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I know that $\bigcup_{\alpha < \omega_1} \Sigma_\alpha^0 = \mathcal{B}(X)$ for any Polish space $X$, but I want to know if it's really necessary to take the union for every ordinal less than $\omega_1$. Is there an example of a Polish space where $\bigcup_{\alpha < \beta} \Sigma_\alpha^0 \varsubsetneq \mathcal{B}(X)$ for every countable ordinal $\beta$? What's the case of $\mathbb{R}$ or $[0, 1]$?

Many thanks in advance.

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    $\begingroup$ The correct term is "stabilize" rather than" collapse". $\endgroup$ – Asaf Karagila Oct 1 '13 at 11:32
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To complement Yann's answer: This is a nice problem, the possible length of Borel hierarchies in different spaces or without assuming the axiom of choice. It has been studied in detail by Arnie Miller. See

Arnold W. Miller. On the length of Borel hierarchies, Ann. Math. Logic, 16 (3), (1979), 233–267. MR0548475 (80m:04003),

Arnold W. Miller. Long Borel hierarchies, MLQ Math. Log. Q., 54 (3), (2008), 307–322. MR2417803 (2009d:03120),

and

Arnold W. Miller. Descriptive set theory and forcing. How to prove theorems about Borel sets the hard way. Lecture Notes in Logic, 4. Springer-Verlag, Berlin, 1995. MR1439251 (98g:03119).

Here is a small sample of his results:

Given a separable metric space $X$, let its Baire order $\mathrm{ord}(X)$ be the least $\alpha$ such that each Borel subset of $X$ is $\Sigma^0_\alpha$ in $X$. For example, $\mathrm{ord}(\mathbb R)=\omega_1$, $\mathrm{ord}(X)=1$ if $X$ is discrete, $\mathrm{ord}(\mathbb Q)=2$, $\mathrm{ord}(X)\le 2$ for any countable metric space $X$.

The result that $\mathrm{ord}(X)=\omega_1$ whenever $X$ is uncountable Polish is due to Lebesgue. Mazurkiewicz asked to identify the ordinals $\alpha$ for which we can find an $X\subseteq\mathbb R$ with $\mathrm{ord}(X)=\alpha$, and Banach conjectured that if $X\subseteq\mathbb R$ is uncountable, then $\mathrm{ord}(X)=\omega_1$. (See On the length of Borel hierarchies for precise references.)

Kunen proved that Banach's conjecture fails in the strong sense that if $\mathsf{CH}$ holds, then for each $\alpha\le\omega_1$ with $2<\alpha$, there is an uncountable $X\subseteq\mathbb R$ with $\mathrm{ord}(X)=\alpha$. Miller improves this by showing that the existence of a Luzin set suffices for this result. He also proves that Banach's conjecture is consistent, and has results under Martin's axiom (for example, there are uncountable subsets $X,Y$ of the Cantor set with $\mathrm{ord}(X)=2$ and $\mathrm{ord}(Y)=3$).

It is consistent with $\mathsf{ZF}$ (without choice) that the Borel hierarchy on the reals has length $\omega_2$ (the relevant models are rather pathological. For example, $\omega_1$ must have countable cofinality). Miller also shows that for any limit $\alpha<\omega_2$ it is consistent that the Borel hierarchy of the reals has length $\alpha+1$. Assuming (significant) large cardinal assumptions, he further shows that for any $\lambda$ (no matter how large) it is consistent that the Borel hierarchy has length at least $\lambda$.

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In

Alexander S. Kechris, Classical descriptive set theory, Springer-Verlag, 1995

which is (one of) the classical reference for (classical) descriptive set theory, you find on page 168 the Theorem 22.4 which says

For any uncountable Polish space, the Borel hierarchy is strict. That is for every countable ordinal $\xi$, $\mathbf{\Delta}^0_\xi \subsetneq \mathbf{\Sigma}^0_\xi \subsetneq \mathbf{\Delta}^0_{\xi+1}$.

In particular, this implies that for every uncountable Polish space $X$ and for every countable ordinal $\xi$, $\bigcup_{\alpha<\xi} \mathbf{\Sigma}^0_\xi \subsetneq \mathcal{B}(X)$.

Here, I assumed you are talking about boldface classes, eventhough you used lightface notations.

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