Is the class of well-ordered sets a pseudo-elementary class?

Question

I know that the class of well-ordered sets is not an elementary class. But, is it at least a pseudo-elementary class? If not, what is the proof?

Answer 1

No, because we can also use the compactness theorem in the context of pseudo-elementary classes.

Let $T'$ be an $L'$-theory, where $L'\supseteq L = \{<\}$, and suppose that the $L$-reducts of models of $T'$ are exactly the well-ordered sets.

Introduce new constant symbols $(c_n)_{n\in \omega}$ and consider $T'' = T'\cup \{c_n < c_m\mid n > m\}$. $T''$ is consistent by compactness, since for any finite subset $\Delta\subseteq T''$, we can find a finite well-order $(W,<)$ satisfying the sentences of the form $c_n < c_m$ in $\Delta$, and then expand $W$ to a model of $T'$.

Let $M$ be a model of $T''$. The $L'$-reduct of $M$ is a model of $T'$, and the $L$-reduct of $M$ has an infinite descending chain, contradiction.

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More abstractly, this result follows from my answer to this previous question of yours. The complement of the class of well-orders is pseudo-elementary. Indeed, let $L = \{<\}$, and let $L' = \{<\}\cup \{c_n\mid n\in \omega\}$. Let $T' = \{\varphi_n\mid n\in \omega\}$, where $\varphi_n$ asserts "if $<$ is a linear order, then $c_{n+1} < c_n$". Then the class of $L$-structures which are not well-orders is the class of $L$-reducts of models of $T'$. So if the class of well-orders were also pseudo-elementary, then it would be a finitely axiomatizable elementary class.