Why do the $-1$-categories form a categeory rather than merely a set? The collection of $n$-categories naturally has the structure of a $(1+n)$-category. For example $\mathbf{Set}$ is a $1$-category and $\mathbf{Cat}$ is a $2$-category. Therefore we would expect the $-1$-categories to form a $0$-category, i.e. a set.
The $-1$-categories are precisely the empty category and the trivial category (these can be seen as the booleans or truth-values, false and true). But there's a functor from the empty category to the trivial category (false implies true). So the $-1$-categories actually form a $1$-category (admitedly a mere partial order) rather than a set as expected.
What's the explanation for this discrepancy in an otherwise straightforward pattern? 
 A: Because sets are the $0$-groupoids; they are the $0$-categories in which every morphism is invertible. In a set, if $\hom(x,y)$ is trivial, then $\hom(y,x)$ is trivial as well.
Its posets that are the $0$-categories.
It may be helpful to think of setoids rather than sets, and preorders instead of posets.
There's a finer classification: an $(n,r)$-category is one where:


*

*$k$-morphisms are equivalences if $k>r$

*Parallel $k$-morphisms are equivalent for $k > n$
Sets are the $(0,0)$-categories, and posets are the $(0,1)$-categories. The usual definition of category is of $(1,1)$-categories.
A: An $(n,r)$-category is an $\omega$-category such that:


*

*Any parallel $k$-morphisms are equivalent for $k>n$

*All $k$-morphisms are equivalences for $k>r$.


This allows us to draw a periodic table of $(n,r)$-categories:
\begin{array}{|c|c|c|c|c|c}
\hline
 r\downarrow \backslash n\rightarrow& -1 & 0 & 1& 2& \dots \\\hline
 0 &\text{boolean}&\text{set}&\text{groupoid}&\text{$2$-groupoid}&\\ \hline
 1  &\text{"}&\text{poset}&\text{category}&\color{grey}{\text{$(2,1)$-category}}&\\ \hline
 2  &\text{"}&\text{"}&\color{grey}{\text{$(1,2)$-category}}&\text{$2$-category}&\\ \hline
 3  &\text{"}&\text{"}&\color{grey}{\text{"}}&\color{grey}{\text{$(2,3)$-category}}&\\ \hline
 \vdots&  & & & &\ddots\\
\end{array}
(in which I have greyed out any cells not corresponding to a preexisting notion).
It's easy to see that the sequence $(\text{boolean}, \text{ set}, \text{ category}, \text{ $2$-category},\dots)$, which I thought was the definition of an $n$-category, actually forms a crooked line in this table.
There are two ways to straighten it:


*

*If we wanted to continue the sequence $(\text{boolean}, \text{ set}, \dots)$ in a straight line, we should continue it with $(\dots,\text{ groupoid}, \text{ $2$-groupoid}, \dots)$.

*If instead we wanted the element before the beginning of the sequence $(\text{set}, \text{ category}, \text{ $2$-category},\dots)$, we should take precisely the $(-1,-1)$-categories. These are off the top of the table, but I believe they include only the empty category (the object of the trivial category cannot be an equivalence, because there is no $0$-morphism "going the other way" to compose it with). The empty category on its own does indeed form a set.
In every case, we have that $\mathbf{(n,r)}$-$\mathbf{Cat}$ is a $(1+n,1+r)$-category (at least in the nontrivial part of the table where $1+n\geq r$). So there are no strange anomalies.
