[proofplan] We take a non-empty subset $S \subseteq X = \bigcup_{i \in I} X_i$ and exhibit its minimum. Because the family is nested, any two $X_i$, $X_j$ satisfy $X_i \leq X_j$ or $X_j \leq X_i$, with the smaller sitting as an initial segment of the larger. We pick any index $i$ for which $S \cap X_i \neq \varnothing$, take the minimum of $S \cap X_i$ in the well-ordering $X_i$, and use the initial-segment property to promote it to the minimum of $S$ in all of $X$. Verifying that the inherited relation is a total order on $X$ and that each $X_i$ embeds as an initial segment of $X$ completes the argument. [/proofplan] [step:Verify that the inherited relation is a total order on $X$] We are given a nested family $\{X_i : i \in I\}$ of well-orderings: for any $i, j \in I$ there exists $k \in I$ with $X_i \leq X_k$ and $X_j \leq X_k$, meaning both $X_i$ and $X_j$ embed as initial segments of $X_k$. Define \begin{align*} X &:= \bigcup_{i \in I} X_i, & {<} &:= \bigcup_{i \in I} {<_i} \subseteq X \times X. \end{align*} For $a, b \in X$ write $a < b$ when there exists $i \in I$ with $a, b \in X_i$ and $a <_i b$. We must check that $<$ is irreflexive, transitive, and trichotomous on $X$. *Irreflexivity.* If $a < a$ then $a <_i a$ for some $i$, contradicting irreflexivity of $<_i$. *Transitivity.* Suppose $a < b$ and $b < c$; pick $i, j$ with $a <_i b$ and $b <_j c$. By the nesting hypothesis there exists $k \in I$ with $X_i \leq X_k$ and $X_j \leq X_k$. Because the embedding of $X_i$ into $X_k$ is order-preserving, $a <_k b$; likewise $b <_k c$. Transitivity of $<_k$ gives $a <_k c$, hence $a < c$. *Trichotomy.* Given $a, b \in X$, pick $i, j$ with $a \in X_i$, $b \in X_j$, and $k$ with $X_i, X_j$ both initial segments of $X_k$. Then $a, b \in X_k$, and trichotomy of $<_k$ gives $a <_k b$, $a = b$, or $b <_k a$. Any of these yields the corresponding relation under $<$. Hence $(X, <)$ is a totally ordered set. [guided] The nested hypothesis means that for any two members $X_i$, $X_j$ of the family we can find $X_k$ in the family that contains both as initial segments. This is exactly the data we need for pairwise comparisons: whenever we must examine two elements $a \in X_i$ and $b \in X_j$, we lift both into a common $X_k$ and reason there. Define \begin{align*} X &:= \bigcup_{i \in I} X_i, & {<} &:= \bigcup_{i \in I} {<_i} \subseteq X \times X, \end{align*} so $a < b$ means there is some $i$ with $a, b \in X_i$ and $a <_i b$. Why is $<$ well-defined (unambiguous)? Suppose $a < b$ is witnessed by $a <_i b$ and also $b <_j a$ for some other $j$. Choose $k$ with $X_i, X_j \leq X_k$. Since the embeddings are order-preserving, we would have both $a <_k b$ and $b <_k a$, violating trichotomy of $<_k$. So the witnesses at different indices cannot disagree. *Irreflexivity.* Suppose $a < a$; then $a <_i a$ for some $i$, but $<_i$ is irreflexive. *Transitivity.* Suppose $a < b$ and $b < c$. Pick $i, j \in I$ with $a <_i b$ and $b <_j c$. We need a single well-ordering in which both comparisons live. The nesting hypothesis supplies $k \in I$ such that $X_i$ and $X_j$ both embed as initial segments into $X_k$. The embeddings are order-preserving, so $a <_k b$ and $b <_k c$. Transitivity of the single well-ordering $<_k$ then gives $a <_k c$, which is the required $a < c$. *Trichotomy.