Example 2 Given $a, b \in \mathbf{R}, h>0$, then “$|a-b|<2 h$” is “$|a-1|<h$ and $|b-1|<h$” ( ).
(A) A sufficient but not necessary condition
(B) A necessary but not sufficient condition
(C) A necessary and sufficient condition
(D) Neither a sufficient nor a necessary condition
Solution: Since $a, b \in \mathbf{R}, h>0$, if we take $a=b=1+h$, then $|a-b|=0<2h$, but $|a-1|=h$ and $|b-1|=h$, meaning $|a-1|<h$ and $|b-1|<h$ do not hold, so the conclusion cannot be derived from the condition;

Conversely, if $|a-1|<h$ and $|b-1|<h$, then $|a-b|=|(a-1)-(b-1)| \leqslant |a-1|+|b-1|<2h$, which means the condition can be derived from the conclusion.
Therefore, the condition is a necessary but not sufficient condition for the conclusion to hold. Choose B.
Note: A correct proposition requires a rigorous proof process; while negating a proposition, only one counterexample is needed. The above two questions are typical examples that need to be understood. Example 2: If considered on the number line, it is more intuitive. $|a-1|<h$ and $|b-1|<h$ indicate points on the number line whose "distance" from 1 is less than $h$, naturally the "distance" between $a$ and $b$ is less than $2h$; conversely, no matter how small the "distance" between $a$ and $b$ is, the "distance" between $a$ or $b$ and 1 may not be less than $h$.