5
$\begingroup$

I remember hearing some time ago that there is a locally compact Hausdorff space $X$ and a non-Borel subset $E$ which intersects every compact set in a Borel set. (This would contradict Lemma 13.9 of Royden, Real Analysis 3rd edition 1988, which is stated without proof).

Is there a reference for this? Can this happen if the space $X$ is perfectly normal?

$\endgroup$

2 Answers 2

4
$\begingroup$

If I understand the question, then you are correct, there is such a space. I'll sketch what I hope is a correct argument.

Take $X=\coprod_AY$ for some fixed locally compact Hausdorff space $Y$ and some index set $A$. As long as $Y$ is sufficiently complicated (probably $Y=\mathbb R$ would work) and $A$ is sufficiently large ($\left|A\right|\geq\aleph_1$ is enough), then you can find a collection of Borel sets $\{B_\alpha\subseteq Y\}_{\alpha\in A}$ which are not all contained in any countable "stage" towards the Borel $\sigma$-algebra of $Y$ (I'm sorry I don't know the standard terminology for this; using the notation of the wikipedia entry on Borel sets, I mean that for any countable ordinal $m$, not all of the $B_\alpha$ are contained in $G^m$). Now the subset: $$\coprod_{\alpha\in A}B_\alpha\subseteq\coprod_{\alpha\in A}Y$$ is not a Borel set. If it were, it would be contained in some $G^m(X)$ for some countable ordinal $m$, but this would imply that every $B_\alpha$ is in $G^m(Y)$, a contradiction.

If we take $Y$ metrizable, then $X$ will be metrizable as well, and hence perfectly normal.

$\endgroup$
3
  • $\begingroup$ If $Y$ is the reals, your $X$ is neither locally compact nor metrizable. $\endgroup$ Aug 5, 2014 at 0:16
  • $\begingroup$ @BillJohnson: coproduct, not product :) $\endgroup$ Aug 5, 2014 at 0:17
  • $\begingroup$ I like both answers, but I can only check one of them. In this answer we take $Y=\mathbb{R}$ and $A=\omega_1$, as suggested. This one is more "concrete" ... and it explicitly answers the perfectly normal query. $\endgroup$ Aug 17, 2014 at 14:06
8
$\begingroup$

Take the ordinal $\omega_{1}$. Then every subset of $\omega_{1}$ intersects each compact subset of $\omega_{1}$ in a Borel set. However, not every subset of $\omega_{1}$ is Borel. If $\mathcal{M}$ is the collection of all sets which are either non-stationary or contains a closed unbounded set, then $(\omega_{1},\mathcal{M})$ is a $\sigma$-algebra that contains each closed set and hence each Borel set. However, $(\omega_{1},\mathcal{M})$ is a proper $\sigma$-subalgebra of $P(X)$.

$\endgroup$
7
  • $\begingroup$ So the only problem is to exhibit a non-Borel subset of $\omega_1$. Can this be done "constructively", e.g. by defining a certain transfinite subsequence of $0\le \alpha<\omega_1$ $\endgroup$ Aug 5, 2014 at 0:23
  • $\begingroup$ I cannot immediately think of a a specific example of a non-Borel subset of $\omega_{1}$. However, the fact that $\mathcal{M}$ is a proper $\sigma$-algebra follows from the fact that the club filter is not an ultrafilter since no non-principal $\sigma$-complete ultrafilters appear until we reach the first measurable cardinal. In fact, Solovay's theorem states that $\omega_{1}$ can be partitioned into $\aleph_{1}$ stationary sets (and each element of this partition is non-Borel). $\endgroup$ Aug 5, 2014 at 0:31
  • 1
    $\begingroup$ @FredDashiell: Without AC, it is possible for the closed unbounded subsets of $\omega_1$ to generate an ultrafilter. This follows from AD, for example. So there is a limit on how "constructive" this can be. $\endgroup$ Aug 5, 2014 at 12:51
  • $\begingroup$ @François: Consistently this can happen without large cardinals, but then countable choice fails. $\endgroup$
    – Asaf Karagila
    Aug 6, 2014 at 0:04
  • $\begingroup$ Do we have a theory of "analytic" subsets of $\omega_1$, i.e., subsets generated from the closed sets by operation A? This is constructive enough for me. $\endgroup$ Aug 6, 2014 at 1:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.