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August 12

Coanalytic counterexample to the continuum hypothesis?

This question comes out of an exchange at talk:aleph number. It follows from ZFC alone that every analytic set (${\mathrm{\Sigma }}_1^1$) of real numbers has the perfect set property, and in particular is either countable or has the cardinality of the continuum. This follows from Borel determinacy (which is the most you can prove in ZFC) via the unfolded perfect set game.

Also, any complement of an analytic set (coanalytic set; ${\mathrm{\Pi }}_1^1$ set) is the union of ${\mathrm{\aleph }}_1$ Borel sets. (I came across this point when searching for the unfolded perfect set game; I don't really have a direct RS but it makes sense for reasons I can explain if anyone's interested, and for that matter I could probably find it in Moschovakis if I had time to look it up.)

So it follows that any coanalytic set is either countable, has cardinality ${\mathrm{\aleph }}_1$, or has cardinality $2^{{\mathrm{\aleph }}_0}$. If the continuum hypothesis holds, the latter two are equal.

Finally get to the question: Is this the best you can do in ZFC alone? Specifically, is it consistent with ZFC+~CH that there is a coanalytic set of reals of cardinality exactly ${\mathrm{\aleph }}_1$? --Trovatore (talk) 21:12, 12 August 2021 (UTC)

It's provable in ZFC that there is a coanalytic set of cardinality exactly ${\mathrm{\aleph }}_1$: the self-constructible reals. A real $X$ is self-constructible if there is an $\alpha <{\omega }_1^{ck}(X)$ with $X\in L_{\alpha }$. It's straightforward to verify that this has cardinality at most ${\mathrm{\aleph }}_1$. Conversely, for any countable $\alpha$, the atomic diagram of $L_{\alpha }$ is self-constructible, so it has cardinality exactly ${\mathrm{\aleph }}_1$. To see that this is ${\mathrm{\Pi }}_1^1$, $X$ is self-constructible iff there is $R\in {\mathrm{\Delta }}_1^1(X)$ a relation on $\omega$ which is well-founded and collapses to an $L_{\alpha }$ containing $X$.--2404:2000:2000:8:D87B:C574:8159:3B76 (talk) 22:36, 12 August 2021 (UTC)

No, that can't be right. A single measurable cardinal gives you ${\mathrm{\Sigma }}_1^1$ (and therefore ${\mathrm{\Pi }}_1^1$) determinacy, and thereby the perfect set property for ${\mathrm{\Sigma }}_2^1$. --Trovatore (talk) 22:47, 12 August 2021 (UTC)

You're right, my argument for uncountability is flawed.--2404:2000:2000:8:D87B:C574:8159:3B76 (talk) 22:49, 12 August 2021 (UTC)

It might work in forcing extensions of L that don't collapse ${\mathrm{\aleph }}_1$. --Trovatore (talk) 22:49, 12 August 2021 (UTC)

Right. Self-constructible reals are cofinal in ${\omega }_1^L$, so start with $V=L$ and add a Cohen generic.--2404:2000:2000:8:D87B:C574:8159:3B76 (talk) 23:12, 12 August 2021 (UTC)