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FIVE MAJOR RESULTS IN ANALYSIS AND TOPOLOGY Aaron ...

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6. THE STONE- ˘ CECH COMPACTIFICATION 35<br />

if there is a ‘biggest’ compactification of the completely regular space X. It turns out that<br />

such a compactification exists, and it was discovered by Marshall Stone and Eduard ˘ Cech.<br />

Theorem (Stone- ˘ Cech Compactification 1 ). Let X be completely regular. Then there<br />

exists a unique compactification β(X) of X such that<br />

(1) Every continuous map of X into a compact Hausdorff space Y extends uniquely to<br />

a continuous mapping of β(X) → Y , and<br />

(2) Every compactification Y of X is equivalent to a quotient space of β(X).<br />

The proof of this theorem is quite large, and will be shown by a sequence of lemmas. We<br />

will introduce the theory of quotient spaces when it is needed.<br />

Definition. Two compactifications Y1 and Y2 of a space X are said to be equivalent if<br />

there is a homeomorphism from Y1 to Y2 that fixes X.<br />

Lemma (1). Let X be a space, and let h be an imbedding of X into the compact Hausdorff<br />

space Z. Then h induces a unique compactification Y of X, such that there is an imbedding<br />

H of Y into Z which agrees with h on X.<br />

Proof. Given h, let X0 = h(X) ⊆ Z, and let Y0 be the closure of h(X) in Z. Then Y0<br />

is a compact Hausdorff space and equals the closure of X0, so Y0 is a compactification of X0.<br />

Choose a set A that corresponds bijectively with Y0 − X0 via the map k: A → Y0 − X0. Let<br />

Y = X ∪ A, and define H : Y → Y0 by H(x) = h(x) for x ∈ X, and H(a) = k(a) for a ∈ A.<br />

H is bijective because h and k are bijections. Let U ⊂ Y be open in Y if, and only if, H(U)<br />

is open in Y0. Then H is automatically a homeomorphism, being a bijective continuous map.<br />

X is a subspace of Y because H agrees with the homeomorphism h when restricted to X.<br />

Hence, H is an imbedding of Y into Z.<br />

Now, suppose Y1, Y2 are two compactifications of X and that H1 : Y1 → Z and H2 :<br />

Y2 → Z are imbeddings that agree with h on X. Then H1(X) = H2(X) = h(X) = X0.<br />

Furthermore, since H1 and H2 are continuous and Y1 and Y2 are closed, H1(Y1) = H2(Y2) =<br />

1 The statements and proofs of lemmas 1 through 5 are taken almost directly from [8]. An alternate<br />

proof for lemma 6 can be found in [5], although the one provided here is much simpler.

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