Kerodon

Proof. Set $S_0 = \{ s \in S: f'(s) < f(s) \} $. We wish to show that $S_0$ is empty. Assume otherwise. Since $S$ is well-ordered, there is a least element $s \in S_0$. Since $f$ is an initial segment embedding, the inequality $f'(s) < f(s)$ implies that we can write $f'(s) = f(t)$ for some $t < s$. Then $t \notin S_0$, so we must have $f(t) \leq f'(t)$. It follows that $f'(s) \leq f'(t)$, contradicting our assumption that the function $f'$ is strictly increasing. $\square$

Proof. For each element $s \in S$, let $S_{\leq s}$ denote the initial segment $\{ s' \in S: s' \leq s \} $. Let $S_0 \subseteq S$ denote the collection of elements $s \in S$ for which there exists an initial segment embedding $f_{\leq s}: S_{\leq s} \hookrightarrow T$. Note that, if this condition is satisfied, then the morphism $f_{\leq s}$ is uniquely determined (Corollary 4.7.1.15). Moreover, if $s' \leq s$, then composite map $S_{\leq s'} \subseteq S_{\leq s} \xrightarrow { f_{\leq s} } T$ is also an initial segment embedding; it follows that $s'$ belongs to $S_0$, and $f|_{\leq s'}$ is the restriction of $f|_{\leq s}$ to $S_{\leq s'}$. Consequently, the construction $s \mapsto f_ s(s)$ determines a function $f: S_0 \rightarrow T$, which is an isomorphism of $S_0$ with an initial segment $T_0 \subseteq T$. If $S_0 = S$, then $f$ is an initial segment embedding from $S$ to $T$. If $T_0 = T$, then $g = f^{-1}$ is an initial segment embedding from $T$ to $S$. Assume that neither of these conditions is satisfied: that is, the sets $S \setminus S_0$ and $T \setminus T_0$ are both nonempty. Let $s$ be a least element of $S \setminus S_0$, and let $t$ be a least element of $T \setminus T_0$. Then $f$ extends uniquely to an initial segment embedding

The existence of $f_{\leq s}$ shows that $s$ belongs to $S_0$, which is a contradiction. $\square$

Proof. The reflexivity of the relation $\leq $ follows from Example 4.7.1.12, and the transitivity follows from Remark 4.7.1.13. Let $\alpha $ and $\beta $ be ordinals, which we identify with the order types of well-ordered sets $(S, \leq )$ and $(T, \leq )$, respectively. Invoking Corollary 4.7.1.17, we deduce that $\alpha \leq \beta $ or $\beta \leq \alpha $. Moreover, if both conditions are satisfied, then Remark 4.7.1.18 shows that $\alpha = \beta $. $\square$

Proof. We will prove existence; uniqueness then follows from Corollary 4.7.1.16. For each $s \in S$, let $\alpha _{s}$ denote the order type of the set $S_{< s}$ (which is well-ordered, by virtue of Remark 4.7.1.5). Note that, since there is an initial segment embedding $S_{< s} \hookrightarrow S$ which is not bijective, we must have $\alpha _{s} < \alpha $ (Remark 4.7.1.18). Consequently, the construction $s \mapsto \alpha _{s}$ determines a function $S \rightarrow \mathrm{Ord}_{< \alpha }$. If $s < t$ in $S$, then there is an initial segment embedding from $S_{< s}$ to $S_{< t}$ which is not bijective, so that $\alpha _{s} < \alpha _{t}$ (again by Remark 4.7.1.18). To complete the proof, it will suffice to show that the function $s \mapsto \alpha _{s}$ is surjective. Let $\beta $ be an ordinal which is strictly smaller than $\alpha $. Then $\beta $ is the order type of some initial segment $S_0 \subsetneq S$. Since $S$ is well-ordered, the set $S \setminus S_0$ has a smallest element $s$. It follows that $S_0 = S_{< s}$, so that $\beta = \alpha _{s}$. $\square$

