[proofplan] We reduce both parts to the [Complete Matching from the Smaller Side](/theorems/2583) theorem applied to the biregular bipartite graph between layers $X^{(r)}$ and $X^{(s)}$. The key observation is that the condition $|n/2 - r| \geq |n/2 - s|$ is equivalent to $\binom{n}{r} \leq \binom{n}{s}$ by the unimodality of binomial coefficients, which determines which layer is smaller. The inclusion graph between the two layers is biregular, so the smaller side admits a complete matching into the larger side. [/proofplan] [step:Construct the biregular bipartite graph between layers $X^{(r)}$ and $X^{(s)}$] Fix $1 \leq r < s \leq n$. Define the bipartite graph $G = (X^{(r)}, X^{(s)}; E)$ where $A \in X^{(r)}$ is adjacent to $B \in X^{(s)}$ whenever $A \subseteq B$. Each $A \in X^{(r)}$ is contained in exactly $\binom{n - r}{s - r}$ members of $X^{(s)}$: to extend $A$ to an $s$-element superset, one must choose $s - r$ elements from the $n - r$ elements of $X \setminus A$. Each $B \in X^{(s)}$ contains exactly $\binom{s}{r}$ members of $X^{(r)}$: one must choose $r$ elements from the $s$ elements of $B$. Therefore $G$ is $\left(\binom{n-r}{s-r},\, \binom{s}{r}\right)$-regular, hence biregular. [guided] We need to set up the bipartite graph that connects the two layers. Given $1 \leq r < s \leq n$, define $G = (X^{(r)}, X^{(s)}; E)$ with the adjacency relation $A \sim B$ if and only if $A \subseteq B$. Why is this graph biregular? We compute the degree of each vertex on both sides. For $A \in X^{(r)}$: we ask how many $s$-element supersets of $A$ exist in $X$. Since $|A| = r$ and we need $|B| = s$ with $A \subseteq B$, we must add exactly $s - r$ elements from $X \setminus A$, which has $n - r$ elements. The number of choices is $\binom{n - r}{s - r}$, so every vertex in $X^{(r)}$ has degree $\binom{n - r}{s - r}$. For $B \in X^{(s)}$: we ask how many $r$-element subsets of $B$ exist. This is $\binom{s}{r}$, so every vertex in $X^{(s)}$ has degree $\binom{s}{r}$. Since every vertex on the left has the same degree and every vertex on the right has the same degree, $G$ is $\left(\binom{n-r}{s-r},\, \binom{s}{r}\right)$-regular. In particular, $G$ is biregular. [/guided] [/step] [step:Translate the distance-from-centre condition into a layer-size comparison] The binomial coefficient $\binom{n}{k}$ is unimodal as a function of $k$: it increases for $k \leq \lfloor n/2 \rfloor$ and decreases for $k \geq \lceil n/2 \rceil$. Therefore $\binom{n}{k_1} \leq \binom{n}{k_2}$ if and only if $k_1$ is at least as far from $n/2$ as $k_2$ is, i.e., $|n/2 - k_1| \geq |n/2 - k_2|$. Applying this to $k_1 = r$ and $k_2 = s$: - The condition $|n/2 - r| \geq |n/2 - s|$ in part (i) is equivalent to $|X^{(r)}| = \binom{n}{r} \leq \binom{n}{s} = |X^{(s)}|$. - The condition $|n/2 - r| \leq |n/2 - s|$ in part (ii) is equivalent to $|X^{(r)}| = \binom{n}{r} \geq \binom{n}{s} = |X^{(s)}|$. [guided] Why does $|n/2 - r| \geq |n/2 - s|$ translate to $\binom{n}{r} \leq \binom{n}{s}$? The binomial coefficient $\binom{n}{k}$ achieves its maximum at $k = \lfloor n/2 \rfloor$ (or at both $\lfloor n/2 \rfloor$ and $\lceil n/2 \rceil$ when $n$ is even). Moving