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**redsoxfan325** We have to find the Mobius function $\displaystyle \mu$ for the set $\displaystyle 2^X$ such that $\displaystyle F(Y) = \sum_{Z\subset Y}f(Z)$ for any $\displaystyle Y\subset X\Longleftrightarrow f(Y)=\sum_{Z\subset Y}\mu(Z)F(Y\backslash Z)$.

I started doing this for a few basic sets. If $\displaystyle X=\{a,b\}$, then

$\displaystyle \mu(\emptyset) = 1$

$\displaystyle \mu(\{a\}) = \mu(\{b\}) = -1$

$\displaystyle \mu(\{a,b\}) = 1$

I know the $\displaystyle \mu$-function for integers is

$\displaystyle \mu(n)=\bigg\{\begin{array}{lr}0:&\mbox{n~is~not~s quarefree}\\(-1)^{p(n)}:&\mbox{n~is~squarefree}\end{array}$

where $\displaystyle p(n)$ is the number of prime factors of $\displaystyle n$.

EDIT: I've had a recent breakthrough (I think) and I believe that $\displaystyle \mu(Y)=(-1)^{|Y|}$. So unless that's incorrect, I don't need any help.