Showing that the equivalence class of partial orders order isomorphic to $(X, \leq)$ is not a set I'm trying to do the following (original image): > EXERCISE 25(R):(a) Argue on the grounds of the architect's view of set theory (which was outlined in the Introduction) that $\mathbf{PO}$ cannot be a set. > (b) Moreover, show that for each given p.o. $\langle X , \preceq_X \rangle$, if $X \neq \emptyset$, then the collection of all p.o.'s $\langle Y , \preceq_y \rangle$ that are order-isomorphic to $\langle X , \preceq_X \rangle$ is not a set. (This is from W. Just and M. Weese, _Discovering Modern Set Theory, vol.1_ , p.23.) (a) was rather easy: Since $\mathbf{PO}$ is itself a partial order with respect to $\subseteq$, $\mathbf{PO}$ would have to contain itself and hence cannot be a set. But now I'm stuck with (b). Can someone show me how to do (b)? Thank you!

Let $(P,\leq)$ be a poset with $P\
eq\emptyset$. Let $x$ be any element of $P$ and $y$ be any element not in $P$. Replace $x$ by $y$ and you have an isomorphic poset containing $y$. So every set is contained in some equivalent poset and the class of equivalent posets cannot be bounded above in the cumulative hierarchy.