There are some models of $\sf ZF$ without the Axiom of choice, where some paradoxical statements hold that are not possible in $\sf ZFC$ (we do not require that all those statements necessarily hold in the same model). Paradoxical means that the statement may seem counterintuitive if one tries to identify the corresponding model with the intended universe of sets as usually conceived (this is, of course, subjective, or may have no sense at all for a pure formalist).

Here are some examples:

  • There is a collection of non-empty sets whose Cartesian product is empty.
  • There is a set that can be partitioned into disjoint nonempty parts, such that the number of parts exceeds the number of elements of the set.
  • There is an infinite set without a countable infinite subset.
  • A countable union of countable sets may be uncountable.
  • The set of reals $\mathbb R$ is a countable union of countable sets.
  • There is a pair of sets such that none of them is equinumerous with a subset of the other.

(some of these, and others are also listed here)

I'm interested in seeing more such examples, particularly those accessible to most people with only basic understanding of naïve set theory, and not too specific to some area like topology or group theory.


closed as too broad by Andrés E. Caicedo, user91500, Asaf Karagila, Claude Leibovici, Mauro ALLEGRANZA Jun 3 '14 at 7:53

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ math.stackexchange.com/questions/120925/… $\endgroup$ – Asaf Karagila Jun 3 '14 at 4:01
  • $\begingroup$ Almost all our intuition is based on choice, so pretty much every counterexample to the axiom of choice is paradoxical. I find the question a bit too broad. $\endgroup$ – Asaf Karagila Jun 3 '14 at 4:02
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    $\begingroup$ @AsafKaragila I think many people with some basic understanding of naïve set theory, but without a knowledge of the role of AC, faced with statements like "not every set can be well-ordered" or "every set of reals is determined" or "not all bounded subsets of $\mathbb R^3$ with non-empty interiors are equidecomposable" would not immediately reject them as counter-intuitive. $\endgroup$ – Vladimir Reshetnikov Jun 3 '14 at 4:38
  • $\begingroup$ Okay, yeah, maybe you're right. But there's still a frightening amount of statements which do break down our basic intuition. Of course, for this you have to have some mathematical intuition, after all the statement "every vector space has a basis" might not be intuitive for some people. But if you expect infinite sets to behave a bit like finite sets, then what I say is true, there are many, many many counterexamples which are paradoxical in nature. $\endgroup$ – Asaf Karagila Jun 3 '14 at 4:41
  • $\begingroup$ @VladimirReshetnikov, you can click on the "reopen" button just below your question in order to re-open it. $\endgroup$ – Mikhail Katz Jul 16 '17 at 11:36