[proofplan] The lower bound $\binom{n}{\lfloor n/2 \rfloor}$ is witnessed by the middle-level family $X^{(\lfloor n/2 \rfloor)}$, which is an antichain because sets of the same size cannot contain one another. For the upper bound, we partition $\mathcal{P}(X)$ into exactly $\binom{n}{\lfloor n/2 \rfloor}$ chains using the [Injections Between Layers of the Boolean Lattice](/theorems/2584). Since any antichain can contain at most one element from each chain, this forces $|\mathcal{A}| \leq \binom{n}{\lfloor n/2 \rfloor}$. [/proofplan] [step:Exhibit the middle-level antichain to establish the lower bound] The family $X^{(\lfloor n/2 \rfloor)}$ consists of all $\lfloor n/2 \rfloor$-element subsets of $X$. No two sets in this family satisfy $A \subsetneq B$, since they have the same cardinality. Therefore $X^{(\lfloor n/2 \rfloor)}$ is an antichain, and its size is $\binom{n}{\lfloor n/2 \rfloor}$. This proves the maximum antichain size is at least $\binom{n}{\lfloor n/2 \rfloor}$. [/step] [step:Observe that an antichain meets each chain in at most one element] [claim:Antichain-chain intersection bound] Let $(S, \leq)$ be a poset. If $\mathcal{C} \subseteq S$ is a chain and $\mathcal{A} \subseteq S$ is an antichain, then $|\mathcal{A} \cap \mathcal{C}| \leq 1$. [/claim] [proof] Suppose for contradiction that $\mathcal{A} \cap \mathcal{C}$ contains two distinct elements $x$ and $y$. Since $\mathcal{C}$ is a chain, $x$ and $y$ are comparable: either $x \leq y$ or $y \leq x$. But since $\mathcal{A}$ is an antichain, no two distinct elements of $\mathcal{A}$ are comparable. This is a contradiction. [/proof] It follows that if $\mathcal{P}(X)$ can be partitioned into $m$ chains $\mathcal{C}_1, \ldots, \mathcal{C}_m$, then for any antichain $\mathcal{A}$: \begin{align*} |\mathcal{A}| = \sum_{j=1}^{m} |\mathcal{A} \cap \mathcal{C}_j| \leq m. \end{align*} [/step] [step:Partition $\mathcal{P}(X)$ into $\binom{n}{\lfloor n/2 \rfloor}$ chains using the layer injections] Set $k = \lfloor n/2 \rfloor$. We construct a partition of $\mathcal{P}(X)$ into chains by composing the injections from the [Injections Between Layers of the Boolean Lattice](/theorems/2584) theorem. **Upper half.** For each $i$ with $k \leq i \leq n - 1$, we have $|n/2 - i| \leq |n/2 - (i+1)|$ (since $i \geq k = \lfloor n/2 \rfloor$ implies $i$ is at least as close to $n/2$ as $i + 1$). By part (ii) of the [layer injection theorem](/theorems/2584) applied with $r = i$ and $s = i + 1$, there exists an injection \begin{align*} g_i: X^{(i+1)} &\to X^{(i)} \end{align*} with $g_i(A) \subseteq A$ for all $A \in X^{(i+1)}$. Composing: for each $A \in X^{(n)}$, the sequence $g_k \circ g_{k+1} \circ \cdots \circ g_{n-1}(A),\, \ldots,\, g_{n-1}(A),\, A$ determines a chain from level $k$ to level $n$. Since $g_k$ maps into $X^{(k)}$, every such chain passes through a unique element of $X^{(k)}$. The injectivity of each $g_i$ ensures these chains partition $X^{(\geq k)}$ into $|X^{(k)}| = \binom{n}{k}$ chains, each containing exactly one element from every level $k, k+1, \ldots$ that it reaches. **Lower half.** For each $i$ with $1 \leq i \leq k$, we have $|n/2 - (i-1)| \geq |n/2 - i|$ (since $i \leq k$ implies $i - 1$ is at least as far from $n/2$ as $i$). By part (i) of the [layer injection theorem](/theorems/2584) applied with $r = i - 1$ and $s = i$, there exists an injection \begin{align*} f_i: X^{(i-1)} &\to X^{(i)} \end{align*} with $A \subseteq f_i(A)$ for all $A \in X^{(i-1)}$. By the same reasoning, these injections partition $X^{(\leq k)}$ into $\binom{n}{k}$ chains, each containing exactly one element of $X^{(k)}$. **Gluing.