non-Borel set which intersects every compact in a Borel set 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?
 A: 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 $P(X)$.
A: 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.
