Graphics Reference
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Exercise 4.2.6 shows that the definitions may have some unexpected consequences in
certain cases.
The boundary and interior of a set change if we change the containing space R n .
For example, the boundary of [0,1] in R 2
is all of [0,1] and its interior is the empty
set.
4.2.3. Proposition
(1) The boundary of a set is a closed set.
(2) If X Õ R n , then bdry( R n - X ) = bdry( X ).
(3) A set X is closed if and only if bdry( X ) Õ X .
(4) The interior of a set is an open set.
(5) int( X ) = X - bdry( X ).
Proof.
Easy.
Definition. A subset X of R n is called a bounded set if X Õ B n (r) for some r > 0. If
X is not bounded, then it is said to be unbounded .
For example, the set of integers is an unbounded set in R . Of course, the whole
set R is an unbounded subset of R . The interval [-1,10] is a bounded set because it
is contained in (-50,50) = B 1 (50).
Definition. Let X be a subset of R n . If S = { U a } is a collection of subsets of R n whose
union contains X , then S is called a cover of X . If all the U a are open sets in R n , then
S is called an open cover of X . If all the U a are closed sets in R n , then S is called a
closed cover of X .
For example, {(1/n,1 - 1/n) | n = 2, 3,...} is an open cover of the subset (0,1) of
R .
Definition. A subset X of R n is said to be compact if every open cover of X has a
finite subcover , that is, there is a finite collection of sets from the cover that already
cover X .
4.2.4. Theorem. (The Heine-Borel-Lebesque Theorem) A subset X of R n is compact
if and only if it is closed and bounded.
Proof. We shall prove half of the theorem, namely, that a compact set is closed and
bounded, and leave the converse to the next chapter (Theorem 5.5.6). One reason for
proving at least part of the theorem here is to show how the property of covers having
finite subcovers gets used. Basically, when one has a finite collection of objects, e.g.,
numbers, then one can talk about the smallest or largest. This is not possible with
infinite collections.
Assume that X is a compact subset of R n . To prove that X is closed we need to
show that R n - X is open. Let p ΠR n - X . For every x ΠX there is a ball neighbor-
hood U x and V x of p and x , respectively, such that U x « V x = f. See Figure 4.3. The
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