# Ordering tuples over (0,1)

To understand the natural ordering of $\{0,1\}^N$ I first thought about $\{0,1\}^3$ which has this Hasse diagram:

$\dpi{150} \large \xymatrix{ &111 \ar[ld] \ar[d] \ar[rd] &\\ 110 \ar[d] \ar[rd] &101 \ar[ld] \ar[rd] &011 \ar[ld] \ar[d] \\ 100 \ar[rd] &010 \ar[d] &001 \ar[ld] \\ &000& }$

There's something interesting going on in the second and third rows. In 3-D it would attach the corners of ▲ to the corners of ▼. But I'm missing if there's a familiar structure (symmetric group?) that can help me decompose the more complicated natural ordering of $\{0,1\}^N$.

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It's simply an hypercube, in which every point of the sequence gives you the $n$-th coordinate
Right, that's the points, but what about the ordering where $(\ldots, 1, \ldots) \succ (\ldots, 0, \ldots)$? Maybe you are suggesting I just need to think of a hypercube turned so that it has a "top corner" and a "bottom corner"... –  isomorphismes Jan 7 at 15:41
Yes, exactly. You put $1^N$ at the top and $0^N$ at the bottom. That will give you a nice structure –  miniBill Jan 7 at 20:05