Number of size 1 partitions of the empty set

Question

Disclaimer: This is a homework problem, but I'm just asking for clarification, not a solution.

We're asked to prove $S(0,1) = 1$, where $S(n,k)$ is "the number of different partitions of [a set of size $n$] into $k$ mutually disjoint subsets", where a partition is defined like so: "Let $X$ be a collection of pairwise disjoint sets and let $Y = \bigcup X$. Then $X$ is called a partition of $Y$ if either (i) $X = Y = \varnothing$; or (ii) $X \neq \varnothing \land \varnothing \notin X$."

As far as I can tell the only set $X$ with $\bigcup X = \varnothing$ and $|X| = 1$ is $X = \{\varnothing\}$, but this does not satisfy either (i) or (ii) of the defintion of a partition. So it seems to me that there are no size 1 partitions of $\varnothing$, and $S(0,1)$ should be $0$, not $1$.

Am I missing something here?

Answer 1

You are right according to that definition. By the way, it can be simplified to "... is called a partition if $\emptyset\notin X$."

Answer 2

I believe you're mistaken when you say that the set $X=\{\varnothing\}$ satisfies $\bigcup X = \varnothing$. It actually satisfies $\bigcup X = \{\varnothing\}$.

The only $X$ satisfying $\bigcup X = \varnothing$ is $X=\varnothing$, the empty set itself.

Does that help?

Answer 3

Condition (i) describes the partition of the empty set into $0$ non-empty parts (the "non-empty" property being vacuously true).

Condition (ii) implies that all parts in a partition are non-empty.

This is consistent with the definition of the Stirling Numbers of the second kind. The number $S(0,1)=0$ since there are no partitions of the empty set into exactly 1 non-empty part.

Answer 4

I think the author intended the "special case" clause (i) to be "$X=\{\emptyset\}$ and $Y=\emptyset$." That would make $S(0,1)=1$ and it would not be subject to Hagen von Eitzen's simplification of what was actually written.