away from the centre in either direction causes $\binom{n}{k}$ to decrease. So the further $k$ is from $n/2$ (in the sense of $|n/2 - k|$), the smaller $\binom{n}{k}$ is. More precisely, for $0 \leq k \leq n - 1$: \begin{align*} \frac{\binom{n}{k+1}}{\binom{n}{k}} = \frac{n - k}{k + 1}, \end{align*} which is $\geq 1$ when $k \leq (n-1)/2$ and $\leq 1$ when $k \geq (n-1)/2$. This confirms the unimodality. Therefore: $|n/2 - r| \geq |n/2 - s|$ means $r$ is at least as far from the centre as $s$, so $\binom{n}{r} \leq \binom{n}{s}$, giving $|X^{(r)}| \leq |X^{(s)}|$. Similarly, $|n/2 - r| \leq |n/2 - s|$ gives $|X^{(r)}| \geq |X^{(s)}|$. [/guided] [/step] [step:Apply the complete matching theorem for biregular graphs to obtain the injections] **Part (i).** Suppose $|n/2 - r| \geq |n/2 - s|$, so that $|X^{(r)}| \leq |X^{(s)}|$. The bipartite graph $G = (X^{(r)}, X^{(s)}; E)$ is biregular with $|X^{(r)}| \leq |X^{(s)}|$. By the [Complete Matching from the Smaller Side](/theorems/2583) theorem, there exists a complete matching from $X^{(r)}$ into $X^{(s)}$. This matching is an injection $f: X^{(r)} \to X^{(s)}$ such that $A \subseteq f(A)$ for all $A \in X^{(r)}$, since edges in $G$ connect $A$ to supersets of $A$. **Part (ii).** Suppose $|n/2 - r| \leq |n/2 - s|$, so that $|X^{(s)}| \leq |X^{(r)}|$. Viewing $G$ with $X^{(s)}$ as the "smaller side," the [Complete Matching from the Smaller Side](/theorems/2583) theorem gives a complete matching from $X^{(s)}$ into $X^{(r)}$. This matching is an injection $g: X^{(s)} \to X^{(r)}$ such that $g(A) \subseteq A$ for all $A \in X^{(s)}$, since edges connect each $s$-element set to its $r$-element subsets. [guided] We now apply the [Complete Matching from the Smaller Side](/theorems/2583) theorem, which states: if $G = (X, Y; E)$ is biregular with $|X| \leq |Y|$, then there is a complete matching from $X$ into $Y$. The theorem requires two conditions: (a) $G$ is biregular, and (b) the side from which we match is the smaller side. We verified (a) in the first step. **For part (i):** The condition $|n/2 - r| \geq |n/2 - s|$ gives $|X^{(r)}| \leq |X^{(s)}|$ by the unimodality argument. We apply the theorem with $X^{(r)}$ as the smaller side. The conclusion is a complete matching from $X^{(r)}$ into $X^{(s)}$: an injection \begin{align*} f: X^{(r)} &\to X^{(s)} \end{align*} with the property that $f(A)$ is a neighbour of $A$ in $G$ for every $A$. By the definition of $G$, this means $A \subseteq f(A)$. **For part (ii):** The condition $|n/2 - r| \leq |n/2 - s|$ gives $|X^{(s)}| \leq |X^{(r)}|$. Now $X^{(s)}$ is the smaller side. The bipartite graph between $X^{(s)}$ and $X^{(r)}$ (with adjacency $B \sim A$ iff $A \subseteq B$) is also biregular — it is the same graph $G$ with sides swapped — so the theorem gives a complete matching from $X^{(s)}$ into $X^{(r)}$: an injection \begin{align*} g: X^{(s)} &\to X^{(r)} \end{align*} with $g(A) \subseteq A$ for every $A \in X^{(s)}$. Note that both parts can hold simultaneously when $|n/2 - r| = |n/2 - s|$, i.e., when $\binom{n}{r} = \binom{n}{s}$: in that case injections exist in both directions, and both are bijections. [/guided] [/step]