
Re: Stone Cech
Posted:
Mar 16, 2013 4:20 AM


On Thu, 14 Mar 2013, fom wrote: > On 3/14/2013 9:49 PM, William Elliot wrote:
> > Let (g,Y) be a Cech Stone compactification of S. > > If (f,X) is a compactification of S, does X embed in Y? > > > > If (g,Y) is a compactification of S and > > for all compactifications (f,X), X embeds in Y > > is (g,Y) a Stone Cech compactification of S?
> Munkres characterizes StoneCech in relation to > one point compactification. > > He says the latter is the "minimal" compactification > of a space whereas the StoneCech compactification > is maximal in a sense described by Exercise 4 in > section 53 > > "Let Y be an arbitrary comactification of X; let B(X) > be the StoneCech compactification. Show there is a > continuous surjective closed map g:B(X) > Y that > equals the identity on X That cannot be as X is not a subspace of B(X) nor is X neccessarily a subspace of Y. For example, { 0, 1/n  n in N } is a compactification of N and B(N) can be a collection of ultrafilters for N.
> [This exercise makes precise what we mean by saying > B(X) is the 'maximal' compactification of X. If you > are familiar with quotient spaces, you will recognize > that g is a quotient map. Thus every compactification > of X is equivalent to a quotient space of B(X).]" Yes, when Y is a (Hausdorff) compactification of X, then by the universal property, there's a continuous g:B(X) > Y, which is a closed map.
IF (f,Y) is a compactification of X, then f = ge where e in an embedding of X into B(X)
Now f(X) = ge(X) = ge(cl X) subset g(cl e(X) = g(B(X)) So Y = cl f(X) subset cl g(B(X)) = g(B(X)).
Thusly g is surjective, hence a quotient map. > The question you ask is precisely Munkres definition: > > "A compactification of a space X is a compact Hausdorff > space Y containing X such that X is dense in Y." > > "In order to have a compactification, X must be > a completely regular space" > > The StoneCech compactification is based on a cube > such that each component of the cube is an interval > > I_a = [glb(f_a(X)),lub(f_a(X))] > > formed from a bounded continuous real valued function > on a completely regular space. > > The cube is the product of all such intervals (all > such functions on the space). > > Define h: X > Pi_a I_a > > where h(x)=(f_a(x)) > > By Tychonoff's theorem, the cube is compact. > Because X is completely regular, the collection of > functions separates points on X. This, makes h an > imbedding. > > Munkres goes on to prove a few uniqueness conditions > that are true for the compactification derived from > this imbedding involving extensions of the original > bounded continuous functions on X to continuous > functions on B(X). Any two compactifications satisfying > these extension properties are equivalent up to > homeomorphim. > > Hope that helps. > > > >

