Question

Suppose that \({R_1}\) and \({R_2}\) are equivalence relations on a set A. Let \({P_1}\) and \({P_2}\) be the partitions that correspond to \({R_1}\) and \({R_2}\), respectively. Show that\({R_1} \subseteq {R_2}\) iff \({P_1}\) is a refinement of \({P_2}\).

Short answer

Hence, \({R_1} \subseteq {R_2}\) if and only if \({P_1}\) is a refinement of \({P_2}\).

Step-by-step solution

Given data

\({R_1}\) and \({R_2}\) are the equivalence relation on a set \(S\).

Concept used of refinement of partition

If all the set in\(A\)is a subset of some set in\(B\), then\(A\)is known as a refinement of the partition\(B\).

The partitions\({P_1}\)and\({P_2}\)correspond with the equivalence relation\({R_1}\)and\({R_2}\)on a set then\({R_1} \subset {R_2}\)iff\({P_1}\)is a refinement of\({P_2}\).

Show the refinement of partition

Assume that \({R_1}\) and \({R_2}\) are the equivalence relation on a set \(S\).

Also, \({P_1}\) and \({P_2}\) are the partitions of corresponding to \({R_1}\) and \({R_2}\) respectively.

Suppose that \({R_1} \subseteq {R_2}\).

Let \({A_1} \in {P_1}\).

Assume \(x \in {A_1}\). Then, \((x,y) \in {R_1}\forall y \in {A_1}\). Here, \({A_1}\) is an equivalence class.

\((x,y) \in {R_2}\)for all \(y \in {A_1}\).

That is, all the elements of \({A_1}\) are in some subset \({B_1}\) in \({P_2}\).

Hence, \({P_1}\) is a refinement of \({P_2}\).

Conversely,

Let \({P_1}\) is a refinement of \({P_2}\).

Suppose \((x,y) \in {R_1}\).

Here, \(x\) and \(y\) are in same subset \(A\) of \({P_1}\).

\(x\)and \(y\) are in same subset \(B\) of \({P_2}\). Also, it is given that \({P_1}\) is a refinement of \({P_2}\).

That is, \((x,y) \in {R_2}\)

Thus, \({R_1} \subseteq {R_2}\)

Hence, \({R_1} \subseteq {R_2}\) if and only if \({P_1}\) is a refinement of \({P_2}\).