb R}}$ such that

$$\forall x\in \mathbb{R}, q_x\in (u(x, 1), u(x,2))\cap {\mathbb Q}$$

However as @HagenvonEitzen pointed out, this argument is not needed here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.