fom
Posts:
1,968
Registered:
12/4/12
|
|
Re: Stone Cech
Posted:
Mar 15, 2013 12:15 AM
|
|
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.
|
|
