  We give a partial answer to the following question of Dobrinen: For a given
topological Ramsey space $\mathcal{R}$, are the notions of selective for
$\mathcal{R}$ and Ramsey for $\mathcal{R}$ equivalent? Every topological Ramsey
space $\mathcal{R}$ has an associated notion of Ramsey ultrafilter for
$\mathcal{R}$ and selective ultrafilter for $\mathcal{R}$ (see
\cite{MijaresSelective}). If $\mathcal{R}$ is taken to be the Ellentuck space
then the two concepts reduce to the familiar notions of Ramsey and selective
ultrafilters on $\omega$; so by a well-known result of Kunen the two are
equivalent. We give the first example of an ultrafilter on a topological Ramsey
space that is selective but not Ramsey for the space, and in fact a countable
collection of such examples.
  For each positive integer $n$ we show that for the topological Ramsey space
$\mathcal{R}_{n}$ from \cite{Ramsey-Class2}, the notions of selective for
$\mathcal{R}_{n}$ and Ramsey for $\mathcal{R}_{n}$ are not equivalent. In
particular, we prove that forcing with a closely related space using
almost-reduction, adjoins an ultrafilter that is selective but not Ramsey for
$\mathcal{R}_{n}$. Moreover, we introduce a notion of finite product among
members of the family $\{\mathcal{R}_{n}: n<\omega\}$. We show that forcing
with closely related product spaces using almost-reduction, adjoins
ultrafilters that are selective but not Ramsey for these product topological
Ramsey spaces.
