
Re: G_delta
Posted:
Jan 19, 2013 2:51 AM


On Jan 19, 1:05 am, William Elliot <ma...@panix.com> wrote: > 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? > > > 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. > > Does not (1) hold because every increasing sequence converges? > Wouldn't every countable set has an upper bound suffice?
Yes, "every countable set has an upper bound" would suffice for (1), but not for (2).
> > 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 S}. Let Z = {y in Y: g({y}) is > > bounded}. > > Is assuming X has no greatest element, an additional premise?
No. The assumption that X has no greatest element is made without loss of generality, because the contrary case is trivial: if X has a greatest element, then every function with domain X is eventually constant.
> Is Z closed?
It will be shown that Y\Z contains a single point c, the eventual value of the function f, and so Z = Y\{c}. To ask whether Z is closed is to ask whether c is an isolated point of Y. It could, but it doesn't have to be.
> > 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.

