Background. Given a poset $(S,<)$ we'll indicate with $\tau(S,<)$ the set of all the ordinals which are isomorphic to a well ordered subset of $(S,<)$. We're in $\mathsf{ZFC}$.
Questions.
- Calculate $\tau(\varepsilon_0,<^*)$, where $\varepsilon_0$ is the smallest $\varepsilon$-number (the least fixed point of $x \mapsto \omega^x$) and $<^*$ is the converse of the usual ordering of ordinals.
- Calculate $\tau(\mathbb Z^2,<_{\textrm{lex}})$, where $<_{\textrm{lex}}$ is the lexicographic order.
- Calculate $\tau(\mathbb Z[x],\prec)$, where $p \prec q$ is defined as $\exists N \in \mathbb Z \ \forall x > N \ p(x) < q(x)$ (and $<$ is the usual ordering of integers).
My (partial) solutions. General remark: $\tau(S,<)$ is always an ordinal itself (it's pretty easy to check directly).
- $\tau(\varepsilon_0,<^*) = \omega$, since every finite ordinal is isomorphic to its converse we've $\omega \subseteq \tau(\varepsilon_0,<^*)$, now I just need to prove that $(\varepsilon_0,<^*)$ doesn't have any countable well ordered subset. If there was a well ordered countable subset $A$ of $(\varepsilon_0,<^*)$, then, with respect to the normal order $<$ of ordinals, every non empty subset of $A$ (included $A$ itself) has a greatest element, so I could construct a $\aleph_0$ descendent chain and, since the ordering of ordinals is given by membership relation $\in$, this would contradict the axiom of foundation (regularity).
I'm pretty sure about the result but not for my last argument (I may be overthinking and there could be a much simpler reason why such an $A$ cannot exists). - $\tau(\mathbb Z^2,<_{\textrm{lex}}) = \omega^2 + 1$, it should be pretty easy to check that $(\omega^2,<) \cong (\mathbb Z_{\geq 0} \times \mathbb Z_{\geq 0}, <_{\textrm{lex}})$, so $\omega^2 \in \tau(\mathbb Z^2,<_{\textrm{lex}}) \implies \omega^2 \subseteq \tau(\mathbb Z^2,<_{\textrm{lex}})$. Therefore if I prove that there's no well ordered subset of $(\mathbb Z^2,<_{\textrm{lex}})$ isomorphic to $\omega^2 + 1$ I'll conclude. My intuition is that there's not enough space in $(\mathbb Z^2,<_{\textrm{lex}})$ for this order type since we'd need a subset like $\mathbb Z_{\geq 0} \times \mathbb Z_{\geq 0}$ with a greatest element, but such a thing doesn't exists in $(\mathbb Z^2,<_{\textrm{lex}})$. I don't know how to write it down properly with a formal argoument though.
- $\tau(\mathbb Z[x],\prec) = \omega^\omega + 1$, with a similar argument as above, firstly I notice that $(\omega^\omega,<) \cong (\mathbb Z_{\geq 0}[x],\prec)$, in fact $\omega^\omega$ identifies with $S = \{f : \omega \to \omega : |\operatorname{supp}(f)|<\aleph_0\}$, which is ordered confronting $f,g \in S$ on the greatest element $x \in \omega$ where $f(x) \ne g(x)$, and then: $$S \to \mathbb Z_{\geq 0}[x] : f \mapsto p_f(x) = \sum_{0 \leq i \leq |\operatorname{supp}(f)|} f(i)\cdot x^i$$ is an isomorphism (all the checks should be easy). Thus $\omega^\omega \in \tau(\mathbb Z[x],\prec)$, and, just like above I need to show that there no well ordered subset of $\tau(\mathbb Z[x],\prec)$ isomorphic to $\omega^\omega + 1$.
Are my answers correct? (I'm pretty confident for 1 and 2, but I may be wrong about 3) If so how can I argument formally to conclude?
