Question
Consider the following sets in the dictionary order. Which arelinear continua? a. Z+ x {0,1) b. [0,1) x Z+ c. [0,1) x [0,1] d. [0,1] x [0,1) Thanks so much. = Consider the following sets in the dictionary order. Which arelinear continua? a. Z+ x {0,1) b. [0,1) x Z+ c. [0,1) x [0,1] d. [0,1] x [0,1) Thanks so much. c. [0,1) x [0,1] d. [0,1] x [0,1) Thanks so much. =
Explanation / Answer
a) The mapping from Z+ x [0, 1) to the interval [1, infinity)of the real line give by (z, x) goes to z + x is a homeomorphism,since it's obviously bijective and open rays [ {x | x a} are open rays ] correspond to each other. Sinceopen rays form a subbase of the topologies on both sides, we seethe mapping gives a correspondence between open sets. Since[1, infinity) is a linear continuum, so is our original set. b) The order type is not dense: there is no element between(0, 1) and (0, 2). c) The proof given in Example 1 goes through exactly, since[0, 1) also has the l.u.b. property. So this is a linearcontinuum. d) The set S = {0} x [0, 1) is bounded above, but has noleast upper bound. An upper bound can't be of the form (0,b), since then (0, b + (1 - b)/2) is an element of S bigger thanit. So an upper bound must have the form (a, b), where a >0. But for any such upper bound (a/2, b) is also an upperbound, which is smaller. So there is no least upper bound, soour set is not a linear continuum.