The avoidance principle for mean curvature flow states as follows.
$\textbf{Theorem.}$ Let $M_0$ and $N_0$ be two smooth closed surfaces and let $M_t$ and $N_t$ be their evolutions under mean curvature flow. Let $T > 0$ be such that $M_t$ and $N_t$ are both defined for all $t \in [0,T]$. Suppose that $M_0$ and $N_0$ are disjoint. Then $M_t$ and $N_t$ are disjoint for all $t \in (0,T]$.
My problem is, can we use this theorem if two surfaces converge to the same point at the same time? To be more specific, given two mean curvature flow $M_t$ and $N_t,$ if they both "shrink" to a point $p$ as $t\to T,$ can we use the avoidance principle to conclude that $M_t$ and $N_t$ intersects for each $t\in[0,T)?$
My concerns are:
I am not quite sure if in the setting above we could say that $M_t$ and $N_t$ are defined on $[0,T]$ (really including $T$?) to apply the theorem.
In the proof of the avoidance principle, it applys the maximum principle to say that if they intersect, then they should be identical. However, it doesn't seem to derive a contradiction since at that time they are really the same (point).
Any comments are welcome!
($\textbf{Edit.}$) When I say "applying the theorem to the point convergent case," I mean the following argument. Suppose $M_t$ and $N_t$ are disjoint for some $t\in[0,T).$ Then by the theorem (if it applies, which is what I would like to make sure in this question), $M_T$ and $N_T$ should be disjoint, contradicting to the assumption ($M_T=N_T=\{p\}$).
There is a more general form of avoidance principle: not only that they remain disjoint, but also that the distance $d_t$ between them $$ d_t = \min_{x \in M_t, y\in N_t} |x-y|$$ is non-decreasing with respect to $t$ (This is shown in the proof of the avoidance principle). Thus it is impossible that the two MCFs shrink to the same point.