Student of Ronald de Wolf---Thijs van Ommen---showed 
that the solution to Proposition 10.2 is  optimal: 
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if F is an antichain, then there is no way to do it with less 
than |b(F)| keys.  
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His proof is as follows.
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We are looking for a set family K (the set of keys) satisfying the 
following:
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 (*) for every subset S of the universe, we have  [S contains (as a 
subset) an element of F iff S blocks K]
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We know by Proposition 10.2 that K=b(F) is one solution to this.  We 
want to show that this is the only solution.  Suppose there is another 
solution K'.  We can assume without loss of generality that K' is an 
antichain (if not, just remove redundant keys).  Define F'=b(K').  Then 
b(F')=b(b(K'))=K' using Proposition 10.1, hence F' satisfies (by 
Proposition 10.2):
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 (**) for every subset S of the universe, we have  [S contains (as a 
subset) an element of F' iff S blocks K']
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Combining (*) (for K') and (**), we get
 for every subset S of the universe, we have  [S contains (as a subset) 
an element of F iff S contains (as a subset) an element of F']
This implies F=F'.  Hence K=b(F)=b(F')=K'.