fom
Posts:
1,968
Registered:
12/4/12
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Re: Stone Cech
Posted:
Mar 15, 2013 12:15 AM
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On 3/14/2013 10:53 PM, 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? >> >> Why in the heck is a compactification an embedding >> function and a compact space? Wouldn't be simpler >> to define a compactification of a space S, as a >> compact space into which S densely embeds? >> > > Munkres characterizes Stone-Cech in relation to > one point compactification. > > He says the latter is the "minimal" compactification > of a space whereas the Stone-Cech compactification > is maximal in a sense described by Exercise 4 in > section 5-3 > > "Let Y be an arbitrary comactification of X; let B(X) > be the Stone-Cech compactification. Show there is a > continuous surjective closed map g:B(X) -> Y that > equals the identity on X > > [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).]" > > 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 Stone-Cech 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. > > >
That extension property is such that the two compactifications map such that F(x)=x for each x in X.
That will probably make it easier to see the relationship to the surjection in the exercise.
The important details, of course, are in the proofs. I would have to review them to say much more.
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