PART II: lim inf and lim sup equal L implies that the limit equals L
4.5 The Arithmetic of Continuous Functions
(c) Œ0; 1 is covered by the collection of open intervals
2j1;2j
for j D1; 2; 3; : : : along with
101;101 .
Write proofs of each of the following statements.
3. The intersection of two compact sets is a compact set.
4. The union of two compact sets is a compact set.
5. If C is a compact set and.a; b/ is an open interval, then the set difference Cn.a; b/
is a compact set.
6. If the function f is uniformly continuous on the intervalŒa; b and uniformly continuous on the intervalŒb; c for a < b < c, then f is uniformly continuous on the intervalŒa; c.
7. If a set A has a cover consisting of a finite number of open intervals, then A has a subcover such that for each x 2 A, x is an element of at most two of the open intervals in the subcover.
4.5 The Arithmetic of Continuous Functions
Chapter3discusses several theorems about how one can calculate limits when faced with the addition, subtraction, multiplication, or division of functions whose limits are known. As one might expect, since continuity and limits are closely related, the proofs of the corresponding theorems about functions continuous at a point are, in fact, very similar. Before starting, it is worth pointing out that if f and g are two functions, then you can define the new functions f C g, f g, f g, and gf at all
118 4 Continuity points in the intersection of the domain of f and the domain of g and, in the case of gf, only where g is not 0. Generally, one is interested in functions that have a common domain, but sometimes this is not the case. Pathological examples do exist.
It could be, for example, that f is only defined for positive real numbers, and g is only defined for negative real numbers as with f.x/ D p1x and g.x/ D p1x. Then f C g has an empty domain and is the empty function, one that contains no ordered pairs,
x; f .x/
. Oddly, the definition of continuity says that the empty function is continuous because it satisfies the definition at each point of its empty domain.
Suppose that functions f and g have a common domain where the point a is an accumulation point of that domain. Also suppose that lim
x!af.x/ D L and lim with minor changes made to match the template for writing proofs about continuity of a function at a point. Of course, the same logic works for proving the continuity of f g, so the two results might as well be combined as follows.
PROOF: Suppose that f and g are functions with common domain containing the point a. If both f and g are continuous at the point a, then so are the functions fC g and f g.
• Let f and g be functions both defined on a set A containing the point a, and assume that f and g are both continuous at a.
• Let > 0 be given.
Now suppose that f and g are functions as discussed above with lim
x!af.x/ D L
jf .x/ Lj is less than 1, ı2> 0 so that jf .x/ Lj is less than 2.jHjC1/ , andı3so that jg.x/ Hj is less than 2.jLjC1/ . Thenı can be set to the least of ı1,ı2, andı3. The proof for continuity of fg at the point a follows this same strategy.
PROOF: Suppose that f and g are functions with common domain containing the point a. If both f and g are continuous at the point a, then so is the function fg.
• Let f and g be functions both defined on a set A containing the point a, and assume that f and g are both continuous at a.
• Let > 0 be given.
• By the definition of continuity, there is aı1 > 0 such that if x 2 A and jx aj < ı1, then jf.x/ f .a/j < 1, and thus, jf .x/j < jf .a/j C 1.
• There is aı2 > 0 such that if x 2 A and jx aj < ı2, then jf.x/ f .a/j <
2.jg.a/jC1/ .
• There is aı3 > 0 such that if x 2 A and jx aj < ı3, then jg.x/ g.a/j <
2.jf .a/jC1/ .
• Letı D min.ı1; ı2; ı3/.
• Then if x 2 A with jx aj< ı,
• jf.x/g.x/ f .a/g.a/j D jf .x/g.x/ f .x/g.a/ C f .x/g.a/ f .a/g.a/j jf .x/j jg.x/ g.a/j C jg.a/j jf .x/ f .a/j <
jf .a/j C 1
2.jf .a/jC1/C jg.a/j 2.jg.a/jC1/ 2C 2 D .
• This shows that fg is continuous at a.
