Usually in set theory, Borel sets are much more nicely behaved than arbitrary sets of reals. One reason for this is Borel determinacy, which immediately yields measurability, Baireness, and the perfect set property for Borel sets; another reason (not totally distinct from the first) is that Borel sets enjoy a number of nice absoluteness properties. General nicenesses like these lead to, for example, the statement that the theory of Borel equivalence relations is largely immune to set-theoretic independence results (Parts of Set Theory immune to independence).

However, in 2008 Arnold Miller (http://arxiv.org/abs/0806.1957) showed that it is consistent with ZF that there is a Borel set which is infinite but Dedekind-finite (that is, admits no non-surjective self-injection). In fact, Miller's example is $F_{\sigma\delta}$, and he shows this is minimal: every Dedekind-finite $G_{\delta\sigma}$ set is finite. Miller notes that it is unknown whether there is an infinite Borel set with no surjection onto it union one additional element, or whether there is an infinite Borel set whose subsets are all countable or co-countable (called "quasi-amorphous").

My question is:

What are some other set-theoretically "bad" behaviors Borel sets can consistently have in $ZF$?

(To be somewhat precise, here "bad" behavior of Borel sets is behavior ruled out by $ZFC+$ some large cardinal axioms. When I say "consistently," I ideally mean "consistently relative to $ZFC$," but I would also be happy with examples which require some extra consistency strength.)

In particular, I suspect that most such results arise from violations of choice. So I would be particularly interested in:

What are some pathological properties Borel sets can have, consistently in $ZFC$?

This latter question of course takes on the pro-large cardinal perspective very forcefully, since in order to argue that a given behavior consistent with $ZFC$ is pathological we need to step outside $ZFC$ for our perspective.

  • $\begingroup$ To my knowledge it's still open, at all, if we can have a quasi-amorphous (I prefer $\aleph_1$-amorphous) set of real numbers. Not just Borel. I'm also not sure if it's consistent to have an infinite Dedekind-finite set of real numbers which cannot be mapped onto itself + 1, so I'm not even sure if we have an example for that as well. $\endgroup$
    – Asaf Karagila
    Apr 7, 2014 at 23:05
  • $\begingroup$ Here is a convoluted way of saying something about Borel sets that is independent over ZFC: If every subset of a set X of reals is (relatively) Borel then X is Borel. $\endgroup$
    – Ashutosh
    Apr 15, 2014 at 0:08

4 Answers 4


Joel speaks on the case where the real numbers are a countable union of countable sets. The Feferman-Levy model is a strange model indeed.

However, I find the Truss construction to be even weirder. Truss repeated the construction of Solovay by starting from an arbitrary limit cardinal, and he proves that the prefect set property holds in that model, and that if we start with a singular cardinal then every set is Borel; if we start with an inaccessible cardinal then we have Solovay's model again.

Consider the Truss model when we start with $\aleph_\omega$, much like the Feferman-Levy. The difference between the two models is delicate.$\DeclareMathOperator{Col}{Col}$

  1. Feferman-Levy begins by collapsing the $\omega_n$'s by using $\prod_{n\in\omega}\Col(\omega,\omega_n)$, and then taking sets hereditarily definable from bounded collapses.

  2. Truss begins by collapsing all the ordinals below $\omega_\omega$ by using $\prod_{\alpha<\omega_\omega}\Col(\omega,\alpha)$, and then taking sets hereditarily definable from bounded collapses.

What is the difference? Truss allows longer and longer sequences of collapsing functions (where a collapse of an ordinal is understood as a generic for the relevant forcing), whereas Feferman-Levy only care about collapsing the cardinals.

The result is stunningly different.

  1. In the Feferman-Levy model, the real numbers is a countable union of countable sets; in the Truss model, the countable union of countable sets of real numbers is countable.

  2. In the Feferman-Levy model, there exists a set whose cardinality is strictly between the cardinality of the continuum and $\omega$; in the Truss model every uncountable set of reals has size continuum.

  3. In both models, every set of real numbers is Borel.

John Truss, Models of set theory containing many perfect sets, Ann. Math. Logic 7 (1974), 197--219.

Other than that, there might be all sort of weird things going on in models of $\sf ZF$, but it's hard to say. Note that Miller's work (on Cohen's famous model) was only done recently, nearly half a century after it was first defined. Unfortunately, there's not much written work on this topic.

  • $\begingroup$ This really is a fantastically weird construction; +1! $\endgroup$ Apr 8, 2014 at 2:42

It is consistent with ZF that the reals $\mathbb{R}$ are a countable union of countable sets. In this case, every set of reals is Borel, and in this model one gets pathological Borel sets.


The following statement is equivalent to the perfect set property for Pi^1_1, hence not decidable in ZFC: Every sigma-compact subset of R^2 which contains uncountably many pairwise disjoint topological circles contains a subset homeomorphic to (2^omega X circle). For a proof, see my paper with Fons van Engelen and Jan van Mill, "Disjoint embeddings of compacta", Mathematika,vol. 41., 1994, 757-784.

Howard Becker

  • $\begingroup$ I was asking specifically about pathologies contradicting choice, but this is still very neat! (Incidentally, which of that statement and its negation is the "pathological" one?) $\endgroup$ Jan 14, 2020 at 20:13

Determinacy implies that for any subset S of any Polish space and any compact K, if S contains an uncountable collection of pairwise disjoint copies of K then it contains a copy of (2^omega X K). This is a generalization of the perfect set property (K is a singleton). Violations of this property can be produced using the axiom of choice; these are the "pathological" cases. This pathology differs from the perfect set property in that it can occur for Borel sets, which I think was your question.

  • $\begingroup$ Lovely, thanks! $\endgroup$ Jan 15, 2020 at 18:23

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