[proofplan] We prove first that deleting one rank from a finite Cohen-Macaulay graded poset preserves Cohen-Macaulayness. On each open interval, this deletion is realized on the order complex by deleting the vertices of one fixed rank. The link of each deleted vertex is the join of the lower and upper interval complexes, hence is Cohen-Macaulay of the correct dimension; the long exact sequence of the pair then shows that the deletion creates no reduced homology below the new top dimension. Repeating the one-rank deletion finitely many times gives the result for an arbitrary rank set $S$. [/proofplan]

[step:Fix the rank-selection convention and reduce to one deleted rank] Let $\rho:P\to \{0,1,\dots,r\}$ be the given rank function, and let $\hat{0}$ and $\hat{1}$ denote the minimum and maximum elements of $P$. For $S\subseteq \{0,1,\dots,r\}$, define the rank-selected subposet \begin{align*} P_S=\{x\in P:\rho(x)\in S\}, \end{align*} with the order inherited from $P$. When $P_S$ is nonempty, its rank function is the order-preserving relabelling of the selected ranks in increasing order. It is enough to prove the following one-rank deletion statement: [claim:Deleting one rank preserves Cohen-Macaulayness] Let $Q$ be a finite bounded Cohen-Macaulay graded poset over $k$ with minimum element $\hat{0}_Q$, maximum element $\hat{1}_Q$, and rank function $\rho_Q:Q\to \{0,1,\dots,m\}$. If $j\in \{1,\dots,m-1\}$ and \begin{align*} T=\{0,1,\dots,m\}\setminus\{j\}, \end{align*} then the rank-selected subposet \begin{align*} Q_T=\{x\in Q:\rho_Q(x)\in T\}, \end{align*} with the inherited order and the relabelled grading on the selected ranks, is Cohen-Macaulay over $k$ in the finite graded poset convention: its whole order complex and the order complexes of all its open intervals are Cohen-Macaulay over $k$, with the void-complex convention in dimension $-1$. [/claim] [proof] The proof occupies the next steps. [/proof] Indeed, given $S\subseteq \{0,1,\dots,r\}$, define the omitted interior-rank set \begin{align*} J_{\mathrm{int}}=(\{1,\dots,r-1\}\setminus S). \end{align*} Starting from $P$, delete the ranks in $J_{\mathrm{int}}$ one at a time. After some interior-rank deletions, the remaining original ranks form a subset $R\subseteq \{0,1,\dots,r\}$ that still contains $0$ and $r$, so the intermediate rank-selected poset is still bounded, with inherited minimum $\hat{0}$ and maximum $\hat{1}$. Its induced rank function is the order-preserving bijection from $R$ to $\{0,1,\dots,|R|-1\}$. To delete the next omitted original rank $j\in J_{\mathrm{int}}\cap R$, apply the one-rank deletion statement to the relabelled rank corresponding to $j$ under this bijection. This relabelled rank is neither the minimum rank nor the maximum rank, so the next intermediate poset remains bounded. Since $P$ is Cohen-Macaulay over $k$, induction gives a bounded Cohen-Macaulay rank-selected subposet $P_R$ whose original rank set is \begin{align*} R=S\cup(\{0,r\}\setminus S). \end{align*} It remains only to remove endpoint ranks if $0\notin S$ or $r\notin S$. If both endpoint ranks remain in $S$, then $P_R=P_S$. If one or both endpoint ranks are omitted, then $P_S$ is obtained from the bounded Cohen-Macaulay poset $P_R$ by taking either an upper open interval, a lower open interval, or the proper part $(\hat{0},\hat{1})_{P_R}$. The whole order complex of that unbounded endpoint-deleted poset is therefore one of the interval complexes already included in the Cohen-Macaulay condition for $P_R$, and its open intervals are also open intervals in $P_R$. Hence $P_S$ is Cohen-Macaulay over $k$ under the finite graded poset convention stated in the theorem. If $S=\varnothing$, then $P_S$ is the empty poset. Its order complex is the dimension-$-1$ complex specified in the theorem statement, and it has no nonempty open intervals. If $S$ has one element, the whole order complex of $P_S$ is a finite discrete pure $0$-dimensional complex and every open interval in $P_S$ is empty, so the Cohen-Macaulay condition holds by the same dimension-$-1$ convention for the interval complexes. [/step]

