I'm trying to understand why the empty function is $\mu$-recursive. I guess I need to use the last condition describing the class of $\mu$-recursive functions:
if $f:N^{k+1}\to N$ is $\mu$-recursive, then so is $g:N^k\to N$ defined by $g(x_1,\dots,x_k)=\mu y. (f(x_1,\dots,x_k,y)=0)$.
In my case, $g$ is supposed to be the empty function. Is it officially a function $N^0\to N$? If so, I need to find a $\mu$-recursive function $f: N^{0+1}\to N$ such that $\mu y. (f(y)=0)$ is undefined (if my understanding is right). So can I just take $f$ to be the constant function $1$?
There's an empty function of each arity: specifically, the empty function of arity $n$ is the partial function $e^n: \subseteq\mathbb{N}^n\rightarrow\mathbb{N}$ whose domain is $\emptyset$. Put another way, its graph $\{(a_1,...,a_n, b): e^n(a_1,...,a_n)\downarrow=b\}$ is $\emptyset$, where "$\downarrow=$" means "is defined and equal to."
The idea for building $e^n$ via $\mu$-recursion is to use an unsatisfiable condition: $e^n(a_1,...,a_n)$ should be the smallest $b$ such that [IMPOSSIBILITY]. This will ensure that $e^n$ is never defined. There are a number of ways to do this, the most natural probably being
$$e^n(a_1,...,a_n)=\mu x.(c_1^{n+1}(a_1,...,a_n,x)=0)$$ where $c_1$ is the $(n+1)$-ary constant function always taking value $1$.
In particular, in the case of the $0$-ary empty function $e^0$, $c_1^0$ is just the number $1$ (remember that a $0$-ary function is just a number).