Five. (20 points) On a piece of paper, there are a line $A B$ and a point $M$ outside the line, and the distance from point $M$ to line $A B$ is $a$. Fold the paper so that point $M$ coincides exactly with a point $M^{\prime}$ on line $A B$. Each such fold leaves a straight line crease on the paper. When $M^{\prime}$ traverses all points on line $A B$, find the set of points on all the crease lines.

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As shown in Figure 5, draw $M H \perp A B$ at $H$, and establish a Cartesian coordinate system with the midpoint $O$ of segment $M H$ as the origin and the line $M H$ as the $y$-axis, then $M\left(0, \frac{a}{2}\right)$.

When point $M^{\prime}$ coincides with $H$, the crease line is the $x$-axis.

When point $M^{\prime}$ does not coincide with $H$, let the slope of line $M M^{\prime}$ be $k(k \neq 0)$, then $M^{\prime}\left(-\frac{a}{k}\right.$, $\left.-\frac{a}{2}\right)$. At this time, the crease line is the perpendicular bisector $P Q$ of segment $M M^{\prime}$. Let $P(x, y)$ be any point on the crease line $P Q$, then $\mid P M$ | $=\left|P M^{\prime}\right|$, i.e.,
$$
\sqrt{x^{2}+\left(y-\frac{a}{2}\right)^{2}}=\sqrt{\left(x+\frac{a}{k}\right)^{2}+\left(y+\frac{a}{2}\right)^{2}}.
$$

Simplifying and rearranging gives
$$
\left(\frac{a}{k}\right)^{2}+2 x\left(\frac{a}{k}\right)+2 a y=0.
$$

This is a quadratic equation in $\frac{a}{k}$, which has real roots, so $\Delta=4 x^{2}-8 a y \geqslant 0$, i.e., $x^{2} \geqslant 2 a y$.
Clearly, the equation of the $x$-axis $y=0$ also satisfies the above inequality.
Therefore, the set of all points on the crease lines is the region outside the parabola $x^{2}=2 a y$ (including points on the parabola).