[step:Translate one-rank deletion to deletion of vertices in an interval complex] Let $Q$ be as in the claim, so $j\in\{1,\dots,m-1\}$, and set \begin{align*} T=\{0,1,\dots,m\}\setminus \{j\}. \end{align*} Define the rank-selected subposet \begin{align*} Q_T=\{x\in Q:\rho_Q(x)\in T\}, \end{align*} ordered by the order inherited from $Q$. Define the rank-difference function \begin{align*} \ell:\{(a,b)\in Q_T\times Q_T:a<b\}\to \mathbb{N} \end{align*} by \begin{align*} \ell(a,b)=\rho_Q(b)-\rho_Q(a). \end{align*} We prove the stronger interval statement by induction on $\ell(a,b)$ for comparable elements $a<b$ of $Q_T$: the order complex of every open interval $(a,b)_{Q_T}$ is Cohen-Macaulay over $k$. The cases $\ell(a,b)\le 1$ have empty open interval and are covered by the void-complex convention. Assume the statement is already known for all smaller rank differences, and fix $a,b\in Q_T$ with $a<b$. Define the open interval in $Q$ by \begin{align*} I=(a,b)_Q=\{x\in Q:a<x<b\}, \end{align*} and define the rank-selected open interval by \begin{align*} I_T=I\cap Q_T. \end{align*} The open interval $(a,b)_{Q_T}$ is exactly $I_T$, with the inherited order. For a finite poset $R$, write $\triangle(R)$ for its order complex, whose vertices are the elements of $R$ and whose simplices are finite chains in $R$. For a simplicial complex $K$, write $\widetilde H_i(K;k)$ for its $i$th reduced simplicial homology group with coefficients in $k$. Since $Q$ is Cohen-Macaulay over $k$, every interval of $Q$ is Cohen-Macaulay over $k$ by [citetheorem:8133]. Hence the order complex \begin{align*} \Delta=\triangle(I) \end{align*} is Cohen-Macaulay over $k$. The complex \begin{align*} \Delta_T=\triangle(I_T) \end{align*} is obtained from $\Delta$ by deleting precisely the vertices \begin{align*} V_j=\{v\in I:\rho_Q(v)=j\}. \end{align*} If $V_j=\varnothing$, then $\Delta_T=\Delta$, so the interval is Cohen-Macaulay. We therefore assume $V_j\ne \varnothing$. [/step]

[step:Identify the link of each deleted rank vertex]Fix a vertex $v\in V_j$. Define the lower and upper open intervals around $v$ inside $I$ by $I_{<v}=\{x\in Q:a<x<v\}$ and $I_{>v}=\{x\in Q:v<x<b\}$. The link of $v$ in $\Delta=\triangle(I)$ is $\operatorname{lk}_{\Delta}(v)=\triangle(I_{<v})*\triangle(I_{>v})$. This is the link formula for a vertex in an order complex, applied to the finite poset $I$ and the vertex $v$; it is exactly [citetheorem:8123]. Because $Q$ is Cohen-Macaulay and intervals of a Cohen-Macaulay poset are Cohen-Macaulay by [citetheorem:8133], the posets $I_{<v}$ and $I_{>v}$ have Cohen-Macaulay order complexes over $k$. The standard join theorem for Cohen-Macaulay complexes says that the join of two Cohen-Macaulay complexes over $k$ is Cohen-Macaulay over $k$, with dimension equal to the sum of the dimensions plus $1$, using the usual convention for a void factor. Therefore $\operatorname{lk}_{\Delta}(v)$ is Cohen-Macaulay over $k$.[/step]

