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Let $g(n,k)$ be the number of chains

$$ A_k \subset A_{k-1} \subset\dots\subset A_1 \subset A_0 $$

of $k$ proper subset inclusions, where $A_k\neq\emptyset$ and $A_0$ is a standard $n$-element set. Then

$$ \sum_{k\ge 0} (-1)^k g(n,k) = (-1)^{n-1}. $$

I can prove this by a fairly boring and unilluminating induction. (And also by a category-theoretic argument involving traces of geometric realizations in derivators, which is how I first noticed it.) Does it have a nicer combinatorial proof, by (say) bijections, generating functions, Mobius inversion, etc.?

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    $\begingroup$ This not an answer to the question as you posed it, but I suspect that your "category-theoretic argument involving traces of geometric realizations in derivators" reduces to a (probably) simpler homotopy theoretic proof. Namely, $g(n,k)$ is the number of $(k-1)$-simplices in the barycentric subdivision of the boundary of the standard $(n-1)$-simplex. Hence your sum (up to the $k=0$ summand and up to sign) computes the Euler characteristic of the $(n-2)$-sphere. $\endgroup$ – Karol Szumiło May 30 '14 at 17:43
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    $\begingroup$ Hall's formula for the Möbius function of a finite bounded ranked poset says that any such poset $P$ satisfies $\mu\left(P\right) = \sum_{k\geq 0} \left(-1\right)^k \left(\text{number of chains } 0 = x_0 < x_1 < \cdots < x_k = 1 \text{ in } P\right)$, where $0$ and $1$ denote the lower bound and the upper bound of $P$. In your case, the poset is the Boolean lattice, but notice that your $g\left(n,k\right)$ counts chains $0 = x_0 < x_1 < \cdots < x_{k+1} = 1$ rather than $0 = x_0 < x_1 < \cdots < x_k = 1$ so you are skipping the $k = 0$ addend of Hall's formula. $\endgroup$ – darij grinberg May 30 '14 at 17:55
  • $\begingroup$ (The $k = 0$ addend, of course, is only relevant for $n = 0$. It is more important that your sign is different because of the $k$ shift.) The Möbius function of the Boolean lattice is easily computed, as it is multiplicative and the Boolean lattice is a product of $2$-chains. $\endgroup$ – darij grinberg May 30 '14 at 17:59
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    $\begingroup$ @darijgrinberg, that looks like an answer; why don't you post it as one? $\endgroup$ – Mike Shulman May 31 '14 at 22:18
  • $\begingroup$ @KarolSzumiło, you're probably right. $\endgroup$ – Mike Shulman May 31 '14 at 23:35
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A chain $$ A_k \subset A_{k-1} \subset\dots\subset A_1 \subset A_0 $$ can be represented by the ordered partition $(B_1, B_2, \dots, B_{k+1})$ of the set $A_0=\{1, 2, \dots, n\}$ where $B_1=A_k$, $B_2=A_{k-1}-A_k$, $\dots,$ $B_{k+1}=A_0-A_1$.

First, a generating function proof. If we wanted to count such chains (or ordered partitions) without signs, the exponential generating function would be $\sum_{j=0}^\infty (e^x-1)^j = 1/(2-e^x)$ (http://oeis.org/A000670). With signs, the generating function is $$\sum_{j=0}^\infty (-1)^{j+1}(e^x-1)^j= -e^{-x} = \sum_{n=0}^\infty (-1)^{n+1} \frac{x^n}{n!}.$$

For a combinatorial proof using a sign reversing involution that changes the parity of the number of blocks, write the entries of each block in increasing order, with a bar between blocks, so $\{2,4,6\}\{1,3\}\{5\}$ would be written as $ 2\, 4\, 6 \,|\, 1\, 3 \,|\, 5 $. Find the first position, if there is one, where a number is followed by a larger number. If there is a bar there, remove it, and if there is no bar there then put one in. So our example would be mapped to $ 2 \,|\, 4 \,6 \,|\, 1 \,3 \,|\, 5 $ . As another example, $4 \,|\, 3 \,|\, 1 \,2\,|\,$ would map to $4 \,|\, 3\,|\, 1 \,|\, 2$ . The only ordered partitions not paired up are of the form $n \,|\, n-1\,|\, \cdots \,|\,2 \,|\, 1\,$ .

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  • $\begingroup$ Thank you! (As a helpful note to other readers, those are exponential generating functions.) $\endgroup$ – Mike Shulman Jun 1 '14 at 4:15
  • $\begingroup$ I added the word "exponential". $\endgroup$ – Ira Gessel Jun 1 '14 at 4:47
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There is quite direct proof from point of view of topological combinatorics.

