The set of $R$-segments of $X$ is well-ordered by the inclusion relation

Let $R$ be an order on a class $X$. A subclass $S$ of $X$ is called an $R$-segment if $$(\forall s \in S) (\forall x \in X) ((x,s) \in R \implies x \in S).$$ If $a \in X$, then the set $S_{X,R} (a) = \{x : x \in X ,\, x \lneq a \}$ is an initial $R$-segment determined by $a$.

Let $R$ be a well-ordering on set $X$. Show that the set of $R$-segments of $X$ is well-ordered by the inclusion relation.

My proof( Is it correct?):

Let $A$ be the set of proper $R$-segments of $X$. Consider a non-empty subset $B$ of $A$.

Let $C = \{x: S_{X,R}(x) \in B\}$. Since every proper $R$-segment is also an initial $R$-segment, $C$ is non-empty. Since $R$ is a well-ordering, $$(\exists x_0) ((x_0 \in C ) \wedge (x \in C \implies x_0\leq x )).$$

Thus, given any $S_{X,R}(x)\in B$, we have $x \in C$ and $x_0 \leq x$.

Given any $x' \in S_{X,R}(x_0)$, we conclude that: \begin{align*} &x' \lneq x_0 \leq x \\ \therefore & x' \in S_{X,R} (x) \\ \therefore & S_{X,R}(x_0) \subseteq S_{X,R}(x) \end{align*}

• It seems to me that it's okay. – pepa.dvorak Jul 28 '13 at 16:25
• Alexy, there is no need to break up each formula into tiny little chunks of MathJax with text in between. Please look at the source code as it stands now, after my edits. – dfeuer Jul 28 '13 at 16:46

Assuming that you’ve already shown that every proper $R$-segment of a well-order is an initial $R$-segment, it’s correct, but it could be made a bit clearer. Specifically, what you’ve actually shown is that $B\,'=\{S\in B:S=S_{X,R}(x)\text{ for some }x\in X\}$ is well-ordered by $\subseteq$, and the desired conclusion follows from the observation that $B\,'=B$.
An alternative approach is to define the map $\varphi:X\to A:x\mapsto S_{X,R}(x)$ and show that $\varphi$ is an order-isomorphism from $\langle X,R\rangle$ to $\langle A,\subseteq\rangle$.