1549901369-Elements_of_Real_Analysis__Denlinger_

3.3 * Compact Sets 157

I

Thus, U is an open cover of (0, 1). Clearly, no finite subcollection of U will

cover (0, ]). Thus, U has no finite subcover of (0, 1). Therefore, (0, 1) is not

compact. D

In fac , it would have been obvious that an open interval is not compact if

we had known the following theorem.

Theorej 3.3.8 Every compact"' is clomJ.

ProJ. (We shall prove the contrapositive.) Suppose A is a set of real

numbers that is not closed. Then, by Theorem 3.2.8, there is at least one

cluster p~int x of A that does not belong to A. Consider the collection U =

{Jn: n EN}, where Vn EN,

J n = (-oo ' x - .1) n U (x + .1 n' +oo).

x-! II x

Figure 3.12

Then U U = JR - { x}, so U covers A since x tJ. A. Thus, U is an open cover of

A. I

To see that no finite subcollection of U covers A , consider any finite sub-

collection of U, say V = { Jn1' Jn 2 , · · · , Jnk }. Then

k

UV= LJ Jn; = (-oo,x - i-r) U (x + i-r,+oo) =JM,

i=l

where M = max{n 1 , n 2 , ... ,nk}. Thus the subcollection V does not contain

any poin of the interval

[x - i-r,x + i-r]·

However, since x is a cluster point of A , this interval must contain a point of

A, call it a. Thus, a tJ. UV. That is, no finite subcollection V of U can cover

A.

Therefore, A is not compact. •