proof writing - Prove that $succeq = bigcap L left(succeq right)$ - understanding elementary order theory

Dear reader of this post,

I am working on a elementary question on order relations. I would be very glad to receive some guidance since I think understanding this question is important for making further progress with order relations. The question is as follows: Let $\left(X,\succeq \right) $ be a preordered set ($\succeq $ being a transitive and reflexive relation) , and define $L \left( \succeq \right)$ as the set of all complete preoders that extend $\succeq$. Prove that $\succeq = \bigcap L \left(\succeq \right)$.

With the two replies by Brian and Hagen (thank you!), I'd like to proceed proving the claim. I am not yet familiar about the usual procedure for answering own questions so I hope I do not violate the etiquette. I would be glad to receive some remarks about my prove. I relegated previous thoughts to the end.

To make use of Szpilrajn's theorem, I will at first create a partially ordered set (poset) out of the preordered set and then extent this poset to a linearly ordered set. ( @ Brian: I expect that the technical details you mention in your post refer to the different elements which can are equivalent in the preordered set but not in the poset. Is this correct?) Let $ \sim $ denote the symmetric part of $\succeq$. Then $ \left( X/~,\succeq^{\star} \right)$ is a poset where $\succeq^{\star} $ is defined on $X/ \sim $ by $[x]_{\sim} \succeq^{\star} [y]_{\sim} \ \text{iff} \ x \succeq y$.

I will no show that $\succeq^{\star} = \bigcap L(\succeq^{\star} ) $. I know that every extension must satisfy $\succeq^{\star} \subseteq \unrhd$. So for each new poset $[x]_{\sim} \unrhd [y]_{\sim} \ \text{iff} \ [x]_{\sim} \succeq^{\star} [y]_{\sim} $ holds. For all other equivalence classes $[h]_{\sim},[g]_{\sim}$ for which $ [h]_{\sim} \not \succeq^{\star} [g]_{\sim}, \ \text{and} \ [g]~ \not \succeq^{\star} [h]~ $ the extended poset differ between whether $[h]_{\sim} \unrhd [g]_{\sim} $ or the other way round. Thus, the intersection of the different extension cannot include any relation which is not stated in the original poset already and $\succeq^{\star} = \bigcap L(\succeq^{\star} ) $.

Can someone give me some hints about the mistakes my prove contains?

Previous thoughts
I know that $\succeq \subseteq \unrhd$ ($\unrhd$ being the extension) so the complete preorder is a larger set than the original preoder. I imagine that extensions generate relations among the previously unrelated elements. For $\succeq = \bigcap L \left(\succeq \right)$ to hold, I know that the extension generates at least one complete preorder. To prove the claims, I expect to show that every relationship contained in the original preoder is an element of each extended preorder. Hence, I basically need to show that both sets (the original preorder and the intersection of the new complete preorders) contain exactly the same relationships.

I would like to ask the following questions in particular:

• I do not understand how one can generate different complete preoders. Could I name them $L_i \left(\succeq \right)$? (Or would $L \left(\succeq_i \right)$ be more precise ?) In what respect do the complete preorders differ?


• I have no idea how to show that both sets contain the same relations. Suppose that $x,y \in X$, could one then have $(x,y) \in \succeq $?

I am looking forward for your replies.