In class I was taking notes about equivalence relations defined as:

Given a generic relation $$R$$ on $$X$$, $$xIy$$ if both $$xRy$$ and $$yRx$$

Now, I don't really understand the following proposition:

If $$R$$ (the original relation) is reflexive and transitive, then $$I$$ (the derived relation) is reflexive, symmetric and transitive.

I understand this is an instance of $$\succsim$$ where the derived relations are $$\sim$$ and $$\succ$$, but the proposition holds only for the former (the strict preference obviously is not symmetric). Why is that so?

Also, how can I prove that $$R$$ having the two properties leads to $$I$$ having the same two properties + symmetry?

Q1. Why is $$\succ$$ not symmetric?

Q2. Why is $$\sim$$ (a) symmetric; (b) reflexive; and (c) transitive?

A1. Suppose $$x \succ y$$. Then by Definition 4 (refer to basic setup and definitions below), $$x \succsim y$$ and $$y \not\succsim x$$. Therefore, by the same definition, $$y \not\succ x$$. Thus, $$\succ$$ is not symmetric.

A2. Let $$x,y,z\in X$$.

A2(a)$$\hspace{10pt}$$ $$x\sim y$$ $$\hspace{10pt}$$ $$\iff$$ $$\hspace{10pt}x\succsim y$$ and $$y\succsim x$$ (Definition 3) $$\hspace{10pt}$$ $$\iff$$ $$\hspace{10pt}y\sim x$$.

A2(b) By Definition 2, $$x \succsim x$$. And so, by Definition 3, $$x \sim x$$.

A2(c) Suppose $$x\sim y$$ and $$y\sim z$$. (We will show that $$x\sim z$$.)

By Definition 2, we have $$x\succsim y$$ and $$y\succsim z$$. And now by the transitivity of $$\succsim$$, we have $$x\overset{1}{\succsim} z$$.

Similarly, we also have $$y\succsim x$$ and $$z\succsim y$$. And so again by the transitivity of $$\succsim$$, we have $$z\overset{2}{\succsim} x$$.

Given $$\overset{1}{\succsim}$$ and $$\overset{2}{\succsim}$$, by Definition 3, we have $$x\sim z$$.

Basic setup and definitions.

Definition 1. A (binary) relation $$R$$ (on a set $$X$$) is:

• Symmetric if for all $$x,y\in X$$, $$xRy \iff yRx$$;

• Reflexive if for all $$x\in X$$, $$xRx$$;

• Transitive if for all $$x,y,z\in X$$, $$xRy$$ and $$yRz$$ implies $$xRz$$.

Let $$X$$ be our set of alternatives.

Definition 2. Let $$\succsim$$ be a reflexive and transitive relation on $$X$$.

We interpret $$\succsim$$ as the weakly-preferred-to relation.

Definition 3. $$x \sim y$$ if $$x \succsim y$$ and $$y \succsim x$$.

We interpret $$\sim$$ as the indifference relation.

Definition 4. $$x \succ y$$ if $$x \succsim y$$ and $$y \not\succsim x$$.

We interpret $$\succ$$ as the strictly-preferred-to relation.

• The defining characteristics of the weak preference relation $\succsim$ is its transitivity and completeness. The reflexivity of $\succsim$ is a consequence. – Kenneth Rios Sep 30 '18 at 1:48
• @KennethRios: I wanted to directly address the question(s), without also having to prove that reflexivity is a consequence. The question(s) included, in particular, the following: "Now, I don't really understand the following proposition: If $R$ (the original relation) is reflexive and transitive, then $I$ (the derived relation) is reflexive, symmetric and transitive." – user18 Sep 30 '18 at 2:03
• Sure. Just preempting confusion in case OP wrongly thinks that all relations that are both reflexive and transitive are preference relations. – Kenneth Rios Sep 30 '18 at 3:19

If both $$xRy$$ and $$yRx$$ $$\forall x,y \in X$$, then the indifference relation $$I$$ is symmetric by definition. Not all symmetric relations are equivalence relations. However, you are given that $$R$$ is also reflexive and transitive. So we have that $$I$$ is symmetric, reflexive, and transitive. Therefore $$I$$, or $$\sim$$, is an equivalence relation.

You can verify that $$\sim$$ is an equivalence relation by reading off the definitions of the three properties. $$R$$, or $$\succsim$$, is a (weak) preference relation, also known as a total preorder. Total preorders are transitive, complete, and therefore reflexive.

Keep in mind that reflexivity and transitivity does not necessarily imply symmetry: $$R = \{(x,x), (y,y), (x,y)\}$$ given $$X = \{x,y\}$$ is clearly reflexive and transitive but not symmetric because $$xRy$$ but not $$yRx$$.