Existence of infimum and supremum in a totally ordered set Problem: Let $M=\langle A,R\rangle$ be a partially ordered set and $C(M)$ is the set of all totally ordered parts of $M$. Prove that each nonempty totally ordered part of $\langle C(M),\subseteq\rangle$ has infimum and supremum.
Any ideas and suggestions on how to approach this problem would be very much welcome!
As usual thanks in advance!
Edit: All terminology is related to Set theory. I also think that the word 'part' could be replaced by 'subset'. 
 A: I will use the term chain instead of totally ordered part. 
Let $\mathscr{C}$ be a non-empty chain in $\langle C(M),\subseteq\rangle$. We want to show that there is an $S\in C(M)$ such that 


*

*$C\subseteq S$ for each $C\in\mathscr{C}$, and  

*if $T\in C(M)$ and $C\subseteq T$ for each $C\in\mathscr{C}$, then $S\subseteq T$.


Whenever the order is $\subseteq$, the natural candidate for the supremum of a collection is the union of that collection, so we try letting $S=\bigcup\mathscr{C}$. 
The first thing to check is that $S\in C(M)$: is it true that $S$ is a chain in $\langle A,R\rangle$? Suppose that $a,b\in S$; then there are $C_a,C_b\in\mathscr{C}$ such that $a\in C_a$ and $b\in C_b$. $\mathscr{C}$ is a chain in $\langle C(M),\subseteq\rangle$, so without loss of generality we may assume that $C_a\subseteq C_b$. But then $a,b\in C_b$, so either $a\mathrel{R}b$ or $b\mathrel{R}a$, and $S$ is indeed a chain in $M$.
I’ll leave it to you to check the two bullet points; the first is obvious, and the second is very nearly so.
There is an equally natural candidate for the infimum of $\mathscr{C}$; I’ll leave its identity in the spoiler-protected block below. Checking that it works is if anything even easier.

 Consider $\bigcap\mathscr{C}$.

