# Thread: Mobius Function for 2^X

1. ## Mobius Function for 2^X

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

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

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

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

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

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

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

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

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

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

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

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

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

EDIT: I've had a recent breakthrough (I think) and I believe that $\mu(Y)=(-1)^{|Y|}$. So unless that's incorrect, I don't need any help.
Should that be $F(Y) = \sum_{Z\subset Y}f(Z)$ for any $Y\subset X\Longleftrightarrow f(Y)=\sum_{Z\subset Y}\mu(Z, Y)F(Z)$, where $\mu(Z, Y)= (-1)^{l(Z,Y)}$ and $l(Z, Y)$ is the length of the interval of [Z,Y] in the order-theoretic term?

In your example, I think if F(Y) is the cardinality of the set Y = {a, b}, then we can apply
$\mu (\emptyset, \{a,b\})= (-1)^2=1$, $\mu(\emptyset, \{a\}) = -1$, $\mu(Y, Y) = 1$, etc.