Show all maximal chains in a power set of a set

Question

Suppose the set $S$, partially ordered by $\subseteq$ respectively, is $S=\{x,y,z\}$. Show all maximal chains in the power set of S.

1st Attempt: I wrote out the power set of S ( $P(S)$ ) as such $P(S)=\{\emptyset,\{x\},\{y\},\{z\},\{x,y\},\{x,z\},\{y,z\},\{x,y,z\}\}$. I think the maximal chain(s) would just be $\{x,y,z\}$ (or $S$ itself) given the definition of a maximal chain.

2nd Attempt:However, this brought me to confusion on $\{\emptyset, x\}$, $\{ \emptyset, y\}$, and so forth on being maximal chains as well despite not being explicitly stated in the power set. Any suggestions on how to move forward?

Answer 1

Hint. One of the maximal chains is $$\emptyset \subseteq \{x\} \subseteq \{x, y \} \subseteq \{x,y,z\}.$$ Can you find the other ones?

Answer 2

The way you state the problem is confusing; I guess it stems from your confusion about the problem.

You write:

Suppose the set $S$, partially ordered by $\subseteq$ respectively, is $S=\{x,y,z\}$.

This would imply that $x$, $y$ and $z$ are sets, so expressions like $x\subseteq y$ make sense. But then, to make any definitive statement about maximal chains, you would need to know the sets $x$, $y$ and $z$, which are not given. Instead, the power set of $S$ is mentioned, which would be pointless if the problem were to find maximal chains in $S$.

Therefore I assume the problem you are supposed to solve is to find maximal chains of the power set $\mathcal P(S)$, partially ordered by $\subseteq$. For the elements of the power set, we can tell the inclusion relation even without knowing $x$, $y$ and $z$.

Note that maximal chains then are sets of elements of $\mathcal P(S)$, which certainly contains the empty set. Note however that it does not contain $x$ or $y$, so e.g.$\{\emptyset,x\}$ is not a chain (neither maximal nor otherwise). However $\{\emptyset,\{x\}\}$ is a chain, because $\emptyset\subseteq\{x\}$.

It is not a maximal chain though, because is is e.g. a subset of the chain $\{\emptyset,\{x\},\{x,y\}\}$.

Indeed, we can immediately say that every maximal chain contains both $\emptyset$ and $S$, since for every $X\in\mathcal P(S)$ we have $\emptyset\subset X\subset\mathcal P(S)$.

I hope that cleared up the confusion.