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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?

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up vote 4 down vote accepted

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.

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If $Y$ is the reals, your $X$ is neither locally compact nor metrizable. – Bill Johnson Aug 5 '14 at 0:16
@BillJohnson: coproduct, not product :) – John Pardon Aug 5 '14 at 0:17
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. – Fred Dashiell Aug 17 '14 at 14:06

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 $P(X)$.

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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 $0\le \alpha<\omega_1$ – Fred Dashiell Aug 5 '14 at 0:23
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). – Joseph Van Name Aug 5 '14 at 0:31
@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. – François G. Dorais Aug 5 '14 at 12:51
@François: Consistently this can happen without large cardinals, but then countable choice fails. – Asaf Karagila Aug 6 '14 at 0:04
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. – Fred Dashiell Aug 6 '14 at 1:37

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