Lecture Notes Topology (2301631) Phichet Chaoha Department of ...

Phichet Chaoha 49
Proof. Let X be a compact Hausdorff space. Let A and B be disjoint
closed subsets of X. By Theorem 4.11, the sets A and B are compact. Also,
by Lemma 4.12, for each a ∈ A, there are disjoint neighborhoods U a of a and V a of
B. Since {U a } a∈A forms an open covering of A and A is compact, there is a finite
subcollection {U ai } n i=1 that covers A. By letting U = ∪n i=1 U a i
and V = ∩ n i=1 V a i
, it
is easy to check that U and V are disjoint neighborhoods of A and B, respectively.
Therefore, X is normal.
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Theorem 4.16. If X is a compact space and f : X → Y is a continuous map,
then f(X) is also compact.
Proof. Suppose X is a compact space and f : X → Y is a continuous map.
Let {U α } α∈Λ be an open covering of f(X). Then, {f −1 (U α )} α∈Λ clearly forms
an open covering of X and, by compactness, it has a finite subcovering, says
{f −1 (U αi )} n i=1 . Hence, {U α i
} n i=1 is the desired finite subcollection of {U α} α∈Λ
that still covers f(X).
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Example 4.17. A quotient space of a compact space is always compact.
Corollary 4.18. Compactness is a topological property.
Theorem 4.19. A bijective continuous map from a compact space onto a Hausdorff
space is a homeomorphism.
Proof. To show that f −1 is continuous, let C be a closed subset of X. Since X
is compact, then so is C (by Theorem 4.11). Since f is continuous, by the previous
theorem, f(C) is compact in Y . Finally, since Y is Hausdorff, it follows directly by
Theorem 4.13 that f(C) is closed in Y .
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Example 4.20. Let C = {(x, y) : x 2 + y 2 = 1} ⊆ R 2 and f : I → C be given
by f(t) = (cos 2πt, sin 2πt). Clearly, f is continuous. Also, let p : I → I /∼ = S 1 be
the quotient map where ∼ is given by 0 ∼ 1. So, f is constant on each fiber of p
and hence it induces a continuous map ˜f : S 1 → C such that ˜f ◦ p = f. It is easy
to see that ˜f is bijective. Since S 1 is compact and C is Hausdorff, ˜f is in fact a
homeomorphism; i.e., S 1 ∼ = C.
Theorem 4.21 (Extreme Value Theorem). If X is a compact space and f :
X → R is a continuous map, then f attains its minimum and maximum on X.
Proof. Suppose that f does not attain its minimum on X; i.e., for each
z ∈ f(X), there exists y ∈ f(X) such that y < z. Then the collection {(y, ∞) | y ∈
f(X)} clearly forms an open covering of f(X). Since f(X) is compact, we obtain
a finite subcovering of f(X), says {(y 1 , ∞), (y 2 , ∞), . . . , (y n , ∞)}. Hence,
f(X) ⊆ (y min , ∞) where y min = min{y 1 , y 2 , . . . , y n }. This is clearly a contradition
since y min ∈ f(X) but y min /∈ (y min , ∞). Therefore, f must attain its minimum.
Similarly, we can show that f attains its maximum as well.
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Definition 4.22. Let (X, d) be a metric space, ∅ ̸= A ⊆ X and x ∈ X. We
define the diameter of A, denoted by diam(A), and the distance between x and A,
denoted by d(x, A), as follows :
diam(A) = sup{d(a, b) | a, b ∈ A},
d(x, A) = inf{d(x, a) | a ∈ A}.