The Real And Complex Number Systems

In addition, part (b) comes from intuition. The reader may think it twice about distance 0.
2. Here is a good exercise to pratice. The statement is that suppose that K and F are
disjoint subsets in a metric space X, K is compact, F is closed. Prove that there exists a
0 such that dp, q if p K, q F. Show that the conclusion is may fail for two
disjoint closed sets if neither is compact.
Proof: Suppose NOT, i.e., for any 0, there exist p K, andq F such that
dp , q . Let 1/n, then there exist two sequence p n K, andq n F such
that dp n , q n 1/n. Note that p n K, andK is compact, then there exists a
subsequence p nk with lim nk p nk p K. Hence, we consider dp nk , q nk n 1 k
to get a
contradiction. Since
dp nk , p dp, q nk dp nk , q nk n 1 , k
then let n k , wehavelim nk q nk p. Thatis,p is an accumulation point of F which
implies that p F. So, we get a contradiction since K F . That is, there exists a
0 such that dp, q if p K, q F.
We give an example to show that the conclusion does not hold. Let
K x,0 : x R and F x,1/x : x 0, then K and F are closd. It is clear that
such cannot be found.
Note: Two disjoint closed sets may has the distance 0, however; if one of closed sets is
compact, then we have a distance 0. The reader can think of them in R n , and note that
a bounded and closed subsets in R n is compact. It is why the example is given.
4.57 In a metric space S, d, letA and B be disjoint closed subsets of S. Prove that
there exists disjoint open subsets U and V of S such that A U and B V. Hint. Let
gx f A x f B x, in the notation of Exercise 4.56, and consider g 1 ,0 and
g 1 0, .
Proof: Let gx f A x f B x, then by Exercixe 4.56, we have gx is uniformly
continuous on S. So,gx is continuous on S. Consider g 1 ,0 and g 1 0, , and
note that A, B are disjoint and closed, then we have by part (b) in Exercise 4.56,
gx 0ifx A and
gx 0ifx B.
So, we have A g 1 ,0 : U, andB g 1 0, : V.
Discontinuities
4.58 Locate and classify the discontinuities of the functions f defined on R 1 by the
following equations:
(a) fx sin x/x if x 0, f0 0.
sinx
Solution: f is continuous on R 0, and since lim x0 x 1, we know that f has a
removable discontinuity at 0.
(b) fx e 1/x if x 0, f0 0.
Solution: f is continuous on R 0, and since lim x0
e 1/x and lim x0
e 1/x 0,
we know that f has an irremovable discontinuity at 0.
(c) fx e 1/x sin 1/x if x 0, f0 0.
Solution: f is continuous on R 0, and since the limit fx does not exist as x 0,
we know that f has a irremovable discontinuity at 0.