Simplicial Structures in Topology

I.1 Topology 19
are closed in Y. Therefore, f −1 (y1) and f −1 (y2) are two disjoint closed subsets
of X (see Theorem (I.1.9)). For every element (x,a) ∈ f −1 (y1) × f −1 (y2) choose
two disjoint open sets Ux,a and Vx,a in X such that x ∈ Ux,a and a ∈ Vx,a. The set
{Vx,a |a ∈ f −1 (y2)} is an open covering of f −1 (y2); since this space is compact (see
Theorem (I.1.23)), there exists a finite subcovering {Vx,a 1 ,...,Vx,an } of f −1 (y2).
And so we have two disjoint open sets
n�
Ux = Ux,a j
j=1
and Vx
n�
=
j=1
Vx,a j
containing x and f −1 (y2), respectively. But {Ux | x ∈ f −1 (y1)} is an open covering
of f −1 (y1) and, since this is a compact space, there is a subcovering {Ux1 ,...,Uxm }
of f −1 (y1). We now note that the sets
m�
m�
U = Ux and V = Vx k k
k=1
k=1
are disjoint open sets of X which contain f −1 (y1) and f −1 (y2), respectively.
The sets W1 = Y � f (X �U) and W2 = Y � f (X �V) are open sets of Y ( f is a
closed map) containing y1 and y2, respectively. We wish to prove that W1 ∩W2 = /0.
Suppose there exists a point y ∈ W1 ∩W2. Then, y �∈ f (X �U) and y �∈ f (X �V),
in other words,
f −1 (y) ∩ (X �U)= /0and f −1 (y) ∩ (X �V)=/0;
hence, f −1 (y) ⊂ U ∩V = /0, which is not possible. �
A first concrete example of a compact space is given by the next theorem.
(I.1.30) Theorem. The unit interval I =[0,1] is compact.
Proof. Let U be a covering of I by open sets of R. The properties of the Euclidean
topology of R ensure that for every x ∈ I we can find a δ(x) > 0andan
element U(x) ∈ U such that the open interval (x − δ(x),x + δ(x)) is contained in
U(x). Theset
I = {I(x)=(x − δ(x),x + δ(x)) | x ∈ I}
is an open covering of I and every interval of I is contained in an open set of U. If
there exists a finite subcovering in I, there is a corresponding finite subcovering in
U; we may, therefore, assume that each open set of U is an open interval of R.
We now consider the set {0,1} with the discrete topology and the function
defined by the following conditions:
f : I →{0,1}
1. f (x) =0 if the closed interval [0,x] is covered by a finite number of elements
of U.
2. Otherwise, f (x)=1.