Saturday, March 17, 2012

Topology of {0,1}^n (part 1)

Def. 00,01,10,11 are points.

S: {00,01,10,11} is a space.

Def. ≤ is a binary relation on S. 00≤01≤11, 00≤10≤11, 00≤11, not 01≤00, not 10≤00, not 01≤10, not 10≤01, not 11≤00, not 11≤01, not 11≤10. For all points p, p≤p.

Th. (S,≤) is a poset.

Proof. By definition, p≤p. Suppose p≤q and q≤p for some p≠q. For 00 and 01,
they contradicts definition. Similar for other combinations. Hence p≤q and q≤p then p=q. For some p,q and r such that p≠q≠r, suppose p≤q and q≤r but not p≤r.
For 00,01 and 11, they contradicts definition. Also for 00,10 and 11. Hence for all p,q and r such that p≤q and q≤r then p≤r. If p=q, p≤q and q≤r then p≤r. If q=r, p≤q and q≤r then p≤r. Hence p≤q and q≤r then p≤r. □