Set cover: To show the number of minimum coverable combination of the power set of universe $U$ is less than $2^n$ Given a set $U=\{1,2,...,n\}$, and its power set $\mathcal S=\{S_1,...,S_{2^n-1}\}$(without emptyset $\emptyset$), a minimum coverable combination is $S_i\cup...\cup S_k=U$ and $S_i,...,S_k$ are pairwise disjoint.

For example, when $U=\{1,2,3\}$, then $\mathcal S=\{\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$ and a minimum coverable combination is $S_4=\{1,2\}$ and $S_3=\{3\}$ then $S_4\cup S_3=U$ and $S_4\cap S_3=\emptyset$.(don't count the permutations so $S_4\cup S_3$ is considered the same as $S_3\cup S_4$)

Now the question is to prove the number of minimum coverable combination of $U$ is less than or equal to $2^n$
Let $U = \{ 1, 2, 3, 4, 5 \}$. Here are some minimum coverable combinations of $U$:

$\begin{align*}
1. \: &amp; \{ 1, 2, 3, 4, 5 \} \\
2. \: &amp; \{ 1 \}, \{ 2, 3, 4, 5 \} \\
3. \: &amp; \{ 2 \}, \{ 1, 3, 4, 5 \} \\
4. \: &amp; \{ 3 \}, \{ 1, 2, 4, 5 \} \\
5. \: &amp; \{ 4 \}, \{ 1, 2, 3, 5 \} \\
6. \: &amp; \{ 5 \}, \{ 1, 2, 3, 4 \} \\
7. \: &amp; \{ 1, 2 \}, \{ 3, 4, 5 \} \\
8. \: &amp; \{ 1, 3 \}, \{ 2, 4, 5 \} \\
9. \: &amp; \{ 1, 4 \}, \{ 2, 3, 5 \} \\
10. \: &amp; \{ 1, 5 \}, \{ 2, 3, 4 \} \\
11. \: &amp; \{ 2, 3 \}, \{ 1, 4, 5 \} \\
12. \: &amp; \{ 2, 4 \}, \{ 1, 3, 5 \} \\
13. \: &amp; \{ 2, 5 \}, \{ 1, 3, 4 \} \\
14. \: &amp; \{ 3, 4 \}, \{ 1, 2, 5 \} \\
15. \: &amp; \{ 3, 5 \}, \{ 1, 2, 4 \} \\
16. \: &amp; \{ 4, 5 \}, \{ 1, 2, 3 \} \\
17. \: &amp; \{ 1 \}, \{ 2 \}, \{ 3, 4, 5 \} \\
18. \: &amp; \{ 1 \}, \{ 3 \}, \{ 2, 4, 5 \} \\
19. \: &amp; \{ 1 \}, \{ 4 \}, \{ 2, 3, 5 \} \\
20. \: &amp; \{ 1 \}, \{ 5 \}, \{ 2, 3, 4 \} \\
21. \: &amp; \{ 2 \}, \{ 3 \}, \{ 1, 4, 5 \} \\
22. \: &amp; \{ 2 \}, \{ 4 \}, \{ 1, 3, 5 \} \\
23. \: &amp; \{ 2 \}, \{ 5 \}, \{ 1, 3, 4 \} \\
24. \: &amp; \{ 3 \}, \{ 4 \}, \{ 1, 2, 5 \} \\
25. \: &amp; \{ 3 \}, \{ 5 \}, \{ 1, 2, 4 \} \\
26. \: &amp; \{ 4 \}, \{ 5 \}, \{ 1, 2, 3 \} \\
27. \: &amp; \{ 1 \}, \{ 2, 3 \}, \{ 4, 5 \}  \\
28. \: &amp; \{ 1 \}, \{ 2, 4 \}, \{ 3, 5 \}  \\
29. \: &amp; \{ 1 \}, \{ 2, 5 \}, \{ 3, 4 \}  \\
30. \: &amp; \{ 2 \}, \{ 1, 3 \}, \{ 4, 5 \}  \\
31. \: &amp; \{ 2 \}, \{ 1, 4 \}, \{ 3, 5 \}  \\
32. \: &amp; \{ 2 \}, \{ 1, 5 \}, \{ 3, 4 \}  \\
33. \: &amp; \{ 3 \}, \{ 1, 2 \}, \{ 4, 5 \}  \\
34. \: &amp; \{ 3 \}, \{ 1, 4 \}, \{ 2, 5 \}  \\
35. \: &amp; \{ 3 \}, \{ 1, 5 \}, \{ 2, 4 \}  \\
36. \: &amp; \{ 4 \}, \{ 1, 2 \}, \{ 3, 5 \}  \\
37. \: &amp; \{ 4 \}, \{ 1, 3 \}, \{ 2, 5 \}  \\
38. \: &amp; \{ 4 \}, \{ 1, 5 \}, \{ 2, 3 \}  \\
39. \: &amp; \{ 5 \}, \{ 1, 2 \}, \{ 3, 4 \}  \\
40. \: &amp; \{ 5 \}, \{ 1, 3 \}, \{ 2, 4 \}  \\
41. \: &amp; \{ 5 \}, \{ 1, 4 \}, \{ 2, 3 \}  \\
\end{align*}$

There are more than $2^5 = 32$, so the claim is false.
Let $B_n$ denote the number of partitions of the set $[n]:=\{1,2,\ldots,n\}$.  This is known as the $n$-th Bell number.  The first few values are $B_0=1$, $B_1=1$, $B_2=2$, $B_3=5$, $B_4=15$, and $B_5=52$.  It can be shown that
$$B_{n+1}=\sum_{k=0}^n\,\binom{n}{k}\,B_k\text{ and }B_n=\sum_{k=0}^n\,\left\{\begin{array}{c}n\\k\end{array}\right\}$$
for all $n\in\mathbb{Z}_{&gt;0}$, where
$$\left\{\begin{array}{c}n\\k\end{array}\right\}=\frac{1}{k!}\,\sum_{j=0}^k\,(-1)^{k-j}\,\binom{k}{j}\,j^n$$
is the Stirling number of the second kind.

It is known that
$$B_n \approx n^{-\frac{1}{2}}\,\left(\frac{n}{W(n)}\right)^{n+\frac{1}{2}}\,\exp\left(\frac{n}{W(n)}-n-1\right)\,,$$
where $W$ is the Lambert $W$-function.  That is, using $W(n)\approx \ln(n)$, we get that 
$$B_n &gt; \left(\frac{n}{\text{e}\,\ln(n)}\right)^n$$
for every positive integer $n$.  Therefore, $B_n$ grows much faster than $2^n$.