reading in "Introductory Combinatorics" about Möbius inversion, some questions have arose:
1) Author defines function $$F(K) - \text{# of elements of S that belong to $exactly$ those sets } A_i\text{with i } \notin K $$
Where $A_i - \text{ some subset of finite set S} $
Note that $K\subseteq{\{1,2,..,n}\}$ and subsets are partially ordered by containment
So element in $S$ is counted iff it belongs to all of the sets $A_{i\notin K}$ and doesn't belong to every set $A_{i\in K}$
Author further defines function $$G(K) = \sum_{L \subseteq K}F(L) $$ And says it's equivalent to $$G(K) = \mid{\cap_{i\notin{K}}A_i}\mid$$
I would like to know how that is the case, as plugging in some subsets into the formula didn't produce the expected outcome. Namely if $S = \{1,2,3,4,5\} $ and $A_1 = \{\emptyset\}, A_i = \{1,2\}, 2 \le {i} \le{5} $ , $K = \{1,3,5\}$
I expect the first formula to produce 1, as the only valid subset that seem to satisfy the conditions is ${ \{1\}}$ Though the second formula produces $2$ as an answer
2) [SOLVED] We define a function that takes in two elements from our partially ordered set and outputs a real number $f$ such that $f(y,y)\ne 0$ for all $y \text{ in } X$
We then define a function $$g(y,y) = \frac{1}{f(y,y)}, y\in X$$
and then letting $$g(x,y) = -\frac{1}{f(y,y)}\sum_{\{z: x\le z \lt y \}}g(x,y)f(z,y), (x\lt y)$$
Now, from the equation above we can get that $$\sum_{\{z:x\le z \lt y\}}g(x,z)f(z,y) = \delta(x,y), (x\le y) $$
Which would seem to me to be correct only if $f(x,y)$ is $0$ for all $(x,y), x\ne y$ how to see that it holds for all pairs of $x,y$ ?
EDIT: Answer to second question was found here - Question about inverse with respect to convolution product.
It would seem that I haven't paid enough attention to set theory, any feedback is welcome, thanks :)
Your equation $A_1=\{\emptyset\}$ is unclear, since $A_1$ is supposed to be a subset of $S$ but $S$ doesn't contain $\emptyset$. I'll take it that you meant $A_1=\emptyset$.
The two expressions for $G(K)$ are indeed equivalent. In your example, there are $8$ subsets $L$ of $K$. We have $F(\{1\})=2$ and $F(S)=3$, and all other values of $F$ are $0$ since all elements belong either to only $A_2$ through $A_5$ (namely $1$ and $2$) or to no subset at all (namely $3$ through $5$). Thus
$$ F(K)=F(\emptyset)+F(\{1\})+F(\{3\})+F(\{5\})+F(\{1,3\})+F(\{3,5\})+F(\{5,1\})+F(\{1,3,5\})\\=0+2+0+0+0+0+0+0\\=2\;. $$