Theorem. A relation $R$ on a set $A$ is reflexive and transitive if and only if there is a set $B$ with a partial order $S$ and a function $f : A \to B$ such that for all $x, y \in A, x R y \iff f(x) S f(y)$.
I think the proof is saying that every reflexive and transitive relation can be mapped into a partial order
I am trying to prove the following
Let $R$ be a partial order on a set $A$. Show that taking $B = A, S = R$, and $f = I_A$ (the identity function on A) satisfies the Theorem.
How do i start this proof. Do I show that $B=A$ is a partial order, $S=R$ is a partial order and $f = I_A$ is a partial order.
This is what i have so far
Reflexive: If $x \in A$, then $f(x) = f(x)$ so $xRx$.
Antisymmetric: If $xRy$, then $f(x) = f(y)$ and $f(y) = f(x)$. It follows that $yRx$. so $x = y$.
Transitive: If $xRy$ and $yRz$, then $f(x) = f(y)$ and $f(y) = f(z)$. It follows that $f(x) = f(z)$ and $xRz$.
I am lost here
Can someone help me on this. Thanks in advance
Suppose $R$ is reflexive and transitive. Then the relation defined by $$ x\mathrel{R_s}y \qquad\text{if and only if}\qquad x\mathrel{R}y\text{ and }y\mathrel{R} x $$ is an equivalence relation. The proof is easy.
Consider $B=A/R_s$ and define, for $[x],[y]\in B$ ($[a]$ denotes the equivalence class with respect to $R_s$ of $a\in A$), $$ [x]\mathrel{S}[y] \qquad\text{if and only if}\qquad x\mathrel{R}y $$
Show that this definition is well posed (it doesn't depend on the representatives of the equivalence class) and that $S$ is an order relation on $B$.
Of course the required map $f\colon A\to B$ is the canonical projection on the quotient set.
The converse is an easy verification.
Note 1. You can't start from “Let $R$ be a partial order”, because you are not given this hypothesis.
Note 2. The statement in bold face is completely obvious: if $R$ is a partial order, $B=A$ and $R=S$, “$x\mathrel{S}y$ if and only if $x\mathrel{R}y$” by definition and $S$ is of course a partial order.