* Given $a, b \in X$, pick indices $i, j$ and $a \in X_i$, $b \in X_j$. Choose $k \in I$ with $X_i, X_j \leq X_k$. Now $a, b \in X_k$, and trichotomy of the single well-ordering $<_k$ delivers exactly one of $a <_k b$, $a = b$, $b <_k a$. Promoting to $<$ via the definition gives the corresponding relation on $X$. Thus $(X, <)$ is a totally ordered set. The argument is entirely the "lift both elements into a common $X_k$" trick. [/guided] [/step] [step:Show that each $X_i$ is an initial segment of $X$] We claim: for every $i \in I$, the inclusion $X_i \hookrightarrow X$ is order-preserving and its image is an initial segment of $X$. *Order-preserving.* For $a, b \in X_i$ with $a <_i b$, we have $a < b$ directly from the definition of $<$. *Initial segment.* Let $a \in X$ and suppose $a < b$ for some $b \in X_i$. We must show $a \in X_i$. Pick $j \in I$ with $a \in X_j$ and choose $k \in I$ with $X_i, X_j \leq X_k$. Then $a, b \in X_k$ and $a <_k b$. Since $X_i$ is an initial segment of $X_k$ and $b \in X_i$, the definition of initial segment forces $a \in X_i$. [guided] To show the inclusion $X_i \hookrightarrow X$ is an initial-segment embedding, we need two things: it preserves the order, and its image is downward closed in $X$. *Order preservation is immediate.* If $a, b \in X_i$ and $a <_i b$, the definition $< := \bigcup_i <_i$ gives $a < b$ directly. *Downward closure.* Take $b \in X_i$ and suppose $a \in X$ satisfies $a < b$. We want $a \in X_i$. Pick $j \in I$ with $a \in X_j$. By the nesting hypothesis there is $k$ with $X_i, X_j \leq X_k$. Both $a$ and $b$ live in $X_k$, and by the unambiguity check from the previous step $a <_k b$. Now comes the payoff of the nesting hypothesis: since $X_i$ embeds into $X_k$ as an *initial segment*, the subset of $X_k$ that is below some element of $X_i$ is contained in $X_i$. Concretely, $X_i \subseteq X_k$ is downward closed, so $a <_k b \in X_i$ forces $a \in X_i$. This is where the nesting hypothesis does work it could not do otherwise: without initial-segment embedding between the family members, $X_i$ would not be downward closed in the union. [/guided] [/step] [step:Extract the minimum of an arbitrary non-empty subset] Let $\varnothing \neq S \subseteq X$. Pick any $a \in S$ and $i \in I$ with $a \in X_i$; then $a \in S \cap X_i$, so $S \cap X_i$ is a non-empty subset of the well-ordering $X_i$. Set \begin{align*} x := \min\nolimits_{X_i}(S \cap X_i), \end{align*} which exists by the well-ordering of $X_i$. We claim $x = \min_X(S)$. Let $y \in S$ be arbitrary. Two cases. *Case 1: $y \in X_i$.* Then $y \in S \cap X_i$, so $x \leq_i y$ by choice of $x$, hence $x \leq y$ in $X$. *Case 2: $y \notin X_i$.* Since $X_i$ is an initial segment of $X$ (by the previous step), $y \not< z$ for any $z \in X_i$ — otherwise downward closure would put $y$ in $X_i$. By trichotomy, for every $z \in X_i$ we have $y > z$ or $y = z$; and $y = z$ is impossible because $z \in X_i$ while $y \notin X_i$. Thus $y > z$ for every $z \in X_i$, and in particular $y > x$, giving $x < y$. In both cases $x \leq y$, so $x$ is the minimum of $S$ in $(X, <)$. Since $S$ was an arbitrary non-empty subset, $(X, <)$ is a well-ordering. [/step] [step:Exhibit each $X_i$ as an initial segment of $X$ to conclude $X_i \leq X$] By the second step, for each $i \in I$ the inclusion $X_i \hookrightarrow X$ is an order-preserving injection whose image is an initial segment of $X$. By definition of the comparison order on well-orderings, this means $X_i \leq X$. Combined with the well-ordering established in the third step, this completes the proof: $X$ is a well-ordering with $X_i \leq X$ for all $i \in I$. [/step]