Proof. Choose an ordinal $\alpha \in S$. If $\alpha $ is a least element of $S$, then we are done. Otherwise, we can replace $S$ by the nonempty subset $S_{< \alpha } = \{ \beta \in S: \beta < \alpha \} $. Note that $S_{< \alpha }$ is a nonempty subset of $\mathrm{Ord}_{< \alpha }$, and therefore has a smallest element by virtue of Corollary 4.7.1.23. $\square$

Proof. Let $\{ S_ q \} _{q \in Q}$ be the collection of all well-ordered subsets of $T$. We regard $Q$ as a partially ordered set, where $q \leq q'$ if the set $S_{q}$ is an initial segment of $S_{q'}$. This partial ordering satisfies the hypotheses of Zorn's lemma, and therefore contains a maximal element $S_{\mathrm{max}}$. To complete the proof, it will suffice to show that the inclusion $S_{ \mathrm{max}} \hookrightarrow T$ is cofinal. Assume otherwise: then there exists an element $t \in T$ satisfying $s < t$ for each $s \in S_{\mathrm{max}}$. Then $S_{\mathrm{max}}$ is an initial segment of the well-ordered subset $S_{\mathrm{max}} \cup \{ t \} \subseteq T$, contradicting the maximality of $S_{\mathrm{max}}$. $\square$

Proof. It is clear that the cofinality $\mathrm{cf}(T)$ satisfies condition $(\ast )$. For the converse, assume that $(S, \leq )$ is a well-ordered set of order type $\alpha $ and that $f: S \rightarrow T$ is an unbounded function. Let us say that an element $s \in S$ is good if, for every element $s' < s$ of $S$, we have $f(s') < f(s)$. Let $S_0$ be the collection of good elements of $S$, and set $f_0 = f|_{ S_0 }$. By construction, the function $f_0$ is strictly increasing. Moreover, the order type of $S_0$ is $\leq \alpha $ (Remark 4.7.1.21). To complete the proof, it will suffice to show that $f_0: S_0 \hookrightarrow T$ is cofinal. Fix an element $t \in T$, and set $S_{\geq t} = \{ s \in S: t \leq f(s) \} $. We wish to show that the intersection $S_{\geq t} \cap S_0$ is nonempty. We first observe that $S_{\geq t}$ is nonempty (by virtue of our assumption that $f$ is unbounded). Since $(S, \leq )$ is well-ordered, the set $S_{\geq t}$ contains a least element $s$. We claim that $s$ belongs to $S_0$. Assume otherwise: then there exists some $s' < s$ satisfying $f(s') \geq f(s)$. It follows that $s'$ belongs to $S_{\geq t}$, contradicting the minimality of $s$. $\square$

Proof. Let $Q$ denote the set of ordered pairs $(T, \leq _{T})$, where $T$ is a subset of $S$ which is closed downward with respect to $\preceq $ and $\leq _{T}$ is a well-ordering of $T$ which refines $\preceq $. We regard $Q$ as a partially ordered set, where $(T, \leq _{T} ) \leq (T', \leq _{T'} )$ if $T$ is an initial segment of $T'$ (with respect to the ordering $\leq _{T'}$), and the ordering $\leq _{T}$ coincides with the restriction of $\leq _{T'}$. The partially ordered set $Q$ satisfies the hypotheses of Zorn's lemma, and therefore contains a maximal element $( T_{\mathrm{max}}, \leq _{ T_{\mathrm{max}} } )$. To complete the proof, it will suffice to show that $T_{\mathrm{max}} = S$. Suppose otherwise. Then the set $S \setminus T_{\mathrm{max}}$ is nonempty, and therefore contains an element $s$ which is minimal with respect to the ordering $\preceq $. Set $T' = T_{\mathrm{max}} \cup \{ s\} $, and extend $\leq _{ T_{\mathrm{max}} }$ to a linear ordering $\leq _{T'}$ of $T'$ by declaring $s$ to be a largest element. Then $(T', \leq _{T'} )$ is an element of $Q$, contradicting the maximality of the pair $( T_{\mathrm{max}}, \leq _{ T_{\mathrm{max}} } )$. $\square$