Construction of amorphous subset of $\Bbb R$

Question

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?

Answer 1

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.

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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.

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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.)

Answer 2

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.