Uncountable set of $\Bbb R^n$ that does not contain any perfect subset

This question is inspired from here:

Every uncountable closed subset of $\Bbb R^n$ contains a perfect subset. This chat confirmed me that every uncountable open subset of $\Bbb R^n$ also contains a perfect subset. I tried set of all irrationals, but they also contain a perfect subset. So my question is :

Is there any uncountable subset of $\Bbb R^n$ that does not have a perfect subset?

• I think you are also mislead as I was: there are sets that are neither closed nor open. – Silent Jan 9 '18 at 10:35
• @MatthiasKlupsch "Every uncountable closed subset" ... – Noah Schweber Jan 9 '18 at 10:36

Assuming the axiom of choice, such sets do indeed exist. This is a straightforward transfinite recursion argument:

• Fix a listing $\{P_\eta: \eta<2^{\aleph_0}\}$ of $\mathbb{R}^n$ the set of perfect subsets of $\mathbb{R}^n$ (crucially, there are only continuum many perfect sets!).

• Now we build a sequence of pairs of sets $(In_\eta, Out_\eta)_{\eta<2^{\aleph_0}}$ as follows:

• At stage $\eta+1$, we pick some real $r\in\mathbb{R}^n\setminus (In_\eta\cup Out_{\eta})$ and some $s\in P_\eta\setminus In_\eta$, and set $$In_{\eta+1}=In_\eta\cup\{r\}, \quad Out_{\eta+1}=Out_\eta\cup\{s\}.$$

• At a limit stage $\lambda$ we set $$In_\lambda=\bigcup_{\alpha<\lambda} In_\alpha,\quad Out_\lambda=\bigcup_{\alpha<\lambda} Out_\alpha.$$

• We set $In_0=Out_0=\emptyset$.

Then it's easy to check that $\bigcup_{\eta<2^{\aleph_0}} In_\eta$ is uncountable but contains no perfect subset.

However, if the axiom of choice fails, this might not work: it is consistent with ZF that every uncountable set has a perfect subset. The most natural way this can happen is if the Axiom of Determinacy (AD) holds. This has the drawback that ZF+AD has strictly greater consistency strength than ZF. A more artificial model of ZF + "every uncountable set has a perfect subset" can be constructed merely from the assumption that ZF is consistent; this was done by Truss.

SEMI-RELATED (but hopefully interesting) CODA:

The necessity of choice suggests that "naturally occurring" sets probably won't be counterexamples. Say that a class $\mathcal{S}$ of sets has the perfect set property if for all $X\in\mathcal{S}$, either $X$ is countable or $X$ contains a perfect subset; can we prove that "large" classes of sets have the perfect set property?

The answer turns out to be yes. ZFC proves that the class of Borel sets - indeed, the class of analytic sets - has the perfect set property. And while ZFC doesn't prove that the class of coanalytic (= complement is analytic) sets has the perfect set property, there are natural axioms which do prove this when added to ZFC - namely, large cardinal axioms. In general, the following is a good (informal) heuristic:

If $X$ is a "nicely definable" set of reals, then - possibly assuming large cardinals - $X$ is either countable or contains a perfect subset.

The study of regularity properties (e.g. perfect set property, Lebesgue measurability, property of Baire, ...) of "nicely definable" sets is (one aspect of) descriptive set theory.

Over time, one of the fundamental aspects of descriptive set theory has turned out to be the importance of determinacy principles and the central role of large cardinal axioms; Larson has a nice paper on the history of this, and for further detail I would recommend Kanamori's book.

• Doesn't the last consistency result require an inaccessible? mathoverflow.net/q/275216/30186 – Wojowu Jan 9 '18 at 11:05
• @Wojowu That needs the assumption that $\omega_1$ is regular in $V$, which need not be true in ZF alone. I believe the result is correct as I've stated it (though I could be wrong). – Noah Schweber Jan 9 '18 at 11:22