Difficulty understanding Hasse Diagrams and total order relations

Question

The part of the answer relating hasse diagram which is total order relation is a bit confusing to me since it says e is not related to a. Is a related to e but e is not related to a?

Answer 1

So, indeed you have show the three partial order diagrams where: $e\mathop {\cal T}b\mathop {\cal T}c\mathop {\cal T}d$ and $a\mathop {\cal T}c$ but $e\mathop {\cal \require{cancel}\cancel T}a$ and the diagram for the chain: $a\mathop {\cal T}e\mathop {\cal T}b\mathop {\cal T}c\mathop {\cal T}d$ is the only total order among them.

The fact of $e\mathop {\cal\cancel T}a$ means that $e$ cannot be a node on a branch leading up toward $a$.   In the first two diagrams $a$ and $e$ are on divergent branches, so neither is in a $\mathop {\cal \cancel T}$ relation to the other (they are incomparable).   In the third diagram we have $a\mathop{\cal T}e$, which does not conflict with the fact that $e\mathop {\cal\cancel T}a$.

Remember: Partial orders relations are anti-symmetric.