
Re: G_delta
Posted:
Jan 18, 2013 4:23 AM


On Tue, 15 Jan 2013, Butch Malahide wrote:
> On Jan 14, 11:10 pm, William Elliot <ma...@panix.com> wrote: > > > > Does this generalize to every uncountable limit ordinal eta, > > that f in C(eta,R) is eventually constant and thusly the Cech > > Stone compactification of of eta is eta + 1? Does eta need > > to have an uncountable cofinality for this generalization? > > [My previous reply was typed in a hurry and had a bunch of typos, some > of which are corrected here.] > > Yes, the same argument that works for omega_1 also works for any > ordinal of uncountable cofinality. > > Let X be a linearly ordered topological space (i.e., a linearly > ordered set with its order topology) in which every increasing > sequence converges. [Examples: any ordinal of uncountable cofinality; > the long line; any countably compact LOTS.] Call a subset of X > "bounded" if it has an *upper* bound in X, "unbounded" otherwise. > Observe that (1) the union of countably many bounded sets is bounded, > and (2) the intersection of countably many unbounded closed sets is > unbounded. > > Let Y be a topological space which is hereditarily Lindelof and such > that, for each point y in Y, the set {y} is the intersection of > countably many closed neighborhoods of y. [Example: any separable > metric space.] > > THEOREM. If X and Y are as stated above, then every function f in > C(X,Y) is eventually constant. > > PROOF. We may assume that X has no greatest element. For S a subset of > Y, let g(S) = {x in X: f(x) is in Y}. Let Z = {y in Y: g({y}) is > bounded}. Did you intend g(S) = f^1(S) = { x in X  f(x) in S }?
> CLAIM. Each point z in Z has a neighborhood V_z such that g(V_z) is > bounded. > > PROOF OF CLAIM. Let {U_n: n in N} be a countable collection of closed > neighborhoods of z whose intersection is {z}. Assume for a > contradiction that each set g(U_n) is unbounded. Since f is > continuous, each g(U_n) is an unbounded closed set. By property (2) > above, g({z}) = /\{g(U_n): n in N} is unbounded, contradicting the > assumption that z is in Z. > > Thus the set Z is covered by open sets V_z such that g(V_z) is > bounded. Since Y is hereditarily Lindelof, it follows that Z is > covered by countably many open sets V such that g(V) is bounded. In > view of property (1) above, it follows that g(Z) is bounded. Since > g(Y) = X is unbounded, Y\Z is nonempty. All we have left to show is > that Y\Z consists of a single point. Assume for a contradiction that Y > \Z contains two distinct points c and d. Thus g({c}) and g({d}) are > unbounded closed subsets of X. By property (2) above, the intersection > of g({c}) and g({d}) is unbounded; in particular, it is nonempty. Let > x be a point in the intersection of g({c}) and g({d}). Then c = f(x) = > d, contradicting the assumption that c and d are two distinct points. >

