Page 203 - DMTH503_TOPOLOGY

Unit 23: The Stone–Cech Compactification

Proof: Notes
1. To show that X is a subspace of Y and X = Y.
Let  be a topology on Y. Let H  , then

H  X = H
and so H  T. Also (Y – C)  X = X – C
and so X – C  T. Conversely any open set in X is of the type (1) and therefore open in Y.
Since X is not compact, each open set Y – C containing  intersects X, meaning thereby 
is a limit point of X, so that X = Y.
2. To show that Y is compact.
Let G be an -open covering of Y. The collection G must contain an open set of the type
Y – C. Also G contains set of the type G, where G  T, each of these sets does not contain the
point . Take all such sets of G different from Y – C, intersect them with X, they form a
collection of open sets in X covering C.
As C is compact, hence a finite number of these members will cover C; the corresponding
finite collection of elements of G along with the elements of Y – C cover all of Y.
Hence Y is compact.
3. To show that y is Hausdorff.
Let x, y  Y.
If both of them lie in X and X is known to be compact so that  disjoint open sets V
in X
s.t. x  , y  V.
On the other hand if

x  X
and y = .
We can chose compact set C and X containing a nbd  of x.

The  and Y – C are disjoint nbds of x and  respectively in Y.
Theorem 3: If (X , T ) be a one point compactification of a non-compact topological space (X, T),
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then (X , T ) is a Hausdorff space iff (X, T) is locally compact.
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Proof: Assuming that X is a Hausdorff space, each pair of distinct points in X , all of which belong
*
to X can be separated by open subsets of X. Thus it is sufficient to show that any pair (x, )  X,
can be separated by open subsets of X . Now X is locally compact
*
 any x  X, has a nbd N whose closure N in X is compact
*
 N and N are disjoint open subsets of X s.t. x  N and   N
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 distinct points x,  of X have disjoint nbds
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 (X , T ) is Hausdorff.
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*
*
Conversely if (X , T ) is Hausdorff, then
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X is a subspace of X  X is Hausdorff, since Hausdorffness is hereditary.

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