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I'm trying to do due diligence and determine whether this is known, trivial, original, etc. I have a proof of:

Theorem: If $S\subseteq \mathbb{N}^{\mathbb{N}}$ then $S=X\cup Y$ for some $X$ which is boldface $\mathbf{\Delta}_2^0$ and some $Y$ with no isolated points, with $X\cap Y=\emptyset$.

Anybody else have a proof of this?

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  • $\begingroup$ The usual Cantor-Bendixson theorem tells you that your set $S$ is the disjoint union of a countable set $X$ and a set $Y$ with no isolated points. The countable set is boldface $\Sigma^0_2$. But the countable set is scattered, i.e., every subset has an isolated point. Does this somehow give you boldface $\Pi^0_2$ as well? Or do you use a different decomposition? $\endgroup$
    – Stefan Geschke
    Jun 4, 2011 at 20:57
  • $\begingroup$ Stefan, isn't every countable set boldface $\Delta^0_1$? (or better?) This would make the result an immediate consequence of the Cantor-Bendixon argument. Just throw out isolated points until there are none left, in a transfinite recursion; one obtains the original set as a union of a countable set (hence $\Delta^0_2$) and a set with no isoltated points, as desired. $\endgroup$
    – Joel David Hamkins
    Jun 4, 2011 at 21:47
  • $\begingroup$ Ahh, that is right. Thanks Joel. I failed to realize the Cantor-Bendixson rank is always countable, but now that I think about it, of course it is. In light of all this, here is an interesting alternate statement of CBT: "Every $S\subseteq\mathbb{N}^{\mathbb{N}}$ is a disjoint union of a countable set $X$ and a set with no isolated points $Y$; and $Y$ can be taken to be closed if $S$ is closed (in which case the union is unique)" $\endgroup$
    – Sam Alexander
    Jun 4, 2011 at 22:19
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    $\begingroup$ @Joel: Boldface $\Delta^0_1$ is the same as clopen, so nonempty countable sets are not $\Delta^0_1$. They are $\Sigma^0_2$ (i.e., $F_\sigma$), as Stefan said, but an additional argument is needed to get $\Pi^0_2$ (i.e., $G_\delta$). That argument will need something like scatteredness, since a countable dense set can't be $G_\delta$ by the Baire category theorem. $\endgroup$
    – Andreas Blass
    Jun 5, 2011 at 2:28
  • $\begingroup$ Yes, I've come to the same conclusion myself. My comment above is not correct. $\endgroup$
    – Joel David Hamkins
    Jun 5, 2011 at 3:43

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Stefan Geschke and Joel David Hamkins pointed out that this is a corollary of the proof of the Cantor-Bendixson theorem. In fact, we can strengthen the conclusion: X is not just $\Delta_2^0$, it is countable and scattered(*). And the proof, like that of Cantor-Bendixson's theorem, is simple: remove isolated points repeatedly until none remain (this may take more than $\omega$ steps and require ordinals, but a simple argument shows it only requires ordinals below $\omega_1$, and thus only countably many isolated points need to be removed).

(*Scattered implies $G_{\delta}$, though this is not trivial. Thus countable+scattered is strictly stronger than $\Delta_2^0$)

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