Finally, suppose that f and g are functions as discussed above with lim
x!af.x/ D L and lim
x!ag.x/ D H and H ¤ 0. This time recall how you prove that lim
x!a f.x/
g.x/ D HL. The idea is the same as with the proof for products, but the algebra took a few more steps. ˇˇˇfg.x/.x/HLˇˇˇ D ˇˇˇf.x/HLg.x/
g.x/H ˇˇˇ D ˇˇˇf.x/HLHCLHLg.x/
g.x/H ˇˇˇ D ˇˇˇˇ.f .x/L/HCL
Hg.x/
g.x/H
ˇˇˇˇ jf .x/Ljjg.x/j C jLjjg.x/Hj
jg.x/jjHj . Then, given an > 0, you can choose ı1 > 0 so that jx aj < ı1 would ensure jg.x/ Hj < jHj2 which, in turn, implies that jg.x/j > jHj2 . Then you choose aı2 > 0 so that jx aj < ı2 gives jf.x/ Lj < jHj4 . Lastly, you choose aı3 > 0 so that jx aj < ı3 gives jg.x/ Hj < 4.jLjC1/H2 . This allowed you to concludeˇˇˇf.x/HLg.x/
g.x/H ˇˇˇ < . Again, the proof for continuity can be constructed by changing the limit L to jf.a/j, the limit H to jg.a/j, and making some other minor wording changes.
120 4 Continuity
PROOF: Suppose that f and g are functions with common domain containing the point a with g.a/ ¤ 0. If both f and g are continuous at the point a, then so is the function fg.
• Let f and g be functions both defined on a set A containing the point a, and assume that f and g are both continuous at a with g.a/ ¤ 0.
• Let > 0 be given.
• Note that jg.a/j > 0. By the definition of continuity, there is a ı1 > 0 such that if x 2 A and jx aj< ı1, then jg.x/ g.a/j < jg.a/j2 . For these x it follows that jg.x/jCjg.a/j2 > jg.x/jCjg.x/g.a/j D jg.x/jCjg.a/g.x/j ˇˇg.x/ C
g.a/ g.x/ˇˇ D jg.a/j which implies that jg.x/j > jg.a/jjg.a/j2 D
jg.a/j 2 .
• By the definition of continuity, there is aı2 > 0 such that if x 2 A and jx aj < ı2, then jf.x/ f .a/j <jg.a/j4 .
• By the definition of continuity, there is aı3 > 0 such that if x 2 A and jx aj < ı3, then jg.x/ g.a/j < 4.jf .a/jC1/g.a/2 .
• Letı D min.ı1; ı2; ı3/.
• Then if x 2 A with0 < jx aj < ı,
• ˇˇˇfg.x/.x/gf.a/.a/ˇˇˇ Dˇˇˇf.x/g.a/f .a/g.x/
g.x/g.a/ ˇˇˇ Dˇˇˇf.x/g.a/f .a/g.a/Cf .a/g.a/f .a/g.x/
g.x/g.a/ ˇˇˇ D ˇˇˇˇ.f .x/f .a//g.a/Cf .a/
g.a/g.x/
g.x/g.a/
ˇˇˇˇ jf .x/f .a/j
jg.x/j Cˇˇˇf.a/.g.x/g.a//
g.x/g.a/ ˇˇˇ <
jg.a/j
4 jg.a/j2 C 4.jf .a/jC1/g.a/2 jg.a/j2jf .a/j2 < 2C 2 D .
• This shows that fg is continuous at a.
4.5.1 Exercises
1. Suppose that f and g are functions that are both uniformly continuous of a set A. Find an example showing that their product need not be uniformly continuous on A.
Write proofs for each of the following statements.
2. The function f.x/ D x52 is continuous for x 0.
3. All polynomials are continuous onR.
4. All rational functions are continuous onR except at points where their denomi-nators are 0.
5. If f and g are uniformly continuous on the set A, then f C g and f g are also uniformly continuous on A.
6. Suppose f and g have common domain A and f C g is continuous at a 2 A. If f is discontinuous at a, then g is discontinuous at a.
7. Suppose f and g have common domain A and fg is continuous at a 2 A. If f is discontinuous at a, then g is either discontinuous at a or g.a/ D 0.