colonel@gloria.UUCP (Col. G. L. Sicherman) (04/28/85)
["A true seaman is never lost!" --Water Polo] > The real trick is mapping the unit line onto the unit square *continuously*. > This was first done by Peano many years ago. I don't think his example is > one-to-one, however. Of course not. If it were one-to-one and continuous, it would be a homeomorphism--the line and the square would be topologically the same. The usual way of showing that A and B have the same number of points is to map A onto B (_not_ necessarily 1-to-1) and then map B onto A (again, not necessarily 1-to-1). Constructing a 1-to-1 map is then routine -- if you know the routine. This can be done with the line and the square, and it won't be continuous! The trick of alternating digits works fine except at the end (0.9999...). -- Col. G. L. Sicherman ...{rocksvax|decvax}!sunybcs!colonel
andrews@yale.ARPA (Thomas O. Andrews) (05/01/85)
Summary: Expires: Sender: Followup-To: Distribution: Keywords: In article <536@gloria.UUCP> colonel@gloria.UUCP (Col. G. L. Sicherman) writes: > >Of course not. If it were one-to-one and continuous, it would be a >homeomorphism--the line and the square would be topologically the same. > . > . > . >Col. G. L. Sicherman >...{rocksvax|decvax}!sunybcs!colonel This statement is accurate, but somewhat misleading. A map is a homeo- morphism if it is 1-1, continuous, and has a continuous inverse. In general 1-1 and continuous does not imply that the inverse is continuous. For example, the map f:[0,2*pi)-->circle defined by f(a)=(cos a,sin a) is 1-1 and continuous, but certainly not a homeomorphism. On the other hand, if f:X-->Y is 1-1 and continuous, and X is compact, then the inverse of f is also continuous, and hence f is a homeomorphism. Since the unit interval is compact, the Col. Sicherman's statement is accurate. -- Thomas Andrews 17? My dear, what worthless, superstitious nonsense!