Consider the set [0, 1], and consider it's partition $A=(0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1)$ and consider a refinement of this partition, B Note that refinement of a partition A contains all points of A and possibly some other points, thus B can be considered as: $B=(0,\frac{1}{8},\frac{1}{4},\frac{1}{2},\frac{3}{4},\frac{7}{8},1)$ Here it can be seen that A is contained in B, i.e. A sub B And if we consider another refinement of A: ${B}_{1}=(0,\frac{1}{8},\frac{1}{4},\frac{3}{8},\frac{1}{2},\frac{3}{4},1),\text{again note that}A\text{sub}{B}_{1}$ i.e. ${B}_{1}$

is an refinement of $A\text{but B}\text{\u29f8}\subset B1\to \text{sub}{B}_{1}$ is not an refinement of B.