An $n$ element set can be written as a union of disjoint sets, in $B_n$ different ways. But what if the sets are not disjoint? If we let $X=\{1,2,3,\ldots n\}$ then what is the value of $\left|\{\mathcal{F}\subseteq\wp(X):\bigcup\mathcal{F}=X\}\right|?$ I know that if one requires the sets in $\mathcal{F}$ be pairwise disjoint this is just the the $n^{\text{th}}$ bell number. However I'm interested in counting all the set covers of an $n$ element set. Could anyone point me to a reference?
 A: I believe the answer is 
$$\displaystyle\sum_{k=0}^{n} {n\choose k} \left(-1\right)^{k} 2^{2^{n-k}}$$
The idea is to think of collections which do not cover S, and subtract them away. The reason is that we take the power set of the power set of S, but then we must subtract off those for which a particular element appears in no subset. The collections of subsets of which 1 isn’t a member is equinumerous with the collections of subsets of n-1 items: subtract this n times, one for each element. Now collections of subsets that don’t include 1 or 2 say, have been subtracted twice, so we add them back in and so on with a inclusion exclusion argument continuing. 
To prove this, for $\mathcal{F}\subseteq S$ let $$\mathcal{T}_{\mathcal{F}}=\mathcal{P}\left(\mathcal{P}\left(S\setminus\mathcal{F}\right)\right)$$
Clearly, $\left|\mathcal{T_{\mathcal{F}}}\right|=2^{2^{n-k}}$ when $\left|\mathcal{F}\right|=k$. 
$$\left\{ \mathcal{F}\subseteq S\vert\bigcup\mathcal{F}=S\right\} =\bigcap_{i=1}^{n}\overline{\mathcal{T}_{\left\{ i\right\} }}$$
So, by inclusion-exclusion, 
$$\left|\left\{ \mathcal{F}\subseteq S\vert\bigcup\mathcal{F}=S\right\} \right|=\left|\mathcal{T_{\emptyset}}\right|-\sum_{i=1}^{n}\left|\mathcal{T}_{\left\{ i\right\} }\right|+\sum_{1\leq i<j\leq n}\left|\mathcal{T}_{\left\{ i,j\right\} }\right|\mp...
 $$
$$={n \choose 0}2^{2^{n}}-{n \choose 1}2^{2^{n-1}}+{n \choose 2}2^{2^{n-2}}\mp...$$
$$=\displaystyle\sum_{k=0}^{n} {n\choose k} \left(-1\right)^{k} 2^{2^{n-k}}$$
