general relativity - Does Schwarzschild metric have cosmological horizon?

Since the space is not expanding in the Schwarzschild metric does that mean that there is no cosmological horizon?

Also, what if the Schwarzschild metric was not asymptotically flat and we replace the familiar $(1−\frac{2GM}{r})$ for $dt^2$ and $dr^2$ by for example $(1−\frac{2GM}{r}+{r^2})$. How will this effect the cosmological horizon?

Answer

Correct, the standard Schwarzschild metric is asymptotically flat and indeed the time co-ordinate $t$ is the local time of an observer infinitely removed from the black hole and sitting in this flat space and so there is no pair of points outside the black hole's event horizon which ultimately cannot causally reach or signal each other. The de Sitter-Schwarzschild metric is the appropriate concept when the cosmological constant is nonzero, and this does have a cosmological horizon.