Principles of Mathematics in Operations Research

10.2 Compact Sets 149

Remark 10.2.14 Here are some remarks:

1. 7/lim„_loo(6n - an) = lim^^ length(In) = 0, => f£°=i In consists of one
   point.


2. If In 's are not closed, conclusion is false, e.g. In = (0, -).


3. If In's are not bounded, conclusion is false, e.g. In = [n,oo].

Definition 10.2.15 Let ax < b\,..., ak < bk be real numbers, then the set of

all points p G Rk 3 p = {x\,... ,xk), a^ < xt < l>i, i = 1,..., k is called a

k-cell. So a k-cell is

[ai,bi] x • • • x [ak,bk].

Theorem 10.2.16 Let k G N be fixed. Let In be a sequence of k-cells in Rfc 3

IiDl 2 D---.Then, ' fl~i W 0.

Theorem 10.2.17 Every k-cells is compact (with d2 metric).

Proof. Let / = [ai,bi] x • • • x \ak,bk} C Rfc be a k-cell. If a\ = b\,... ,ak = bk,
then / consists of one point. Then, / is compact. So assume for at least one
j, dj < bj, j G {1,... k}. Let 5 = [Yll=i(°i — di)^2 }^1 > 0. Suppose / is not
compact. So, there is an open cover {Ga,a G A} of / 3 {Ga} does not have
any finite subcollection the union of whose elements covers /.
Let c% = ^|^. Then, \aubi\ = [a^c,] n [ci:bi.

This way / can be divided into 2fc k-cells Qj 3 |JLi Qc = I-

Also, Vj we have p,q G Qj, d(p,q) < ^S.

Since / cannot be covered by a finite number of Ga's, at least one of the Qj's,

say /] cannot be covered by a finite number of Ga's. Subdivide I\ into 2 cells

by halving each side. Continue this way ... We eventually get a sequence {/„}

of k-cells such that

a) /, C h C • • •;

b) /„ cannot be covered by any finite subcollection of {Ga,x G A} , Vn;

c) p,q £ In => d(p, q) < ^-S, Vn.

By a) |XLi In + $• Let P e 0^=1 In C I, then 3a 0 G A 3 p G Gao. Since
Gan is open, 3r > 0 3 Br{p) c GQ(). Find n 0 G N 3 % < 2n" [i.e. ^ < r].
Show /,,,„ c G„„ : pGfir=i Ai c ^n„- Let p G Jno, by c) d{p,p) < ^5 < r.
=> p G Br(p) C G„„ => /„,„ C Gao and this contradicts to b). Thus, / is
compact. •

Theorem 10.2.18 Consider Rk with d 2 metric, let E C Rfc. Then, the fol-
lowing are equivalent:

(a) E is closed and bounded.
(b) E is compact.
(c) Every infinite subset of E has a limit point which is contained in E.