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