For a poset with ground set P, isn’t every set containing a single element of $X$ a chain of $P$

Question

For example, consider this partially ordered set from my textbook solely through its Hasse diagram:

My textbook says that one of the chains of the poset $P$ above is ${1}$. This makes sense. Explicitly using definitions of partial orders and their chains, here is the logic:

Posets are transitive, therefore $1$ is related to itself through the partial order relation, therefore $1$ is comparable to itself, therefore every pair of distinct elements in $\{1\}$ are comparable to each other, therefore $\{1\}$ is a chain of $P$

But apparently $\{3\}$ is not a chain, but instead is an antichain of $P$. Why is this? I could follow the same logic as I did for $\{1\}$ to conclude that $\{3\}$ should also be a chain…

Answer 1

For a poset with ground set $P$, isn't every set containing a single element of $X$ a chain of $P$?

Yes. Your reasoning is correct.

$1$ is comparable to itself, therefore every pair of distinct elements in $\{1\}$ are comparable to each other

A minor nitpick here: $1$ is not distinct from itself. So you don't need to say that $1$ is comparable to itself to say that every pair of distinct elements is comparable. In fact, there are no pairs of distinct elements in $\{1\}$, and therefore "vacuously", every pair of distinct elements is comparable.

But apparently $\{3\}$ is not a chain, but instead is an antichain of $P$. Why is this?

It is both a chain and an antichain! In general, a set of elements in a Poset can be a chain, an antichain, neither, or both.