[proofplan] The diagonal case is exactly the defining normalization of the Möbius function. For $x<y$, we expand the reduced Euler characteristic of the order complex as an alternating sum over chains in the open interval. We then prove, directly from the Möbius recurrence, that the Möbius function is given by the identical alternating sum over endpoint-fixed chains from $x$ to $y$. The two sums match term by term, including the empty-chain contribution when $(x,y)$ has no elements. [/proofplan]

[step:Handle the diagonal case from the defining normalization] By the defining normalization of the Möbius function of a finite poset, as stated in [citetheorem:8139], \begin{align*} \mu_P(u,u)=1 \end{align*} for every $u\in P$. Taking $u=x$ gives \begin{align*} \mu_P(x,x)=1. \end{align*} This proves the asserted formula in the case $x=y$. [/step]

[step:Expand the reduced Euler characteristic as an alternating sum over interior chains]Assume from now on that $x<y$. Define, for every integer $r\ge 0$, the finite set \begin{align*} \mathcal C_r(x,y):=\{(z_0,\dots,z_{r+1})\in P^{r+2}:x=z_0<z_1<\cdots<z_r<z_{r+1}=y\}. \end{align*} Thus $\mathcal C_0(x,y)$ consists of the single endpoint-fixed chain $x<y$ with no interior elements. A nonempty simplex of $\Delta((x,y))$ with $r$ vertices is exactly a strict chain \begin{align*} z_1<\cdots<z_r \end{align*} in $(x,y)$, and this is equivalent to an element \begin{align*} x=z_0<z_1<\cdots<z_r<z_{r+1}=y \end{align*} of $\mathcal C_r(x,y)$. Such a simplex has dimension $r-1$. The empty face of the order complex contributes the reduced Euler characteristic term $-1$, and it corresponds to the unique element of $\mathcal C_0(x,y)$. Therefore \begin{align*}\widetilde{\chi}\bigl(\Delta((x,y))\bigr)=\sum_{r\ge 0}(-1)^{r+1}|\mathcal C_r(x,y)|.\end{align*} The sum is finite because $P$ is finite.[/step]

[guided]We now translate topology into combinatorics. The order complex $\Delta((x,y))$ is the abstract simplicial complex whose nonempty simplices are finite strict chains in the open interval \begin{align*} (x,y)=\{z\in P:x<z<y\}. \end{align*} If such a chain has $r$ elements, \begin{align*} z_1<\cdots<z_r, \end{align*} then it is a simplex of dimension $r-1$, because a simplex with $r$ vertices has dimension $r-1$. For comparison with the Möbius function, we add the fixed endpoints $x$ and $y$. This gives a bijection between $r$-element chains in $(x,y)$ and endpoint-fixed chains \begin{align*} x=z_0<z_1<\cdots<z_r<z_{r+1}=y. \end{align*} The set of all such endpoint-fixed chains is denoted \begin{align*} \mathcal C_r(x,y):=\{(z_0,\dots,z_{r+1})\in P^{r+2}:x=z_0<z_1<\cdots<z_r<z_{r+1}=y\}. \end{align*} The reduced Euler characteristic includes the empty face with sign $-1$. This convention is essential when $(x,y)$ is empty, because then the order complex has no nonempty simplices and its reduced Euler characteristic is $-1$. The empty face corresponds exactly to the case $r=0$, namely the endpoint-fixed chain \begin{align*} x=z_0<z_{1}=y. \end{align*} For $r\ge 1$, each $r$-element chain contributes sign \begin{align*} (-1)^{r-1}. \end{align*} Since $(-1)^{r-1}=(-1)^{r+1}$, the same sign formula also matches the $r=0$ contribution: \begin{align*} (-1)^{0+1}=-1. \end{align*} Hence the reduced Euler characteristic is \begin{align*} \widetilde{\chi}\bigl(\Delta((x,y))\bigr) = \sum_{r\ge 0}(-1)^{r+1}|\mathcal C_r(x,y)|. \end{align*} The finiteness of $P$ ensures that only finitely many chains occur, so this alternating sum is finite.[/guided]

[step:Derive the same chain formula from the Möbius recurrence]Define a function \begin{align*} M:\{(a,b)\in P\times P:a\le b\}\to \mathbb Z \end{align*} by \begin{align*} M(a,a)=1 \end{align*} and, for $a<b$, \begin{align*} M(a,b)=\sum_{r\ge 0}(-1)^{r+1}|\mathcal C_r(a,b)|, \end{align*} where \begin{align*} \mathcal C_r(a,b):=\{(w_0,\dots,w_{r+1})\in P^{r+2}:a=w_0<w_1<\cdots<w_r<w_{r+1}=b\}. \end{align*} Because $P$ is finite, every displayed sum is finite. We verify the recurrence in [citetheorem:8139]. Fix $a,b\in P$ with $a<b$. Write \begin{align*} c_s(a,b):=|\mathcal C_s(a,b)| \end{align*} for each integer $s\ge 0$. Since $a<b$, the set $\mathcal C_0(a,b)$ consists of the single chain $a<b$, so $c_0(a,b)=1$. Expanding the left recurrence gives \begin{align*} \sum_{a\le z\le b}M(a,z)=1+\sum_{a<z<b}M(a,z)+M(a,b). \end{align*} The term $M(a,b)$ is \begin{align*} M(a,b)=\sum_{s\ge 0}(-1)^{s+1}c_s(a,b). \end{align*} It remains to reindex the middle sum. For $a<z<b$ and $r\ge 0$, appending $b$ to a chain in $\mathcal C_r(a,z)$ produces a unique chain in $\mathcal C_{r+1}(a,b)$ whose final interior element is $z$. Conversely, every chain in $\mathcal C_s(a,b)$ with $s\ge 1$ has a unique final interior element $z=w_s$, and deleting the terminal $b$ gives an element of $\mathcal C_{s-1}(a,z)$. Therefore \begin{align*} \sum_{a<z<b}M(a,z)=\sum_{s\ge 1}(-1)^s c_s(a,b). \end{align*} Substituting the two expansions yields \begin{align*} \sum_{a\le z\le b}M(a,z)=1+\sum_{s\ge 1}(-1)^s c_s(a,b)+\sum_{s\ge 0}(-1)^{s+1}c_s(a,b). \end{align*} The $s=0$ term in the final sum is $-c_0(a,b)=-1$, and for each $s\ge 1$ the two coefficients $(-1)^s$ and $(-1)^{s+1}$ cancel. Hence \begin{align*} \sum_{a\le z\le b}M(a,z)=0. \end{align*} We have proved $M(a,a)=1$ and \begin{align*} \sum_{a\le z\le b}M(a,z)=0 \end{align*} for every $a<b$. By the uniqueness part of the [interval recursion for the Möbius function](/theorems/8139) in [citetheorem:8139], \begin{align*} M(a,b)=\mu_P(a,b) \end{align*} for all $a\le b$ in $P$. In particular, for $x<y$, \begin{align*} \mu_P(x,y)=\sum_{r\ge 0}(-1)^{r+1}|\mathcal C_r(x,y)|. \end{align*}[/step]

[guided]The point is to prove the Möbius recurrence by an exact cancellation, not by an informal statement about chains. We have defined \begin{align*} M:\{(a,b)\in P\times P:a\le b\}\to \mathbb Z \end{align*} with $M(a,a)=1$ and, for $a<b$, \begin{align*} M(a,b)=\sum_{r\ge 0}(-1)^{r+1}|\mathcal C_r(a,b)|. \end{align*} The theorem [citetheorem:8139] says that, for a finite poset, the Möbius function is uniquely determined by the diagonal condition and by the recurrence \begin{align*} \sum_{a\le z\le b}\mu_P(a,z)=0 \end{align*} whenever $a<b$. Thus it is enough to prove the same equation with $M$ in place of $\mu_P$. Fix $a,b\in P$ with $a<b$, and define \begin{align*} c_s(a,b):=|\mathcal C_s(a,b)| \end{align*} for every integer $s\ge 0$. The case $s=0$ counts endpoint-fixed chains with no interior elements. Since $a<b$, there is exactly one such chain, namely $a<b$, so \begin{align*} c_0(a,b)=1. \end{align*} Now expand the recurrence sum by separating the three possibilities $z=a$, $a<z<b$, and $z=b$: \begin{align*} \sum_{a\le z\le b}M(a,z)=1+\sum_{a<z<b}M(a,z)+M(a,b). \end{align*} The last term is already expressed in terms of chains from $a$ to $b$: \begin{align*} M(a,b)=\sum_{s\ge 0}(-1)^{s+1}c_s(a,b). \end{align*} We next rewrite the middle term in the same language. Suppose $a<z<b$ and choose a chain in $\mathcal C_r(a,z)$: \begin{align*} a=w_0<w_1<\cdots<w_r<w_{r+1}=z. \end{align*} Appending $b$ gives \begin{align*} a=w_0<w_1<\cdots<w_r<w_{r+1}=z<b, \end{align*} which is a chain from $a$ to $b$ with $r+1$ interior elements. Conversely, if a chain from $a$ to $b$ has $s\ge 1$ interior elements, \begin{align*} a=w_0<w_1<\cdots<w_s<w_{s+1}=b, \end{align*} then its final interior element is the unique element $z=w_s$ with $a<z<b$, and deleting the terminal $b$ gives a chain in $\mathcal C_{s-1}(a,z)$. This gives a bijective reindexing of all summands in the middle sum, with $s=r+1$. Therefore \begin{align*} \sum_{a<z<b}M(a,z)=\sum_{s\ge 1}(-1)^s c_s(a,b). \end{align*} The sign changes because the original sign is $(-1)^{r+1}$ and $r=s-1$, so $(-1)^{r+1}=(-1)^s$. Substituting the two chain expansions gives \begin{align*} \sum_{a\le z\le b}M(a,z)=1+\sum_{s\ge 1}(-1)^s c_s(a,b)+\sum_{s\ge 0}(-1)^{s+1}c_s(a,b). \end{align*} The $s=0$ term in the final sum equals $-c_0(a,b)=-1$, so it cancels the initial $1$. For each fixed $s\ge 1$, the chain-count $c_s(a,b)$ appears twice, once with coefficient $(-1)^s$ from the truncated chain ending at its final interior element and once with coefficient $(-1)^{s+1}$ from the full chain ending at $b$. These coefficients add to zero. Hence \begin{align*} \sum_{a\le z\le b}M(a,z)=0. \end{align*}[/guided]

[step:Identify the two alternating sums] From the Euler characteristic expansion, \begin{align*}\widetilde{\chi}\bigl(\Delta((x,y))\bigr)=\sum_{r\ge 0}(-1)^{r+1}|\mathcal C_r(x,y)|.\end{align*} From the Möbius chain formula just derived, \begin{align*}\mu_P(x,y)=\sum_{r\ge 0}(-1)^{r+1}|\mathcal C_r(x,y)|.\end{align*} The right-hand sides are identical finite sums. Therefore \begin{align*} \mu_P(x,y)=\widetilde{\chi}\bigl(\Delta((x,y))\bigr). \end{align*} Together with the diagonal case already proved, this establishes the theorem. [/step]