SOLUTION OF HW6
MINGFENG ZHAO
April 10, 2013
1. [10 Points] Let f be a function defined on a closed domain. Show that f is continuous if and only
if the inverse image of every closed set is a closed set.
Proof. (=⇒) Assume that f is continuous on a closed domain A. For any closed subset B of f (A).
Let x0 be a limit point of f −1 (B), then there exists some sequence x1 , x2 , · · · ∈ f −1 (B) such that
lim xn = x0 .
n→∞
Since A is closed, then x0 ∈ A. Since f is continuous, then we know that
lim f (xn ) = f (x0 ).
n→∞
Since B is closed, then f (x0 ) ∈ B, that is, x0 ∈ f −1 (B). Hence we know that f −1 (B) is closed.
(⇐=) Let A be a closed domain, and assume that for any closed set B, we have f −1 (B) is closed. If
f is not continuous on A, then there exists some x0 ∈ A such that f is not continuous at x0 , that is,
there exists some m0 ∈ N such that for any n ∈ N, there exists some xn ∈ A such that |xn − x0 | <
1
n,
but we have
|f (xn ) − f (x0 )| ≥
Let B =
−∞, [f (x0 ) −
1
m0
1
.
m0
[
1
f (x0 ) −
, ∞ , then B is closed. By the assumption, we know
m0
that f −1 (B) is closed in A. Since xn ∈ f −1 (B) and lim xn = x0 , then x0 ∈ f −1 (B), that is, we
n→∞
have
0 = |f (x0 ) − f (x0 )| ≥
1
.
m0
We get a contradiction. Hence f must be continuous on A.
1
2
MINGFENG ZHAO
2. [10 Points] Let A be the set defined by the equations f1 (x) = 0, f2 (x) = 0, · · · , fn (x) = 0, where
f1 , · · · , fn are continuous functions defined on the whole line. Show that A is closed. Must A be
compact?
Proof. For any k = 1, · · · , n, define
Ak = {x ∈ R : fk (x) = 0} = fk−1 ({0}).
Since fk is continuous on R, by the result of Problem 1, we know that Ak is closed for all k =
1, · · · , n. On the other hand, it is easy to see that
A = {x ∈ R : f1 (x) = · · · = fn (x) = 0} =
n
\
Ak
k=1
So we know that A is closed.
Counterexample I: A may be not compact. Let f1 (x) = · · · fn (x) = sign x, then we know that
A = {x ∈ R : sign x = 0} = (−∞, 0].
So A is closed, but A is not compact.
3. [10 Points] Let A be the set defined by the inequalities f1 (x) ≥ 0, f2 (x) ≥ 0, · · · , fn (x) ≥ 0, where
f1 , · · · , fn are continuous defined on the whole line. Show that A is closed. Show that the set defined
by f1 (x) > 0, · · · , fn (x) > 0 is open.
Proof. For any k = 1, · · · , n, define
Ak = {x ∈ R : fk (x) ≥ 0} = fk−1 ([0, ∞)).
SOLUTION OF HW6
3
Since fk is continuous on R, by the result of Problem 1, we know that Ak is closed for all k =
1, · · · , n. On the other hand, it is easy to see that
A = {x ∈ R : f1 (x) = · · · = fn (x) = 0} =
n
\
Ak
k=1
So we know that A is closed.
Let B = {x ∈ R : f1 (x) > 0, · · · , fn (x) > 0}. For any k = 1, · · · , n, define
Bk = {x ∈ R : fk (x) > 0} = fk−1 ((0, ∞)).
Since fk is continuous on R, then we know that Ak is open for all k = 1, · · · , n. On the other hand,
it is easy to see that
B = {x ∈ R : f1 (x) > 0, · · · , fn (x) > 0} =
n
\
Bk
k=1
So we know that B is open.
4. [10 Points] Let f have a jump discontinuity at x0 . Show that if x1 , x2 , · · · is any sequence of points
in the domain of f converging to x0 , with no xj equal to x0 , then the sequence f (x1 ), f (x2 ), · · · has
at most two limit-points.
Proof. Assume that the sequence f (x1 ), f (x2 ), · · · has three different limit-points y1 , y2 , y3 , which
∞
implies that there are three subsequences {xi,k }∞
k=1 , i = 1, 2, 3 of {xk }k=1 such that
lim f (xi,k ) = yi ,
k→∞
i = 1, 2, 3.
