Corollary of Inclusion-Exclusion: Count elements in at most $k$ sets?

Question

I'm trying to prove the Corollary$1$ in a good answer about the Generalized Inclusion-Exclusion Principle, which I'm studying these days: https://math.stackexchange.com/a/362516/390226, for easy reading I use some of the definitions there. (Apologies for not mention this when I posted!)

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Let $\{S(i)\}_{i=1}^m$ be a collection of sets from a finite universe, and $A$ a subset of $\{1,2,\dots,m\}$, then $N(j)$ defined as \begin{align}N(j)=\sum_{|A|=j}\left|\bigcap_{i\in A} S(i)\right|\end{align}

What I want to get is the number of elements in at most $k$ of $S(i)$. My progress:
(Edit for note: The $\color{red}{\textrm{red}}$ part is what I did wrong!) \begin{align} \textrm{Atmost(k)}&=\sum_{l=0}^k\textrm{Exact(l)}\\ &=\sum_{l=0}^k\left[\sum_{j=l}^m(-1)^{j-l}\binom{j}{l}N(j)\right]\\ &=\sum_{j=0}^m(-1)^j\sum_{l=0}^{\color{red}{j+[j>k](-j-1)}}(-1)^l\binom{j}{l}N(j)\\ &=\sum_{j=0}^m(-1)^j\sum_{l=0}^{j}(-1)^l(-1)^{k-l}\binom{-1}{k-l}\binom{j}{l}N(j)\\ &=\sum_{j=0}^m(-1)^{j-k}\binom{j-1}{k}N(j)\tag{*}\\ &{\large=^?}\sum_{j=k+1}^m(-1)^{j-k}\binom{j-1}{k}N(j) \end{align}

The problem of $\textrm{(*)}$ is that it has to compute all $N(j), 0\le j\le m$, but the following line seems counter-intuitive since its first term is $$(-1)^1\binom{k}{k}N(k+1),$$

is there any wrong in my proof?

Answer 1

In the third equation of $\text{Atmost}(k)$, $$\require{cancel} \begin{align} &\sum_{l=0}^k\sum_{j=l}^m(-1)^{j-l}\binom{j}{l}N(j)\\ &=\sum_{j=0}^m\sum_{l=0}^{\color{#C00}{\min(j,k)}}(-1)^{j-l}\binom{j}{l}N(j)\tag1\\ &=\sum_{j=0}^k\cancelto{[j=0]}{\sum_{l=0}^j(-1)^{j-l}\binom{j}{l}}N(j)+\sum_{j=k+1}^m\color{#090}{\sum_{l=0}^k}(-1)^{j-k}\color{#090}{\binom{-1}{k-l}\binom{j}{l}}N(j)\tag2\\[3pt] &=N(0)+\sum_{j=k+1}^m(-1)^{j-k}\color{#090}{\binom{j-1}{k}}N(j)\tag3 \end{align} $$ Explanation:
$(1)$: change the order of the summation
$(2)$: in the left sum $j\le k$ so limit of the inner sum is $j$
$\phantom{\text{(2):}}$ in the left sum $j\gt k$ so limit of the inner sum is $k$
$(3)$: in the left sum, the inner sum is $(1-1)^j=[j=0]$
$\phantom{\text{(3):}}$ in the right sum, we apply Vandermonde's Identity

As is marked in red in $(1)$, the upper index is $\min(j,k)$. The upper index in your answer leaves out all terms for $j\gt k$, but the next line seems to ignore this, and, without justification, puts the upper index at $j$, which makes it identical to line $(5)$ below. From there, you get the correct answer. The next line then removes all terms for $j\le k$, which is almost okay, since $\binom{j-1}{k}=0$ for all $j$ except $j=0$. Since your answer leaves out the $j=0$ term, your answer is short by $N(0)$.

However, an easier approach might be $$ \begin{align} \sum_{l=0}^k\sum_{j=l}^m(-1)^{j-l}\binom{j}{l}N(j) &=\sum_{l=0}^m\sum_{j=l}^m(-1)^{j-k}\binom{-1}{k-l}\binom{j}{l}N(j)\tag4\\ &=\sum_{j=0}^m\sum_{l=0}^j(-1)^{j-k}\binom{-1}{k-l}\binom{j}{l}N(j)\tag5\\ &=\sum_{j=0}^m(-1)^{j-k}\binom{j-1}{k}N(j)\tag6 \end{align} $$ Explanation:
$(4)$: $(-1)^{j-k}\binom{-1}{k-l}=(-1)^{j-l}[l\le k]$ so we can extend the outer sum to $m$
$(5)$: change the order of summation (easier since we extended the outer sum)
$(6)$: apply Vandermonde's Identity

Note that $(6)$ is the same as $(3)$; in $(6)$, the coefficient of $N(0)$ is $1$, and for $1\le j\le k$ the coefficient of $N(j)$ is $0$.

Answer 2

If I’m not mistaken, your final result is actually the negative of $\operatorname{Atleast}(k+1)$; adding the size of the universe to it would give you $\operatorname{Atmost}(k)$. Since $N(0)$ is the size of the universe, your final result is missing an $N(0)$ term.

The problem seems to arise in going from the third line to the fourth. The third line can be written

$$\sum_{j=0}^m(-1)^j\sum_{\ell=0}^{\min\{j,k\}}(-1)^\ell\binom{j}\ell N(j)\;,$$

and painstakingly reversing the order of summation makes it

$$\sum_{j=0}^k(-1)^j\sum_{\ell=0}^j(-1)^\ell\binom{j}\ell N(j)+\color{red}{\sum_{j=k+1}^m(-1)^j\sum_{\ell=0}^k\binom{j}\ell N(j)}\,,$$

where the red term ultimately reduces to your final line.

The black term still has an $N(0)$ term when $j=0$, but it disappears when one rewrites $\sum_{\ell=0}^j(-1)^\ell\binom{j}\ell$ as $\binom{j-1}j$, which is equivalent to your manipulation. The problem is that here we’re taking the alternating sum across the entire $j$-th row of Pascal’s triangle, and while the sum is $0$ for all $j\ge 1$, for $j=0$ it’s $1$. Thus, the black term actually reduces to the missing $N(0)$, and your final result should be

$$N(0)+\sum_{j=k+1}^m(-1)^{j-k}\binom{j-1}k N(j)\,.$$