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Let $I_n$ be a totally ordered set ${1<2<...<n}$. We need to calculate $\mu(m,n)$.

We shall use the recursive definition of $\mu$. We have the following: $$\mu(m,m) = 1$$ $$\mu(m,m+1) = -\sum_{m < y< m+1} \mu(m,y) = 0$$

Let by strong induction $\mu(m,m+1), \mu(m,m+2),..., \mu(m,m+k-1) =0$ Then:

$$\mu(m,m+k) = -\sum_{m < y< m+1} \mu(m,y) = -(\mu(m,m+1)+....\mu(m,m+k-1)=0$$

Thus we have: $\mu(m,n) = 1$ if $m=n$. Else, $\mu(m,n) = 0$ if $m \neq n$


Here are my doubts:

  1. Is my solution correct?
  2. Is the inequalities in the sum correct?
  3. Whats the right recursive relation for $\mu$
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1 Answer 1

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For $x<y$ the correct recurrence is

$$\mu(x,y)=-\sum_{x\le z<y}\mu(x,z)\,,$$

so you should have

$$\mu(m,m+1)=-\sum_{m\color{red}{\le}y<m+1}\mu(m,y)=-\mu(m,m)=-1\,.$$

Then

$$\begin{align*} \mu(m,m+2)&=-\sum_{m\le y<m+2}\mu(m,y)\\ &=-\mu(m,m)-\mu(m,m+1)\\ &=-1-(-1)\\ &=0\,. \end{align*}$$

Now try an induction on $k$ to get $\mu(m,m+k)$ for $k\ge 2$.

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