There are many amazing results about amorphous sets. However, I have yet to find one actual construction. Can an amorphous subset of $\Bbb R$ be explicitly constructed, assuming the negation of choice?

  • $\begingroup$ What is amorphous set? $\endgroup$
    – nonuser
    Sep 16, 2017 at 18:45
  • 1
    $\begingroup$ @johnnobody An infinite set that is not the disjoint union of two infinite sets. $\endgroup$
    – Kenny Lau
    Sep 16, 2017 at 18:46

2 Answers 2


There is no amorphous set of reals - in fact, no amorphous set can be linearly ordered.

(For further discussion of what kind of structure amorphous sets, and more generally Dedekind-finite sets can have, see this paper of Truss or Agatha Walczak-Typke's Ph.D. thesis.)

To see this, suppose $A$ is linearly ordered. Let $L$ be the set of elements of $A$ with only finitely many things to the left, and $R$ be the set of elements of $A$ with only finitely many things to the right. If $L$ is infinite, then $L$ has ordertype $\omega$; similarly, if $R$ is infinite, $R$ has ordertype $\omega^*$. (Why? It's a nice induction argument which I'll leave as an exercise; for an outline, see my comment below to Holo.)

Either possibility yields a countably infinite subset of $A$, which can't happen since $A$ is amorphous. So $L$ and $R$ are each finite, and since $A$ is infinite this means $L\cup R\subsetneq A$.

Pick $a\in A\setminus(L\cup R)$, and think about $\{b\in A: b<a\}$ and $\{c\in A: c\ge a\}$.

To make this perhaps easier to visualize, note that $17+\mathbb{Z}+17$ gives an example of an infinite linear order where both $L$ and $R$ are finite and nonempty. So the situation described above isn't actually too weird.

EDIT: The situation with Dedekind-finiteness is quite different. A set of reals can be infinite but Dedekind-finite; indeed, this happens in Cohen's original model of ZF+$\neg$AC (the generic set of reals he adds is Dedekind-finite in that model), and such a set can even be Borel.

FURTHER EDIT: I can't help but add a very silly fact; or rather, the fact isn't particularly silly, but the only proof I know looks a lot like "nuking a mosquito:"

Suppose there is an amorphous set. Then there are Dedekind-finite sets of incomparable cardinality.

Proof: An amorphous set is clearly neither strongly even (= can be partitioned into two equinumerous sets) or strongly odd (= can be partitioned into two equinumerous sets plus one singleton). But if the Dedekind-finite cardinalities are linearly ordered, they form a nonstandard model of true arithmetic, and every natural number is either (strongly) even or (strongly) odd. :P

(Note that a Dedekind-finite set can be e.g. weakly even, that is, partitionable into pairs.)

  • $\begingroup$ Can an infinite subset of $\Bbb R$ be Dedekind-finite? $\endgroup$
    – Kenny Lau
    Sep 16, 2017 at 18:59
  • $\begingroup$ @KennyLau See my edit. $\endgroup$ Sep 16, 2017 at 19:02
  • $\begingroup$ @_@ strongly even and weakly even $\endgroup$
    – Kenny Lau
    Sep 16, 2017 at 19:15
  • $\begingroup$ @KennyLau There are two different notions of "even" for arbitrary sets - partitionable into pairs, and partitionable into two equinumerous sets. With choice these are the same, but without choice the former is strictly weaker than the latter (this is just the idea of Russell's shoes and socks). The arithmetic structure on Dedekind-finite sets, assuming that they are linearly ordered, corresponds to the strong notion of evenness (etc.), so every Dedekind-finite set in such a model is either strongly even or strongly odd. $\endgroup$ Sep 16, 2017 at 19:19
  • $\begingroup$ Noah, you're confusing the poor lad. :) $\endgroup$
    – Asaf Karagila
    Sep 16, 2017 at 19:34

One cannot "explicitly construct" any counterexample of the axiom of choice. If one could explicitly construct it, then the construction would have gone through in $\sf ZFC$ as well. Not to mention that even if you work in $\sf ZF+\lnot AC$, then you still don't know how much choice is failing and where.

It could be that there are no Dedekind-finite sets; and it could be that there are Dedekind-finite sets, but no amorphous sets; and it could be that there are amorphous sets, but choice fails so high up in the von Neumann hierarchy that the real numbers and any set a "working mathematician" would ever dream of using can still be well-ordered.

On top of this, of course, there is what Noah explains. An amorphous set cannot be linearly ordered, whereas any subset of the real numbers can.

Alas, I feel that you might be asking about this in relation to a previous question from today about a set where every permutation has a fixed point, where I pointed out that amorphous sets can provide a counterexample.

In this case, fear not. In Cohen's original model of $\sf ZF+\lnot AC$, where he effectively proved that $\sf ZF$ does not prove the axiom of choice, there is a Dedekind-finite set of reals, which while being very far from being amorphous, it does have a very interesting property: every finitary partition (i.e. a partition into finite sets) will necessarily have all but finitely many parts as singletons. Therefore any permutation of this set will also move only finitely many points.

  • $\begingroup$ "there is a Dedekind-finite set of reals" which set? Also, the reason I asked this is to provide an explicit example of a function sequentially continuous at a point but not epsilon-delta continuous at such point. $\endgroup$
    – Kenny Lau
    Sep 16, 2017 at 19:04
  • $\begingroup$ math.stackexchange.com/questions/126010/… $\endgroup$
    – Asaf Karagila
    Sep 16, 2017 at 19:05
  • $\begingroup$ @KennyLau Like I said at the end of my answer, it's exactly the set of reals Cohen adds. That is, the forcing in question produces a sequence $a_i$ of reals, $i\in\omega$; in the symmetric submodel, we lose the sequence but keep the set $\{a_i: i\in\omega\}$, and this is Dedekind-finite in that model. $\endgroup$ Sep 16, 2017 at 19:05
  • $\begingroup$ Can anyone tell me which number is actually in that set? $\endgroup$
    – Kenny Lau
    Sep 16, 2017 at 19:07
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    $\begingroup$ @KennyLau I think you don't understand. This set of reals, and each real in this set, is extremely undefinable - e.g. each real in the set is Cohen generic over the whole model. $\endgroup$ Sep 16, 2017 at 19:08

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