Given a set $X$ with a strict partial order $<$, do the following conditions correctly characterize a set $C$ as a maximal chain in $X$? (Where $x\approx y$ iff $x<y$ or $y<x$ or $x=y$.)
- $C$ is a non-empty subset of $X$.
- For all $x$ and $y$ in $C$, $x\approx y$.
- For all $x$ in $X$, if $x\approx y$ for every $y$ in $C$, then $x$ is in $C$.
Thanks.
I'm not sure wether you mean that those conditions, alone, characterize maximal chains, or that the conjunction of them does.I will respond to both, just in case:
None of them alone characterizes maximal chains in a poset:
For (1), if we take $X=\{ 0,1\}$, and $(\mathcal{P} (X), \subset)$ as the poset, we have $C=\{\{0\},\{1\}\}$ as a candidate for a maximal chain but $\{ 0\}\not\approx\{1\}$, with the $\approx$ relation you defined. Therefore, it isn't even a chain, let alone a maximal chain. Also, the void set $\phi$ is a trivial maximal chain in $(\phi, \phi)$, (as it is trivially totally ordered and there aren't any elements in $\phi$ strictly greater than all the elements in $\phi$), and it's a void subset of $\phi$, so none of the implications work.
For (2), it is clear that any chain from a poset satisfies that, as it is the definition of totality, but that doesn't grant us that it is maximal. For example, in $(\mathcal{P} (X), \subset)$, where $X=\{0,1\}$, $C_0 =\{\phi ,\{0\}\}$ complies (2), but it's not maximal, since $\phi\subset \{0,1\}$ and $\{0\}\subset \{0,1\}$ but $\{0,1\}\notin \{\phi ,\{0\}\}$.
For (3), every maximal chain of a poset satisfies that, sure. But not every $C\subseteq X$ that satisfies (3), where $(X,<)$ is a poset, is a maximal chain. A counter-example is, with the same $(\mathcal{P} (X), \subset)$ as before, $C_0 =\mathcal{P} (X)$ complies it but it's not even a chain.
However, if you mean all three together in conjunction, yes, they characterize all maximal chains of posets EXCEPT for the trivial case $(X,\subset)=(\phi, \phi)$ where $\phi$ is a maximal chain but it doesn't satisfy (1)