= Solution
{c}
Take the expanding flat branch $a(t)=Ae^{H_\Lambda t}$, with $A>0$. A <radial null geodesic in FLRW spacetime> satisfies $d\rho/dt=\pm1/a(t)$. For the incoming <past light cone> of the <spacetime event> $(t_0,\rho=0)$, integration gives
$$
\boxed{\rho(t;t_0)=\int_t^{t_0}\frac{ds}{a(s)}
=\frac1{H_\Lambda}\left(\frac1{a(t)}-\frac1{a(t_0)}\right),\qquad t\le t_0,}
$$
with arbitrary angular direction. Equivalently, <conformal time> $\eta=-1/(H_\Lambda a)$ gives $\rho=\eta_0-\eta$.
Let the reception time tend to infinity. The largest <comoving radial distance> from which a signal emitted at time $t$ can reach the central <comoving observer> is
$$
\rho_{\mathrm h}(t)=\int_t^\infty\frac{ds}{a(s)}=\frac1{H_\Lambda a(t)}.
$$
Thus the <cosmological event horizon> has constant <proper distance in cosmology>:
$$
\boxed{\rho=\frac1{H_\Lambda a(t)},\qquad D_{\mathrm h}=a(t)\rho_{\mathrm h}=H_\Lambda^{-1}.}
$$
An emitted <spacetime event> is seen at a finite time if and only if $a(t)\rho<H_\Lambda^{-1}$. Equality would require $t_0=\infty$, while an event with $a(t)\rho>H_\Lambda^{-1}$ never reaches the observer. This is a bound on events at their emission time; a fixed <comoving observer> can have earlier visible events and later invisible ones.
The boundary really is a <null hypersurface>. For $f(t,\rho)=\rho-[H_\Lambda a(t)]^{-1}$, its normal has components $\partial_tf=1/a$ and $\partial_\rho f=1$, so
$$
g^{\mu\nu}\partial_\mu f\partial_\nu f=-\frac1{a^2}+\frac1{a^2}=0.
$$
The finite future integral, and hence this <observer event horizon>, uses the expanding branch. For the contracting flat solution $a=Ae^{-H_\Lambda t}$, the same integral diverges and there is no future <observer event horizon> of this kind; the static zero-density solution likewise has none.
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