** Each element of $X^{(k)}$ is the endpoint of exactly one chain from the upper half and exactly one chain from the lower half. Concatenating these two chains at their shared element of $X^{(k)}$ produces a single chain in $\mathcal{P}(X)$. This yields a partition of $\mathcal{P}(X)$ into exactly $\binom{n}{k} = \binom{n}{\lfloor n/2 \rfloor}$ chains. [guided] The strategy is to build chains that run vertically through $\mathcal{P}(X)$, with each chain passing through the middle level $X^{(k)}$ where $k = \lfloor n/2 \rfloor$. We construct the chains in two halves — above and below the middle — and then glue them together. **Upper half construction.** For each pair of consecutive levels $X^{(i)}$ and $X^{(i+1)}$ with $i \geq k$, we need an injection from the higher level down to the lower level. Since $i \geq k = \lfloor n/2 \rfloor$, the level $X^{(i)}$ is at least as close to the centre as $X^{(i+1)}$, so $|X^{(i)}| \geq |X^{(i+1)}|$. The [layer injection theorem](/theorems/2584), part (ii), with $r = i$ and $s = i + 1$ (so that $|n/2 - r| \leq |n/2 - s|$) provides an injection $g_i: X^{(i+1)} \to X^{(i)}$ with $g_i(A) \subseteq A$. How do these injections assemble into chains? Start with any element $A \in X^{(i+1)}$ for some $i \geq k$. The injection $g_i$ sends $A$ down to $g_i(A) \in X^{(i)}$. Then $g_{i-1}$ sends $g_i(A)$ down to $X^{(i-1)}$, and so on until we reach $X^{(k)}$. The resulting sequence is a chain (each set contains the next as a subset) from level $i+1$ down to level $k$. The injectivity of each $g_i$ ensures that different elements at the same level are sent to different elements one level down, so no two chains share an element. Since $g_k: X^{(k+1)} \to X^{(k)}$ is injective and $|X^{(k)}| \geq |X^{(k+1)}|$, the image of $g_k$ covers at most $|X^{(k+1)}|$ elements of $X^{(k)}$. Elements of $X^{(k)}$ not in any chain coming from above form singleton chains. In total, we get $|X^{(k)}| = \binom{n}{k}$ chains partitioning $X^{(\geq k)}$. **Lower half construction.** Symmetrically, for $1 \leq i \leq k$, the level $X^{(i-1)}$ is at least as far from the centre as $X^{(i)}$, so $|X^{(i-1)}| \leq |X^{(i)}|$. Part (i) of the [layer injection theorem](/theorems/2584) gives an injection $f_i: X^{(i-1)} \to X^{(i)}$ with $A \subseteq f_i(A)$. These injections push elements upward from $X^{(0)}$ toward $X^{(k)}$, producing $\binom{n}{k}$ chains that partition $X^{(\leq k)}$. **Gluing at the middle level.** Each chain in the upper partition contains exactly one element of $X^{(k)}$, and each chain in the lower partition contains exactly one element of $X^{(k)}$. Moreover, both partitions use all $\binom{n}{k}$ elements of $X^{(k)}$. Matching chains by their shared element of $X^{(k)}$ and concatenating gives $\binom{n}{k}$ chains that together partition all of $\mathcal{P}(X)$. [/guided] [/step] [step:Conclude the upper bound] By the chain partition constructed above, $\mathcal{P}(X)$ is the disjoint union of $\binom{n}{\lfloor n/2 \rfloor}$ chains. By the antichain-chain intersection bound, any antichain $\mathcal{A}$ satisfies \begin{align*} |\mathcal{A}| \leq \binom{n}{\lfloor n/2 \rfloor}. \end{align*} Combined with the lower bound from the first step, the maximum antichain size is exactly $\binom{n}{\lfloor n/2 \rfloor}$, achieved by $X^{(\lfloor n/2 \rfloor)}$. [/step]