I am trying to find a chain with six elements as I am told it exists as it is the Dedekind number for the power set of $\{1,2,3,4\}$
But I keep only being able to come up with 5
for example
$\{ \emptyset, \{1\},\{1,2\},\{1,2,3\},\{1,2,3,4\}\}$
I cant add in another singleton because then it wont be able to be ordered with $\{1\}$ for same reason I cant add a two element one as it wont be ordered with $\{1,2\}$
If it helps to provide context, I am trying to find the minimum number k of chains it takes to write the partial order as a decomposition of said chains.
Also it appears no such chain exists. So where is the mistake in what I have previously said? I used the binomial formula on n choose floor(n/2) ie 4 choose 2 which is 6
Any advice /hints/help please?
Looking at wikipedia, the dedeking number is actually the number of elements in an antichain (not chain).
In the case in which your base set is just the powerset of $\{1,2,3,\dots n\}$ the dedekind number is known to be $\binom{n}{\lfloor n/2 \rfloor}$ by a beautiful theorem due to Sperner.
So in our case the Dedekind number is $\binom{4}{2}=6$.
The only antichain of this size is $\{ \{1,2\}, \{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\}\}$