# Are suborders of pseudo-well-ordered sets themselves pseudo-well-ordered?

The class $$W$$ of well-ordered sets is not first-order axiomatizable. However, it does have an associated first-order theory $$Th(W)$$. I define a pseudo-well-ordered set to be an ordered set $$(S;\leq)$$ that satisfies $$Th(W)$$. My question is, is the class of pseudo-well-ordered set closed under taking substructures? If so, I want a proof. If not, I would like an explicit pseudo-well-ordered set $$(S;\leq)$$ and a suborder $$(T;\leq)$$ of $$(S;\leq)$$ such that $$(T;\leq)$$ is not pseudo-well-ordered.

• This is an interesting question! Can you describe how it came up? Jul 13, 2022 at 4:26
• @TomKern I don't really know, but it just popped into my brain yesterday night. That is all there is to it. Jul 13, 2022 at 14:54

$$\omega+\zeta$$ satisfies all the first-order sentences satisfied by $$\omega$$, but $$\zeta$$ does not satisfy "there is a smallest element".

• And $\zeta$ here is? Jul 13, 2022 at 4:32
• The order type of the integers Jul 13, 2022 at 4:35
• What is the proof that it satisfies the first-order sentences satisfied by $\omega$? Jul 13, 2022 at 14:51

Your question has a strong negative answer. Let $$S$$ be any linear order which is not a well-order (for example, any pseudo-well-order which is not a well-order). Then $$S$$ has an infinite descending chain $$s_0>s_1>s_2>\dots$$. Let $$T=\{s_i\mid i\in \omega\}$$. Then $$T$$ is a sub-order with no minimal element, so $$T$$ is not a pseudo-well-order.

So we've shown: every sub-order of $$S$$ is a pseudo-well-order if and only if $$S$$ is actually a well-order.

However, there is a positive answer to a related question. A definable sub-order of $$S$$ is one of the form $$\varphi(S)=\{s\in S\mid S\models \varphi(s)\}$$ for some formula $$\varphi(x)$$ with one free variable. Let $$S$$ be a pseudo-well-order. Then every definable sub-order of $$S$$ is a pseudo-well-order.

The proof is not hard. Fix a pseudo-well-order $$S$$ and a formula $$\varphi(x)$$ with one free variable. Let $$\psi\in \mathrm{Th}(W)$$. Write $$\psi^\varphi$$ for the relativization of $$\psi$$ to $$\varphi(x)$$. So for any linear order $$L$$, $$L\models \psi^\varphi$$ if and only if $$\varphi(L)\models \psi$$. Now for any well-order $$L$$, the suborder $$\varphi(L)$$ is also a well-order, so $$\varphi(L)\models \psi$$. Thus $$L\models \psi^\varphi$$, so $$\psi^\varphi\in \mathrm{Th}(W)$$. Since $$S$$ is a pseudo-well-order, $$S\models \psi^\varphi$$, so $$\varphi(S)\models \psi$$. We've shown that $$\varphi(S)\models \mathrm{Th}(W)$$, so this definable sub-order is a pseudo-well-order.

Take any countable non-standard model of $$\sf PA$$ which is elementarily equivalent to the standard one, it is known that the order type of such model has the form $$\Bbb{N+Q\times Z}$$, therefore it contains a suborder isomorphic to $$\Bbb Q$$, which is clearly not satisfying the theory of pseudo-well-orders.