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?
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What is amorphous set? – nonuser Sep 16 '17 at 18:45
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1@johnnobody An infinite set that is not the disjoint union of two infinite sets. – Kenny Lau Sep 16 '17 at 18:46
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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.)
Noah Schweber
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@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. – Noah Schweber Sep 16 '17 at 19:19
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Why would the order type of $L$ be omega?(same question about $R$ and omega*). Can't it be that $L$ is infinite dedekind finite itself? – Holo Oct 21 '18 at 05:23
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@Holo No, that can't occur. I'll describe the $L$-case first. First, for each point $p$ in $L$ let $n_p$ be the number of predecessors of $p$, and note that by definition of $L$ that $n_p$ is always finite. Now show that for $p,q\in L$ we have $n_p=n_q\iff p=q$ (this uses the fact that $n_p,n_q$ are each finite; the analogous fact fails for e.g. $\omega^*$). Finally show that if $L$ is infinite then the map $L\rightarrow\mathbb{N}:p\mapsto p_n$ is in fact an order-preserving bijection (we already have injectivity and order-preservingness is pretty trivial so you're just proving surjectivity). – Noah Schweber Oct 21 '18 at 15:51
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@Holo: Here's a nice exercise I used to give my students, $\Bbb N$ is the unique linear order without a maximal that every proper initial segment is finite (I then used this to prove that the finite subsets of $\Bbb N$ is also a countable set a week later). So if infinitely many points have finitely many predecessors, there is a least one with infinitely many, and the initial segment below it is countable. If there are only finitely many with finitely many predecessors, then by removing them and using a similar argument $\omega^*$ is embedded into the order. – Asaf Karagila Oct 21 '18 at 15:59
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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.
Asaf Karagila
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"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. – Kenny Lau Sep 16 '17 at 19:04
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https://math.stackexchange.com/questions/126010/continuity-and-the-axiom-of-choice – Asaf Karagila Sep 16 '17 at 19:05
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@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. – Noah Schweber Sep 16 '17 at 19:05
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1@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. – Noah Schweber Sep 16 '17 at 19:08
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@Kenny: No, because the construction is of a generic subset, so all of its elements are "transcendental beyond your wildest dreams". And also because you can always add $0$ to the set and it is still Dedekind-finite. So if it bothers you that much, $0$ is in that set. – Asaf Karagila Sep 16 '17 at 19:08
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Forcing is a technique which only adds undefinable objects, in a very precise sense. So most of the time, forcing won't add "explicit" examples of the things you care about. (This second sentence is a bit vague, and there are counterexamples - e.g. in Cohen's model $\mathbb{R}$ is a perfectly definable set which is not well-orderable - but the general point holds.) – Noah Schweber Sep 16 '17 at 19:10
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(Unrelated question for Asaf: do you know of a way to prove that if there's an amorphous set, then there are incomparable Dedekind-finite cardinalities without invoking Ellentuck's work?) – Noah Schweber Sep 16 '17 at 19:13
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@Noah: If I recall correctly, you have that the ordered and unordered pairs of an amorphous sets are incomparable. But I'm not quite sure of that. I need to look something up, but that should be the direction. – Asaf Karagila Sep 16 '17 at 19:15