Since x0 is a jump discontinuity at x0 , then there exists A, B ∈ R such that
lim f (x) = A,
x→x+
0
and
lim f (x) = B.
x→x−
0
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MINGFENG ZHAO
For any i = 1, 2, 3, then there exists a subsequence of {xi,k } such that this subsequence should be
contained in either (−∞, x0 ) or (x0 , ∞), which implies that either yi = A or yi = B. Hence we know
that {y1 , y2 , y3 } ⊂ {A, B}, which contradicts with the assumption that y1 , y2 and y3 are different.
5. [10 Points] Is the inverse image of a convergent sequence under a continuous function necessarily a
convergent sequence?
Proof. The inverse image of a convergent sequence under a continuous function may not be a convergent sequence.
Counterexample I: Let f (x) = x2 on R, then f is continuous on R. Let yk = 1 for all k ≥ 1, then
lim yk = 0. On the other hand, we know that
k→∞
f −1 ({yk : k ≥ 1}) = {1, −1},
which is not convergent.
6. [10 Points] Show that f −1 (A
S
B) = f −1 (A)
S
f −1 (B) and f −1 (A
T
B) = f −1 (A)
T
f −1 (B) for
any function f . Is the same true of images?
Proof. Claim I: f −1 (A
Since A ⊂ A
S
S
B) = f −1 (A)
B and B ⊂ A
S
S
f −1 (B).
B, then f −1 (A) ⊂ f −1 (A
S
B) and f −1 (B) ⊂ f −1 (A
S
B), which
implies that
f −1 (A)
On the other hand, for any x ∈ f −1 (A
[
S
f −1 (B) ⊂ f −1 (A
[
B), then f (x) ∈ A
S
B).
B, which implies that either f (x) ∈ A
or f (x) ∈ B. If f (x) ∈ A, then x ∈ f −1 (A). If f (x) ∈ B, then x ∈ f −1 (B). In any case, we have
SOLUTION OF HW6
x ∈ f −1 (A)
S
5
f −1 (B). Hence we get
f −1 (A
[
B) ⊂ f −1 (A)
[
f −1 (B).
f −1 (A
[
B) = f −1 (A)
[
f −1 (B).
Hence we have
Claim II: f −1 (A
Since A
T
T
B) = f −1 (A)
B ⊂ A and A
T
T
f −1 (B).
B ⊂ B, then f −1 (A
T
B) ⊂ f −1 (A) and f −1 (A
T
B) ⊂ f −1 (B), which
implies that
f −1 (A
\
B) ⊂ f −1 (A)
On the other hand, for any x ∈ f −1 (A)
T
\
f −1 (B).
f −1 (B), then x ∈ f −1 (A) and x ∈ f −1 (B), that is,
f (x) ∈ A and f (x) ∈ B, which implies that either f (x) ∈ A
T
B. So we have x ∈ f −1 (A
T
B). Hence
we get
f −1 (A)
[
f −1 (B) ⊂ f −1 (A
\
B).
Hence we have
f −1 (A
Claim I: f (A
S
Since A ⊂ A
B) = f (A)
S
S
\
B) = f −1 (A)
\
f −1 (B).
S
B) and f (B) ⊂ f (A
f (B).
B and B ⊂ A
S
B, then f (A) ⊂ f (A
S
B), which implies
that
f (A)
On the other hand, for any y ∈ f (A
S
[
f (B) ⊂ f (A
[
B).
B), that is, there exists some x ∈ A
S
B such that y = f (x). If
x ∈ A, then y = f (x) ∈ f (A). If x ∈ B, then y = f (x) ∈ f (B). In any case, we have y ∈ f (A)
Hence we get
f (A
[
B) ⊂ f (A)
[
f (B).
S
f (B).
6
MINGFENG ZHAO
Therefore, we get
f (A
Counterexample I: It is not true that f (A
[
T
B) = f (A)
B) = f (A)
[
T
f (B)
f (B), but we have f (A
Let f (x) = x2 , A = {1} and B = {−1}, then we know that A
T
T
B) ⊂ f (A)
T
f (B).
B = ∅, and f (A) = f (B) = {1}. So
we know that
f (A
\
B) = ∅ =
6 f (A)
\
f (B) = {1}.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu