[proofplan] We prove both directions of the equivalence. The forward direction shows that any colex initial segment is $(U,V)$-compressed for every valid pair, by observing that $C_{UV}$ moves sets earlier in the colex order. The converse shows that if $\mathcal{A}$ is not a colex initial segment, we can exhibit a valid pair $(U,V)$ for which $\mathcal{A}$ is not compressed, by extracting $U$ and $V$ from the symmetric difference of a witness pair $B \in \mathcal{A}$, $C \notin \mathcal{A}$ with $C <_{\mathrm{colex}} B$. [/proofplan] [step:Show that every colex initial segment is $(U,V)$-compressed for all valid pairs] Let $\mathcal{A}$ be an initial segment of $X^{(r)}$ in the colex order, and let $U, V \subseteq X$ with $|U| = |V|$, $U \cap V = \varnothing$, and $\max V > \max U$. Take any $A \in \mathcal{A}$ with $A \cap (U \cup V) = V$ (so that the compression acts nontrivially). Define $A' = C_{UV}(A) = (A \setminus V) \cup U$. The symmetric difference $A \triangle A' = U \cup V$ (since $A$ contains $V$ but not $U$, and $A'$ contains $U$ but not $V$). The largest element of $A \triangle A'$ is $\max(U \cup V) = \max V$, and $\max V \in V \subseteq A$, so $\max(A \triangle A') \in A$. By the definition of the colex order, this means $A' <_{\mathrm{colex}} A$. Since $\mathcal{A}$ is an initial segment and $A' <_{\mathrm{colex}} A \in \mathcal{A}$, we have $A' \in \mathcal{A}$. Therefore $C_{UV}(A) \in \mathcal{A}$ for every $A \in \mathcal{A}$, which means $C_{UV}(\mathcal{A}) = \mathcal{A}$. [guided] Let $\mathcal{A}$ be an initial segment of $X^{(r)}$ in the colex order, and let $U, V \subseteq X$ with $|U| = |V|$, $U \cap V = \varnothing$, and $\max V > \max U$. We need to show $C_{UV}(\mathcal{A}) = \mathcal{A}$. Take any $A \in \mathcal{A}$ satisfying $A \cap (U \cup V) = V$ (so the compression acts nontrivially on $A$). Define $A' = C_{UV}(A) = (A \setminus V) \cup U$. We claim $A' <_{\mathrm{colex}} A$, so that $A' \in \mathcal{A}$ by the initial segment property. To compare $A'$ and $A$ in the colex order, we examine $A \triangle A'$. Since $A \supseteq V$ and $A \cap U = \varnothing$, while $A' \supseteq U$ and $A' \cap V = \varnothing$, the symmetric difference is $A \triangle A' = (A \setminus A') \cup (A' \setminus A) = V \cup U$. The largest element of $A \triangle A'$ is $\max(U \cup V) = \max V$ (since $\max V > \max U$ by hypothesis). This largest element belongs to $V \subseteq A$, so by the definition of the colex order ($B <_{\mathrm{colex}} A$ iff $\max(A \triangle B) \in A$), we conclude $A' <_{\mathrm{colex}} A$. Why does this finish the forward direction? Because $\mathcal{A}$ is an initial segment: every set that precedes a member of $\mathcal{A}$ in colex also belongs to $\mathcal{A}$. Since $A' <_{\mathrm{colex}} A$ and $A \in \mathcal{A}$, we get $A' \in \mathcal{A}$. This means $C_{UV}(A)$ already belongs to $\mathcal{A}$ for every member $A$ on which the compression acts nontrivially. For members on which $C_{UV}$ is the identity, the membership is automatic. Therefore $C_{UV}(\mathcal{A}) = \mathcal{A}$. [/guided] [/step] [step:Prove the converse: if $\mathcal{A}$ is not a colex initial segment, exhibit a valid pair $(U,V)$ for which $\mathcal{A}$ is not compressed] Suppose $\mathcal{A}$ is $(U,V)$-compressed for every pair $U, V \subseteq X$ with $|U| = |V|$, $U \cap V = \varnothing$, and $\max V > \max U$, but $\mathcal{A}$ is not a colex initial segment. Then there exist $B \in \mathcal{A}$ and $C \notin \mathcal{A}$ with $C <_{\mathrm{colex}} B$. Define \begin{align*} U &:= C \setminus B, \\ V &:= B \setminus C. \end{align*} Since $|B| = |C| = r$, we have $|U| = |C \setminus B| = |B \setminus C| = |V|$. Moreover $U \cap V = (C \setminus B) \cap (B \setminus C) = \varnothing$. Since $C <_{\mathrm{colex}} B$, the largest element of $B \triangle C = U \cup V$ belongs to $B$, i.e., $\max(U \cup V) \in B \setminus C = V$. Therefore $\max V = \max(U \cup V) > \max U$, so $(U, V)$ is a valid pair. Now verify that the compression $C_{UV}$ acts nontrivially on $B$. We have $B \cap (U \cup V) = B \cap ((C \setminus B) \cup (B \setminus C)) = B \cap (B \setminus C) = B \setminus C = V$ (using $B \cap (C \setminus B) = \varnothing$). Therefore \begin{align*} C_{UV}(B) = (B \setminus V) \cup U = (B \setminus (B \setminus C)) \cup (C \setminus B) = (B \cap C) \cup (C \setminus B) = C. \end{align*} Since $C \notin \mathcal{A}$, the compression $C_{UV}$ maps $B \in \mathcal{A}$ to $C \notin \mathcal{A}$, so $\mathcal{A}$ is not $(U,V)$-compressed. This contradicts our assumption. [guided] Suppose $\mathcal{A}$ is $(U,V)$-compressed for every valid pair but is not a colex initial segment. Being a colex initial segment means: for every $A \in \mathcal{A}$ and every $C \in X^{(r)}$ with $C <_{\mathrm{colex}} A$, we have $C \in \mathcal{A}$. Since $\mathcal{A}$ fails this, there exist $B \in \mathcal{A}$ and $C \notin \mathcal{A}$ with $C <_{\mathrm{colex}} B$. How do we extract a compression pair from $B$ and $C$? The natural choice is to let $U$ contain the elements that $C$ has but $B$ does not, and $V$ the elements $B$ has but $C$ does not: \begin{align*} U &:= C \setminus B, \\ V &:= B \setminus C. \end{align*} We verify the three requirements for a valid compression pair: 1. **Equal size**: $|U| = |C \setminus B|$ and $|V| = |B \setminus C|$. Since $|B| = |C| = r$ and $|B \cap C|$ is the same from both sides, $|B \setminus C| = |C \setminus B|$, so $|U| = |V|$. 2. **Disjointness**: $U \cap V = (C \setminus B) \cap (B \setminus C) = \varnothing$, since an element cannot simultaneously be in $C$ but not $B$ and in $B$ but not $C$. 3. **Directional condition**: The symmetric difference is $B \triangle C = (B \setminus C) \cup (C \setminus B) = V \cup U$. Since $C <_{\mathrm{colex}} B$, the definition of colex gives $\max(B \triangle C) \in B$. This largest element belongs to $B \setminus C = V$ (it cannot belong to $B \cap C$, which is outside $B \triangle C$). So $\max V = \max(B \triangle C) > \max U$, since $\max(B \triangle C)$ is the global maximum and it lies in $V$, not $U$. Now we show the compression $C_{UV}$ acts nontrivially on $B$ and maps it to $C$. We compute $B \cap (U \cup V)$: since $U = C \setminus B$ is disjoint from $B$, we have $B \cap U = \varnothing$. And $B \cap V = B \cap (B \setminus C) = B \setminus C = V$. Therefore $B \cap (U \cup V) = V$, which is exactly the condition for $C_{UV}$ to act on $B$. Applying the compression: \begin{align*} C_{UV}(B) &= (B \setminus V) \cup U = (B \setminus (B \setminus C)) \cup (C \setminus B) = (B \cap C) \cup (C \setminus B) = C. \end{align*} Since $B \in \mathcal{A}$ and $C_{UV}(B) = C \notin \mathcal{A}$, the family $\mathcal{A}$ is not $(U,V)$-compressed, contradicting our assumption. This contradiction shows that $\mathcal{A}$ must be a colex initial segment. [/guided] [/step] [step:Combine both directions to conclude the equivalence] The forward direction (step 1) shows: $\mathcal{A}$ a colex initial segment $\implies$ $\mathcal{A}$ is $(U,V)$-compressed for all valid pairs. The contrapositive of step 2 shows: $\mathcal{A}$ is $(U,V)$-compressed for all valid pairs $\implies$ $\mathcal{A}$ is a colex initial segment. Together, these establish the equivalence: $\mathcal{A}$ is an initial segment of $X^{(r)}$ in colex order if and only if $\mathcal{A}$ is $(U,V)$-compressed for every pair of disjoint sets $U, V \subseteq X$ with $|U| = |V|$ and $\max V > \max U$. [/step]