0
$\begingroup$

I'm currently reading Mac Lane's "Categories for the Working Mathematician", and on page 18 I found a construction which seems to me to use a universal choice function.

Namely, we work with the categories Finord and Setf, the latter being the category of finite sets and the former the category of sets of the form $$ n = \{0, 1, \ldots, n-1\} $$ as a full subcategory of Setf. And in constructing an equivalence of these categories, we choose a function $\theta_X: X \to \#X$ for each finite set $X$, where $\# X$ denotes the cardinality of $X$ (which is, of course, an ordinal number in a canonical way).

Here is my question:

Is there any way this may be accomplished using the axiom of choice, or do we need to assume the existence of a global choice function (or a different axiom stronger than the axiom of choice)?

$\endgroup$
1
$\begingroup$

You need global choice.

Add using an Easton support product two sets of Cohen subsets to each regular $\kappa$, namely force with $\operatorname{Add}(\kappa,2\times\kappa)$. In the resulting model there is no uniform way of assigning sets of size $2$ a bijection with $\{0,1\}$.

$\endgroup$
  • $\begingroup$ Thank you very much indeed! I'll have to read up on model theory, but as soon as that is done, I'll accept the answer. For now, an upvote will have to do. $\endgroup$ – AlgebraicsAnonymous Feb 22 '19 at 14:40
  • $\begingroup$ You might want to clarify that the pairs from which you can't define choices are not the pairs of Cohen subsets of cardinals that you added (you can choose the lexicographically earlier element from such pairs) but rather pairs of suitable equivalence classes. $\endgroup$ – Andreas Blass Feb 22 '19 at 14:46
  • $\begingroup$ @Andreas: You are right. I managed to somehow convince myself that just pairs of Cohen reals is enough. It felt odd, but I blame the beer festival going on in Norwich. I'll edit as soon as I am home! $\endgroup$ – Asaf Karagila Feb 22 '19 at 15:02

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.