= Solution
For a homogeneous isotropic metric, radial light propagation covers comoving distance $\int c\,dt/a(t)$. The physical <particle horizon> is $d_p(t)=a(t)\int_{t_i}^t c\,dt'/a(t')$, when this is finite. A future <event horizon> exists when $d_e(t)=a(t)\int_t^{t_f}c\,dt'/a(t')$ is finite. The <Hubble radius> is the instantaneous scale $c/H$, with $H=\dot a/a$; it is not generally the same as either causal horizon.
For an <Einstein-de Sitter universe>, $H=2/(3t)$, so $c/H=3ct/2$, while $d_p=3ct$ from a beginning at zero. The future integral diverges, so there is no event horizon. For an eternal-past <de Sitter spacetime>, $H$ is constant and the particle-horizon integral from $-\infty$ diverges. Its event horizon and Hubble radius both equal $c/H$. Thus \b[a Hubble radius is not automatically a particle horizon], even in these simple isotropic models.
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