Let $G$ be a topological group, $B \subseteq G$ Borel and $C \subseteq G$ closed.
Is it true that $BC$ is Borel?
Because left and right multiplication are homeomorphisms, it should suffice to prove this separately for $B$ open or closed.
Suppose $B$ is open, then

\begin{equation} BC = \bigcup \limits_{c \in C} Bc = \bigcup \limits_{c \in C} \rho_c(B) \end{equation}

where $\rho$ denotes right multiplication. So if $B$ is open $BC$ is also open and hence obviously Borel.

Now, what happens when $B$ is closed? I know that the product of two closed sets is not necessarily closed (for that one factor has to be compact), but how about Borel?
I suspect it will not work in general (I would be happy to be proven wrong!), so that one has to resort to compact $B$'s instead of closed ones.
These should however be sufficient in Hausdorff $\sigma$-compact topological groups.

  • 4
    $\begingroup$ This has been answered on MO. Erdös and Stone gave an example of a compact set $C \subset \mathbb{R}$ and a $G_{\delta}$ set $D \subset \mathbb{R}$ such that $C+D$ is not Borel. $\endgroup$ – Dominique R.F. Jun 20 '17 at 21:44

Evidently my intuition was completely off...
This question has been answered negatively here on mathoverflow, which also completely kills my suspicions mentioned in the question.


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