In a homogeneous and isotropic Universe (even if recent observations challenge this hypothesis), you can derive the Friedmann equations, which describe the evolution of the Hubble constant with time:
$frac{dot{a}}{a} = H(t) = frac{8 pi G}{3}rho - frac{k}{a^2} + frac{Lambda}{3}$ (with $c=1$) (Equation $1$)
where $a=a(t)$ is the scale factor, $dot{a}$ its derivative, $G$ the gravitational constant, $rho$ the matter density, $frac{k}{a^2}$ the spatial curvature (a parameter that describes the metric of the Universe), and $Lambda$ the cosmological constant (an integration constant added by Einstein).
It could be useful to rewrite the equation as:
$H^2 = frac{8 pi G}{3}(rho + rho_{Lambda}) - frac{k}{a^2}$
where $rho_{Lambda} = frac{Lambda}{8 pi G}$ is the "density of cosmological constant".
We can also expand the matter density as $rho = rho_{matter} + rho_{radiation}$.
So we have a "total" density $rho_{tot} = rho_{matter} + rho_{radiation} + rho_{Lambda}$. The destiny of the Universe depends on this amount.
In case of $rho_{tot} > rho_{crit}$, or equivalently a closed Universe ($k=+1$), the equation $(1)$ becomes:
$dot a^2 = frac{8 pi G}{3}rho a^2 -1$
Which points out that the scale factor must have an upper limit $a_{max}$ ($dot a^2$ must be positive).
This in turn means that the second derivative $ddot a$ of the scale factor must be negative, when approaching $a_{max}$, that is the scale factor function inverts its behavior:
Look at here and here if you want to go deeper.
@Bardathehobo This figure shows what I mean when I say that a currently accelerating Universe can still crunch. This is because we are basically ignorant upon the dark energy issue.
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