I'm reading my textbook and I just found that the Hasse diagram for the subset-equal ordering, $\subseteq,$ of the power set $P(S),$ which $|S|=3,$ looks like a cube
Since I'm curious about the number of edges it has, then I found the formula
For the Hasse Diagram of $(P(S), \subseteq),$ which $|S|=n,$ the number of edges of it is $$\sum_{k=1}^{n}{k\choose1}{n\choose k}.$$
And the terms in the summation is symmetric, e.g. for $n=4$, it's
$$4+12+12+4=32,$$
and since for $n=3$ the graph looks like a cube, I also checked that for $n=4,$ the number of edges, $32$, does be the number of edges of a 4-d hypercube($12\times2+8$)
I have no idea why the subset-equal relation is related to a hypercube, any idea?
The number of subsets of an $n$ element set $A$ is $2^n$. So the number of vertices in the Hasse diagram of $(P(A), \subseteq)$ is $2^n$ as is the number of vertices in an $n$ dimensional hypercube whose vertices are labeled as length $n$ bit strings.
Two vertices of Hasse diagram are connected by an edge if and only if there is exactly one element new in one subset than in the other subset. Two vertices in a hypercube are adjacent if and only if the labeling of the vertices have changed in exactly one bit position. So if you denote by $0$ the absence of an element in a subset and $1$ by its presence, then this correspondence will give exactly the same edges as in the Hasse diagram.