[guided]We need to understand exactly what local piece of $\Delta$ is removed when the vertex $v$ is deleted. A simplex of $\Delta=\triangle(I)$ is a finite chain in $I$. Such a chain contains $v$ precisely when all elements below $v$ in the chain lie in \begin{align*} I_{<v}=\{x\in Q:a<x<v\}, \end{align*} and all elements above $v$ in the chain lie in \begin{align*} I_{>v}=\{x\in Q:v<x<b\}. \end{align*} Thus a simplex in the link of $v$ is exactly the union of a chain below $v$ and a chain above $v$. This is the simplicial join description \begin{align*} \operatorname{lk}_{\Delta}(v)=\triangle(I_{<v})*\triangle(I_{>v}), \end{align*} which is [citetheorem:8123] applied to the finite poset $I$. Now we verify the homological input. The lower interval $I_{<v}$ and the upper interval $I_{>v}$ are intervals in $Q$. Since $Q$ is Cohen-Macaulay over $k$, intervals in $Q$ are Cohen-Macaulay over $k$ by [citetheorem:8133]. Hence $\triangle(I_{<v})$ and $\triangle(I_{>v})$ are Cohen-Macaulay complexes over $k$. The standard join theorem for Cohen-Macaulay complexes says that the join of Cohen-Macaulay complexes over $k$ is again Cohen-Macaulay over $k$, with the expected dimension shift and with the usual convention for a void factor. Applying that theorem to the two complexes above gives that \begin{align*} \operatorname{lk}_{\Delta}(v) \end{align*} is Cohen-Macaulay over $k$.[/guided]

[step:Delete the vertices of the omitted rank without creating low-dimensional homology]Enumerate the finite set $V_j$ as \begin{align*} V_j=\{v_1,\dots,v_N\}. \end{align*} For $0\le q\le N$, define the induced subcomplex \begin{align*} \Delta_q=\Delta-\{v_1,\dots,v_q\}, \end{align*} where deleting vertices means taking the induced subcomplex on the remaining vertices. Thus $\Delta_0=\Delta$ and $\Delta_N=\Delta_T$. Vertices of the same rank in a graded poset are incomparable. Therefore no chain in $I$ contains two distinct vertices of $V_j$. For each $q\in\{1,\dots,N\}$, every simplex of the closed star of $v_q$ in $\Delta_{q-1}$ contains no previously deleted vertex, and its boundary inside $\Delta_q$ is exactly the link of $v_q$. Hence \begin{align*} \operatorname{lk}_{\Delta_{q-1}}(v_q)=\operatorname{lk}_{\Delta}(v_q), \end{align*} and the pair $(\Delta_{q-1},\Delta_q)$ is the cone on $\operatorname{lk}_{\Delta}(v_q)$ modulo its base. Let $d=\dim \Delta$. Since $V_j\ne\varnothing$, the rank $j$ lies strictly between $\rho_Q(a)$ and $\rho_Q(b)$. Every maximal chain in the graded interval $[a,b]_Q$ contains exactly one element of each rank between $\rho_Q(a)$ and $\rho_Q(b)$ by the [Jordan-Dedekind chain condition](/theorems/8087) [citetheorem:8087]. Therefore every maximal chain in $I$ contains exactly one rank-$j$ element. Conversely, a maximal chain in $I_T$ can be extended inside $[a,b]_Q$ by inserting the missing rank-$j$ element between the consecutive selected ranks that surround $j$; gradedness and maximality force the extended chain to have one element in each intermediate rank. Thus every maximal chain in $I_T$ has exactly one fewer vertex than a maximal chain in $I$, and $\Delta_T$ is pure of dimension $d-1$. For each $q\in\{1,\dots,N\}$, the star-link relative homology calculation for the cone pair gives \begin{align*} \widetilde H_s(\Delta_{q-1},\Delta_q;k)\cong \widetilde H_{s-1}(\operatorname{lk}_{\Delta}(v_q);k) \end{align*} for every integer $s$. The link $\operatorname{lk}_{\Delta}(v_q)$ is Cohen-Macaulay over $k$ and has dimension $d-1$. Indeed, because $j$ lies strictly between $\rho_Q(a)$ and $\rho_Q(b)$, every maximal chain in $[a,b]_Q$ contains $v_q$ precisely when it is obtained by concatenating a maximal chain in $[a,v_q]_Q$ and a maximal chain in $[v_q,b]_Q$; after deleting $a$, $v_q$, and $b$, this gives a maximal simplex of $\operatorname{lk}_{\Delta}(v_q)$ with exactly one fewer vertex than a maximal simplex of $\Delta$. This remains valid when one of the lower or upper open intervals is empty, using the dimension-$-1$ convention for the corresponding join factor. Therefore \begin{align*} \widetilde H_{s-1}(\operatorname{lk}_{\Delta}(v_q);k)=0 \end{align*} whenever $s-1<d-1$, equivalently whenever $s<d$. We prove by induction on $q$ that \begin{align*} \widetilde H_i(\Delta_q;k)=0 \end{align*} for every $i<d-1$. The case $q=0$ follows because $\Delta$ is Cohen-Macaulay of dimension $d$, so $\widetilde H_i(\Delta;k)=0$ for every $i<d$, hence for every $i<d-1$. Assume the assertion holds for $q-1$, and fix $i<d-1$. The long exact sequence for the reduced simplicial homology of the pair $(\Delta_{q-1},\Delta_q)$ contains \begin{align*} \widetilde H_{i+1}(\Delta_{q-1},\Delta_q;k)\to \widetilde H_i(\Delta_q;k)\to \widetilde H_i(\Delta_{q-1};k). \end{align*} Here $i+1<d$, so the relative group on the left vanishes by the preceding star-link calculation, and the group on the right vanishes by the induction hypothesis. Exactness gives $\widetilde H_i(\Delta_q;k)=0$. Taking $q=N$ gives \begin{align*} \widetilde H_i(\Delta_T;k)=0 \end{align*} for every $i<\dim\Delta_T$.[/step]

[guided]The delicate point is the degree range in the long exact sequence. After deleting one rank, the target complex has dimension $d-1$, so we only need vanishing below $d-1$, not below $d$. This is why we fix $i<d-1$ throughout the induction on the deleted vertices. First we verify the dimension statement. Since $j$ lies strictly between $\rho_Q(a)$ and $\rho_Q(b)$, the Jordan-Dedekind chain condition [citetheorem:8087] implies that every maximal chain from $a$ to $b$ in $Q$ has one element in each intermediate rank, hence one element of rank $j$. Removing that one rank-$j$ element from such a chain gives a maximal chain in $I_T$, and inserting the unique missing rank position into a maximal chain in $I_T$ gives a maximal chain in $I$. Thus every maximal simplex of $\Delta_T$ has exactly one fewer vertex than a maximal simplex of $\Delta$, so $\Delta_T$ is pure of dimension $d-1$. For a fixed deleted vertex $v_q$, same-rank vertices are incomparable, so a chain containing $v_q$ cannot contain any other vertex of $V_j$. Therefore previous deletions do not change the local star of $v_q$: the link of $v_q$ in $\Delta_{q-1}$ is still \begin{align*} \operatorname{lk}_{\Delta}(v_q). \end{align*} The part removed when passing from $\Delta_{q-1}$ to $\Delta_q$ is the closed star of $v_q$, which is the cone on this link with cone point $v_q$. Its intersection with $\Delta_q$ is exactly the base of the cone, namely the link. Hence the relative pair is the cone on $\operatorname{lk}_{\Delta}(v_q)$ modulo its base, and reduced relative homology gives \begin{align*} \widetilde H_s(\Delta_{q-1},\Delta_q;k)\cong \widetilde H_{s-1}(\operatorname{lk}_{\Delta}(v_q);k) \end{align*} for every integer $s$. The link $\operatorname{lk}_{\Delta}(v_q)$ is Cohen-Macaulay over $k$ of dimension $d-1$: its lower and upper interval factors are Cohen-Macaulay, their join is Cohen-Macaulay, and the dimension count follows from the same maximal-chain calculation, with the dimension-$-1$ convention when a factor is empty. Therefore its reduced homology vanishes below degree $d-1$, which means the relative group above vanishes for $s<d$. Now choose $i<d-1$. The relevant segment of the long exact sequence is \begin{align*} \widetilde H_{i+1}(\Delta_{q-1},\Delta_q;k)\to \widetilde H_i(\Delta_q;k)\to \widetilde H_i(\Delta_{q-1};k). \end{align*} The left group is zero because $i+1<d$, and the right group is zero by the induction hypothesis on $q$. Exactness then forces $\widetilde H_i(\Delta_q;k)=0$. Iterating this from $q=0$ to $q=N$ proves the required reduced homology vanishing for $\Delta_T$ below its top dimension $d-1$.[/guided]

