0
$\begingroup$

One can show that the axiom of choice is equivalent to: "the product of a family of non-empty sets $\{ X_i \}_{i \in I}$ is never empty". Now, I've always seen a relation being defined via $R \subset X \times X$ and $x \leq y \Leftrightarrow (x, y) \in R$. But doesn't one use the axiom of choice implicitly for the existence of $X \times X$? Or is there another way to define a relation?

$\endgroup$
  • 1
    $\begingroup$ Finite products are a non-issue. Besides, $X$ could be empty anyway. $\endgroup$ – Zhen Lin Sep 24 '15 at 12:05
  • 1
    $\begingroup$ A missing axiom of choice does not mean that you couldn't happen to be able to do the same operation in certain cases. You could for example (manage to) prove that $X\times X$ is non-empty without the axiom of choice (given that $X$ is non-empty). $\endgroup$ – skyking Sep 24 '15 at 12:06
  • $\begingroup$ @Zhen Lin: Thanks, edited it. $\endgroup$ – Steven Sep 24 '15 at 12:08
4
$\begingroup$

Relations as subsets of binary products exist regardless of the axiom of choice: they're just sets of ordered pairs, and ordered pairs can be constructed from the axiom of pairing by identifying $(a,b)$ with $\{ \{ a \}, \{ a, b \} \}$ for example.

The key word in your question is 'never', which refers to a hidden universal quantifier - the equivalent form of the axiom of choice you're thinking of is: for all sets $I$, and all families of non-empty sets $\{ X_i \}_{i \in I}$ indexed by $I$, the product $\prod_{i \in I} X_i$ is non-empty.

When $I$ is finite, the statement is true. But the truth for arbitrary $I$ depends on (and is equivalent to) the axiom of choice.

$\endgroup$

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.