Let $\Delta$ be the standard $(n-1)$-simplex with vertex set $A_0$. Considering the barycentric subdivision $\Delta'$ of $\Delta$, the vertices of $\Delta'$ are nonempty subsets of $A_0$ and faces of $\Delta'$ are subsets $\{A_k, \dots, A_1\}$ of vertices of $\Delta'$ such that $$ \emptyset \neq A_k \subset A_{k-1} \subset \cdots \subset A_1 \subseteq A_0. $$ (Note that the last inclusion needn't be proper.)

In this correspondence, your chains correspond to $(k-1)$-faces of $\Delta'$ which do not contain the barycentre $A_0$. These are the faces of the barycentric subdivision of the boundary of $\Delta$, which is topologically an $(n-2)$-sphere. There, if you were summing in your sum for $k \geq 1$, the value you obtain is the minus Euler characteristic of the $(n-2)$-sphere, which is $-(1 + (-1)^{n-2})$. Since you sum for $k \geq 0$, you get $(-1)^{n-1}$.

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  • $\begingroup$ Well, this answer now only explains the comment of Karol S. (I did not refresh the comments before posting an answer.) $\endgroup$ – Martin Tancer May 30 '14 at 18:06
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    $\begingroup$ That's okay; there is a distressing tendency on MO to post answers as comments, so I'm glad to have it here as an answer. $\endgroup$ – Mike Shulman May 31 '14 at 23:37
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This can indeed be seen as a consequence of the properties of the Möbius function of a poset. We recall two important properties of this function (see §7.2.1 in Vic Reiner's and my Hopf Algebras in Combinatorics (arXiv:1409.8356v5), which is the first reference that comes into my mind because it is currently open in my editor):

Property P1. If $P$ is a finite bounded poset, then $\mu\left(P\right) = \sum_{k\geq 0} \left(-1\right)^k \left(\text{number of chains } 0 = x_0 < x_1 < \cdots < x_k = 1 \text{ in } P\right)$.

Property P2. If $P$ and $Q$ are two finite bounded posets, then the Cartesian product $P \times Q$ (with componentwise order) satisfies $\mu\left(P \times Q\right) = \mu\left(P\right) \cdot \mu\left(Q\right)$.

Now, let's return to the question at hand. We assume that $n$ is positive, because otherwise your claim is only valid under a very creative interpretation of the $g\left(n,k\right)$ and the sum. Let $B_n$ be the poset of all subsets of $\left\{1,2,\ldots ,n\right\}$, ordered by inclusion. Then, $B_n$ is the $n$-fold Cartesian product $\underbrace{B_1 \times B_1 \times \cdots \times B_1}_{n \text{ times}}$; thus, by iterated application of Property P2, we obtain $\mu\left(B_n\right) = \mu\left(B_1\right)^n = \left(-1\right)^n$ (since $B_1$ is a $2$-element chain and thus has $\mu\left(B_1\right) = -1$). Hence,

$\left(-1\right)^n = \mu\left(B_n\right) $

$= \sum_{k\geq 0} \left(-1\right)^k \left(\text{number of chains } 0 = x_0 < x_1 < \cdots < x_k = 1 \text{ in } B_n\right)$ (by Property P1)

$= \sum_{k\geq 0} \left(-1\right)^k \left(\text{number of chains } \emptyset = A_0 \subsetneq A_1 \subsetneq \cdots \subsetneq A_k = \left\{1,2,\ldots,n\right\}\right)$

$= \sum_{k\geq 1} \left(-1\right)^k \underbrace{\left(\text{number of chains } \emptyset = A_0 \subsetneq A_1 \subsetneq \cdots \subsetneq A_k = \left\{1,2,\ldots,n\right\}\right)}_{\substack{=g\left(n,k-1\right) \\ \text{(not }g\left(n,k\right)\text{, since your chain does not start at }\emptyset\text{)}}}$ (we got rid of the $k = 0$ addend here, since this addend is $0$)

$= \sum_{k\geq 1} \left(-1\right)^k g\left(n,k-1\right) = \sum_{k\geq 0} \left(-1\right)^{k+1} g\left(n,k\right)$.

Dividing by $-1$, we transform this into

$\left(-1\right)^{n-1} = \sum_{k\geq 0} \left(-1\right)^k g\left(n,k\right)$,

qed.

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