Let $R$ be well founded and set like on $A$. I want to show that $R^{*}$ is a total order iff $rank_{A,R}(y)=\sup\{rank(x)+1|$ $xRy$ $\wedge{x\in{A}}\}$ is injective. Here $R^{*}$ is the transitive closure of $R$ in $A$.
I proved the forward direction. For the backward direction I looked at $X= \{a\in{A}|\exists{y}\in{A}(y\neq{a}$ $\wedge$ $\neg{yR^{*}a}$ $\wedge$ $\neg{aR^{*}y})\}$. If $X$ is empty then we are done. So suppose not. Since $R^{*}$ is also well founded it has a $R^{*}$ minimal element, say $a$. Now fix $b$ s.t. $a\neq{b}$, $\neg{bR^{*}a}$, $\neg{aR^{*}b}$.
Since $a\neq{b}$, we have that $rank(a)\neq{rank(b)}$. Now we know that $rank(a)$, ${rank(b)}$ are ordinals. I showed that $rank(a)>{rank(b)}$ leads to a contradiction. However I can't figure out how to show that $rank(a)<{rank(b)}$ also leads to contradiction.
Any ideas on how I can get a contradiction?
Here's a similar, but slightly more direct approach. Instead of assuming towards contradiction, assume the contrapositive. Suppose that $R^*$ is not a linear order. Let $a,b$ two elements incomparable by $R^*$, such that:
Now if $\alpha>\beta$ then this is a contradiction to the choice of $a$. If $\alpha<\beta$, then for every $b'$ such that $b'\mathrel{R^*}b$ we have that $a$ and $b'$ are comparable. Pick any such $b'$ of rank $\geq\alpha$, then $a\mathrel{R^*}b'$ and therefore $a\mathrel{R^*}b$. This is a contradiction to the choice of $b$.
Therefore $\alpha=\beta$. $\qquad\square$