[step:Verify all links in the rank-selected interval are Cohen-Macaulay] To prove that $\Delta_T=\triangle(I_T)$ is Cohen-Macaulay over $k$, we must also verify reduced-homology vanishing for links of faces. Let $F$ be a face of $\Delta_T$, so $F$ is a finite chain in $I_T$. If \begin{align*} F=\{x_1<\cdots<x_s\}, \end{align*} set $x_0=a$ and $x_{s+1}=b$. For each $h\in\{0,\dots,s\}$, the gap \begin{align*} (x_h,x_{h+1})_{Q_T}=\{y\in Q_T:x_h<y<x_{h+1}\} \end{align*} is the rank-selected subposet of the open interval $(x_h,x_{h+1})_Q$ with the same selected rank set $T$. Its rank difference satisfies \begin{align*} \rho_Q(x_{h+1})-\rho_Q(x_h)<\rho_Q(b)-\rho_Q(a)=\ell(a,b), \end{align*} because the chain $F$ inserts at least one element between $a$ and $b$ unless the corresponding gap is the whole interval, and in the latter case $F$ is empty and there is no link condition beyond the homology vanishing already proved for $\Delta_T$. For nonempty $F$, the link of $F$ in the order complex decomposes as \begin{align*} \operatorname{lk}_{\Delta_T}(F)=\triangle((x_0,x_1)_{Q_T})*\triangle((x_1,x_2)_{Q_T})*\cdots*\triangle((x_s,x_{s+1})_{Q_T}), \end{align*} where empty gap factors are interpreted using the void-complex convention. By the induction hypothesis on the rank difference, each gap factor is Cohen-Macaulay over $k$. Applying the standard join theorem for Cohen-Macaulay complexes, with the same void-factor convention, gives that $\operatorname{lk}_{\Delta_T}(F)$ is Cohen-Macaulay over $k$. In particular its reduced homology over $k$ vanishes below its top dimension. The previous step proved the reduced homology vanishing for $\Delta_T$ itself, which is the face-link condition for the empty face. Thus every face link in $\Delta_T$ has the required reduced homology vanishing, and $\Delta_T$ is pure of dimension $d-1$. Therefore $\Delta_T$ is Cohen-Macaulay over $k$. [/step]

[step:Conclude the one-rank deletion and iterate over all omitted ranks] The preceding steps show that for every pair $a<b$ in $Q_T$, the order complex \begin{align*} \triangle((a,b)_{Q_T}) \end{align*} is Cohen-Macaulay over $k$. Since the repaired one-rank deletion claim has $j\in\{1,\dots,m-1\}$, both $\hat{0}_Q$ and $\hat{1}_Q$ belong to $Q_T$. Thus $Q_T$ is bounded, and the proper part of $Q_T$ is the interval $(\hat{0}_Q,\hat{1}_Q)_{Q_T}$ already handled. The whole order complex $\triangle(Q_T)$ is a cone with apex $\hat{0}_Q$, so it is Cohen-Macaulay over $k$ exactly when the links of its faces are Cohen-Macaulay; these links are either cones again or joins of the interval complexes already proved Cohen-Macaulay. Therefore the whole order complex of $Q_T$ is Cohen-Macaulay over $k$, and the open-interval checks have already been proved. This proves the one-rank deletion claim under the finite graded poset convention stated in the theorem. The initial reduction has already applied this claim only to the finitely many omitted interior ranks, keeping both endpoint ranks present at every intermediate stage. After those deletions, any endpoint ranks not belonging to $S$ are removed by the separate endpoint argument in the first step: deleting only $\hat{0}$, only $\hat{1}$, or both endpoints turns the relevant whole order complex into a cone over, or the proper part of, an interval complex already known to be Cohen-Macaulay, and all open intervals remain open intervals of the bounded Cohen-Macaulay intermediate poset. Hence the rank-selected subposet \begin{align*} P_S=\{x\in P:\rho(x)\in S\} \end{align*} is Cohen-Macaulay over $k$ for every subset $S\subseteq \{0,1,\dots,r\}$. This is the Baclawski rank-selection theorem. [/step]