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Unit 23: The Stone–Cech Compactification
Let , be respectively the families of all continuous functions from X, Y respectively to the Notes
1 2
2
1
unit interval [0, 1] and let e, e be the embedding of X, Y into [0,1] and [0,1] respectively. For
2
any g letg : [0,1] [0, 1] be the corresponding projection.
2
Then o e of is a map from X into [0, 1] and so it has an extension say to (X). Then
g g
o e o e o f.
g g
Now consider the family { = g } of maps from (X) into [0, 1]. Let : (X)[0, 1] be the
g 2
evaluation map determined by this family. We claim that o e = e o f. Let xX. Them (e(x)) is
2
an element of [0,1] given by
(e(x))(g) = g(e(x)) [by the definition of the evaluation functions]
But (e(x)) = (e’f(x)) = e’(f(x))(g)
g g
Thus for all g
2
[ e(x) (g) o ] = e f(x) (g) and so
é
ù
û
ë
o
o g = e f as claimed.
o
Now (e(x)) = e (f(X))e(Y).
2
Since Y is compact, e(Y) compact and hence a closed subset of [0,1] .
)
So (e(X) e(Y).
But since is continuous,
(B(X)) = (e(X) ) (e(X) )
Thus we see that maps (X) into e(Y). Since e is an embedding, there exists a map e : e(Y) Y
1
which is an inverse to e regarded as a map from Y onto e(Y). Then e o e o f = f.
1
Uniqueness of the extension is immediate in view of the fact that Y is a Hausdorff space and e(X)
is dense in (X).
23.2 Summary
A compactification of a space X is a compact Hausdorff space Y containing X as a subspace
such that X = Y.
Two compactifications Y and Y of X are said to be equivalent if there is a homeomorphism
1 2
h : Y Y such that h(x) = x for every xX.
1 2
If X has a compactification Y, then X must be completely regular, being a subspace of
completely regular space Y.
If X is completely regular, then X has a compactification.
The pair (e, (X)), where X is a Tychonoff space and (X) ( e(x)= ) is called Stone-Cech
compactification of X, e is a map from X into (x).
The Stone-Cech compactification is defined for all Tychonoff spaces.
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