Example 1 For any $n(n \geqslant 2)$ points given in space. If the midpoint of each pair of points connected by a line segment is painted red, find the minimum number of red points.
Let the minimum number of distinct red points (midpoints) in the case of $n$ points be denoted as $f(n)$.
Obviously, $f(2)=1, f(3)=3, f(4)=5$.
Based on this, we conjecture that generally $f(n)=2 n-3$.
(1) Consider the case where $n$ points are collinear.
Take the line they share as the number line, from left to right, let these $n$ points be $A_{1}, A_{2}, \cdots, A_{n}$. Use the notation $A<B$ to mean “point $A$ precedes point $B$”, i.e., point $A$ is to the left of point $B$ on the number line, then
$A_{1}<A_{2}<\cdots<A_{n}$.
Also, let the midpoints of $A_{1} A_{i}(i=2,3, \cdots, n)$ be $P_{i}$, and the midpoints of $A_{j} A_{n}$ $(j=1,2, \cdots, n-1)$ be $Q_{j}$. Then it is clear that
$$
P_{2}<P_{3}<\cdots<P_{n}=Q_{1}<Q_{2}<\cdots<Q_{n-1} \text {. }
$$

This yields at least $2 n-3$ distinct red points:
$$
P_{2}, P_{3}, \cdots, P_{n}\left(=Q_{1}\right), Q_{2}, \cdots, Q_{n-1} \text {. }
$$

Thus, $f(n) \geqslant 2 n-3$, and this minimum value can be achieved (for example, if points $A_{1}, A_{2}, \cdots, A_{n}$ are taken as $\left.1,2, \cdots, n\right)$.
(2) Consider the case where $n$ points are coplanar.
Let $A_{1}, A_{2}, \cdots, A_{n}$ be any $n$ points in the plane. Draw the midpoints (red points) of the line segments between each pair, and denote the midpoint of $A_{i} A_{j}$ as $P_{i j}$. The set of all points $\left\{A_{k}\right\}$ and $\left\{P_{i j}\right\}$ together is finite, and the number of lines connecting any two points is also finite, so there must exist a line $l$ that is not perpendicular to any of these lines.

Now project all these points onto the line $l$, then the projections of any two different points on $l$ will not coincide. If the projections of points $\left\{A_{k}\right\}$ and $\left\{P_{i j}\right\}$ on $l$ are denoted as $\left\{A_{k}^{\prime}\right\}$ and $\left\{P_{i j}^{\prime}\right\}$, then $P_{i j}^{\prime}$ is also the midpoint of the line segment $A_{i}^{\prime} A_{j}^{\prime}$, i.e., this mapping is one-to-one, which returns to case (1), at this time,
$$
f(n) \geqslant 2 n-3 \text {. }
$$
(3) Similarly, for any $n$ points $A_{1}$, $A_{2}, \cdots, A_{n}$ in space, if the midpoints (red points) of the line segments between each pair are drawn, and the midpoint of $A_{i} A_{j}$ is denoted as $P_{i j}$, the set of all points $\left\{A_{k}\right\}$ and $\left\{P_{i j}\right\}$ together is finite, and the number of lines connecting any two points is also finite, so there must exist a plane $\alpha$ that is not perpendicular to any of these lines.

Project all these points onto the plane $\alpha$, the projections of any two different points on $\alpha$ will not coincide. If the projections of points $\left\{A_{k}\right\}$ and $\left\{P_{i j}\right\}$ on $\alpha$ are denoted as $\left\{A_{k}^{\prime}\right\}$ and $\left\{P_{i j}^{\prime}\right\}$, then $P_{i j}^{\prime}$ is also the midpoint of the line segment $A_{i}^{\prime} A_{j}^{\prime}$, i.e., this mapping is also one-to-one, which returns to case (2), at this time,
$$
f(n) \geqslant 2 n-3 \text {. }
$$

From this example, we see that the special often contains the general.