Accetping axiom of choice gives rise to all sort of paradoxes. What if we assume the contrary?


marked as duplicate by Asaf Karagila axiom-of-choice Aug 1 '17 at 1:08

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  • $\begingroup$ It depends which contrary you mean. If you allow say dependent choice then you can at least do most of basic real analysis. $\endgroup$ – Ian Jul 31 '17 at 22:03
  • $\begingroup$ Well, I definitely find the cardinals not being strictly ordered to be counterintuitive. Or the emptiness of some Cartesian products of non-empty sets. $\endgroup$ – Malice Vidrine Jul 31 '17 at 22:14
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    $\begingroup$ If you just assume the negation of AC, you're not committing to any particular violation of AC, just some unspecified one. Maybe it just fails for sets of some enormous cardinality, so that for everything you could possibly be interested in it is still true. $\endgroup$ – Robert Israel Jul 31 '17 at 22:33

I find these two to be the most "obviously true" equivalents to the axiom of choice:

  1. Empty cartesian products: The axiom of choice is equivalent to the assumption that every cartesian product of non-empty sets is non-empty. So if the axiom of choice is false there is a collection of nonempty sets whose cartesian product is empty. (This condition is sometimes the defintion of the axiom of choice, though.)
  2. Cardinal trichotomy: The axiom of choice is equivalent to the assumption that for any two sets $A$ and $B$, either there exists and injective map from $A$ into $B$ or there exists an injective map from $B$ into $A.$ In other words, without the axiom of choice, certain pairs of sets are incomparable with regard to cardinality.

It is also surprising that some even more innocuous and obviously true statements require some weak form of choice to prove:

  1. A countable union of finite (or countable) sets is countable (requires countable choice).
  2. Any set which is not finite (i.e. not equipotent to a natural number) has a countably infinite subset (requires countable choice)

Others have mentioned that the axiom of choice gives us a lot of things we like. I would also add that it makes the cardinalities of infinite sets into a nice orderly hierarchy (The only problem is that it's impossible to figure out where cardinalities go in that hierarchy... it's consistent with ZFC that the cardinality of the reals could be pretty much anywhere.)

You are free to not assume the axiom of choice, but if you want to "assume the contrary" you must state what you are assuming. If you work without assuming the axiom of choice but without any replacement (i.e. in ZF), the world of infinite sets is kind of a difficult-to-navigate mess (whereas as I mentioned before, the axiom of choice gives it a nice structure). Just assuming the bare negation, you know for instance that some sets are incomparable but have no idea which ones. So it's more fruitful to replace the axiom of choice with a positive assertion that gives some information about the nature of sets that exist. The most popular "anti-choice axiom" is the axiom of determinacy.


Well, I don't know a ton on this subject. But without the Axiom of Choice, a number of commonly used mathematical theorems can no longer be proven - for example:

-Tychonoff's Theorem - which states that under the product topology, any product of compact spaces is compact.

-Every ring has a maximal ideal.

-The existence of a non-Lebesgue measurable set.

If you assume the negation of choice, you lose these commonly cited and useful results, and many I haven't mentioned or do not know. However, undoubtedly there are other things which you could prove, since that assuming there is a model of ZF, there is a model of ZF $\cup \neg $ AC.

For example, the Axiom of Determinacy is not consistent with the Axiom of Choice, and you can look at some of the things proven with the Axiom of Determinacy as things that could not be proven in ZFC.


From the article, an answer to your exact question in the heading:

The axiom of determinacy is inconsistent with the axiom of choice (AC); the axiom of determinacy implies that all subsets of the real numbers are Lebesgue measurable, have the property of Baire, and the perfect set property.


Maybe problematic question: You need AC to prove that every vector space has base. You need countable AC to prove that continuity defined vie $\delta, \epsilon$ and $\lim x_{n}$ are the same. You need AC to prove that every set dedekind finite is finite.

  • $\begingroup$ (i) countable AC is strictly weaker than AC, so the continuity definition isn't a convincing point (ii) it's certainly not particularly intuitive that every vector space has a base, since many vector spaces aren't that intuitive (iii) Dedekind finite $\Longleftrightarrow$ finite doesn't require the full strength of AC, only a weaker form $\endgroup$ – MathematicsStudent1122 Jul 31 '17 at 22:37

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