ALL 8 CHAPTERS COVERED
Solutions to Problems
,Christopher Heil
Metrics, Norms, Inner Products
and Operator Theory
Solutions to Selected Problems
November 9, 2018
c 2018 by Christopher Heil
,Solutions c 2018 Christopher Heil 1
Solutions to Problems
These are my solutions to the exercises from my text “Metrics, Norms, Inner
Products, and Operator Theory.” Of course, many problems have solutions
other than the ones I sketch here (in particular, there could very well be easier
solutions than the ones I give). These solutions have not been proofread as
carefully as has the text proper, so the probability of errors is correspondingly
higher. Please send comments and corrections to “”.
S
1.2.2 If x ∈ lim sup Ek , then x belongs to ∞ k=j Ek for every j ∈ N. Therefore,
for each j, there must exist some k ≥ j such that x ∈ Ek . If x belonged
to only finitely many Ek , then this couldn’t happen, so x must belong to
infinitely many of the set Ek . This reasoning is reversible, so we conclude
that lim sup Ek precisely consists of the points that lie in infinitely many
T Ek .
If x ∈ lim inf Ek , then there is some j such that x belongs to ∞ k=j k .
E
Hence for that j we have x ∈ Ek for every k ≥ j. Again, this reasoning is
reversible.
1.3.2 If x ∈ f −1 (B C ), then f (x) ∈ B C . Therefore f (x) ∈ Y but f (x) ∈
/ B, so
x∈ / f −1 (B). Therefore x ∈ f −1 (B)C . This shows that f −1 (B C ) ⊆ f −1 (B)C .
For the other inclusion, suppose that x ∈ f −1 (B)C . Then x ∈ X but
x∈ / f −1 (B). Therefore f (x) ∈ / B, so f (x) ∈ B C . Hence x ∈ f −1 (B C ). This
shows that f (B) ⊆ f (B C ).
−1 C −1
The other equalities are proved similarly.
To show that f (f −1 (B)) ⊆ B, suppose that y ∈ f (f −1 (B). Then, by
definition of direct image, there exists a point x ∈ f −1 (B) such that f (x) =
y. By definition of inverse image, the fact that x ∈ f −1 (B) implies that
f (x) ∈ B. Hence y = f (x) ∈ B. This shows that f (f −1 (B)) ⊆ B.
Assume in addition now that f is surjective, and choose any point y ∈ B.
Since f is surjective, there is some x ∈ X such that f (x) = y. By definition
of the inverse image, we have x ∈ f −1 (B). Consequently, by definition of the
direct image, y = f (x) ∈ f (f −1 (B)). This shows that B ⊆ f (f −1 (B)). The
converse inclusion was proved above, so equality holds.
Now consider f : N → N defined by f (x) = 2x. Let E be the set of pos-
itive even integers and O the set of odd integers. Then N = f −1 (N), but
f (f −1 (N)) = f (N) = E 6= N. Hence equality need not hold if f is not surjec-
tive.
1.3.3 Unions. Choose any point
S
y ∈ f Ai .
i∈I
Then y = f (x) for some x ∈ ∪Ai . Hence x ∈ Ai for some i, so y ∈ f (Ai ) for
that i, so S
y ∈ f (Ai ).
i∈I
, 2 Solutions c 2018 Christopher Heil
Conversely, if S
y ∈ f (Ai ),
i∈I
Then y ∈ f (Ai ) for some i, so there is some point x ∈ Ai such that y = f (x).
Since x ∈ ∪Ai , it follows that
S
y ∈ f Ai .
i∈I
Intersections. Choose any point
T
y ∈ f Ai .
i∈I
Then y = f (x) for some x ∈ ∩Ai . This point x belongs to every Ai , so
y ∈ f (Ai ) for every i. Therefore
T S
f Ai ⊆ f (Ai ).
i∈I i∈I
On the other hand, suppose that f is injective and choose any point
S
y ∈ f (Ai ).
i∈I
Then y ∈ f (Ai ) for every i, so for each i there exists some point xi ∈ Ai such
that f (xi ) = y. However, since f is injective there is a unique point x such
that f (x) = y. Therefore we must have xi = x for every i. Hence y = f (x)
and x = xi ∈ Ai , so y ∈ f (Ai ) for every i. Therefore, for injective functions
we have
S T
f (Ai ) ⊆ f Ai .
i∈I i∈I
However, if f is not injective then the preceding inclusion need not hold.
For example, define f : R → R by f (x) = |x|. Then
f [−2, 1] ∩ [−1, 2] = f [−1, 1] = [0, 1],
while
f [−2, 1] ∩ f [−1, 2] = [0, 2] ∩ [0, 2] = [0, 2].
Complements. Choose any point y ∈ f (X)\f (A). Then y = f (x) where
x ∈ X, but y 6= f (z) for any z ∈ A. If we had x ∈ A, then we would have
y = f (x) with x ∈ A, which is a contradiction. Therefore x ∈
/ A, so we have
y = f (x) where x ∈ AC . Thus y ∈ f (AC ). Therefore
f (X)\f (A) ⊆ f (AC ).
Solutions to Problems
,Christopher Heil
Metrics, Norms, Inner Products
and Operator Theory
Solutions to Selected Problems
November 9, 2018
c 2018 by Christopher Heil
,Solutions c 2018 Christopher Heil 1
Solutions to Problems
These are my solutions to the exercises from my text “Metrics, Norms, Inner
Products, and Operator Theory.” Of course, many problems have solutions
other than the ones I sketch here (in particular, there could very well be easier
solutions than the ones I give). These solutions have not been proofread as
carefully as has the text proper, so the probability of errors is correspondingly
higher. Please send comments and corrections to “”.
S
1.2.2 If x ∈ lim sup Ek , then x belongs to ∞ k=j Ek for every j ∈ N. Therefore,
for each j, there must exist some k ≥ j such that x ∈ Ek . If x belonged
to only finitely many Ek , then this couldn’t happen, so x must belong to
infinitely many of the set Ek . This reasoning is reversible, so we conclude
that lim sup Ek precisely consists of the points that lie in infinitely many
T Ek .
If x ∈ lim inf Ek , then there is some j such that x belongs to ∞ k=j k .
E
Hence for that j we have x ∈ Ek for every k ≥ j. Again, this reasoning is
reversible.
1.3.2 If x ∈ f −1 (B C ), then f (x) ∈ B C . Therefore f (x) ∈ Y but f (x) ∈
/ B, so
x∈ / f −1 (B). Therefore x ∈ f −1 (B)C . This shows that f −1 (B C ) ⊆ f −1 (B)C .
For the other inclusion, suppose that x ∈ f −1 (B)C . Then x ∈ X but
x∈ / f −1 (B). Therefore f (x) ∈ / B, so f (x) ∈ B C . Hence x ∈ f −1 (B C ). This
shows that f (B) ⊆ f (B C ).
−1 C −1
The other equalities are proved similarly.
To show that f (f −1 (B)) ⊆ B, suppose that y ∈ f (f −1 (B). Then, by
definition of direct image, there exists a point x ∈ f −1 (B) such that f (x) =
y. By definition of inverse image, the fact that x ∈ f −1 (B) implies that
f (x) ∈ B. Hence y = f (x) ∈ B. This shows that f (f −1 (B)) ⊆ B.
Assume in addition now that f is surjective, and choose any point y ∈ B.
Since f is surjective, there is some x ∈ X such that f (x) = y. By definition
of the inverse image, we have x ∈ f −1 (B). Consequently, by definition of the
direct image, y = f (x) ∈ f (f −1 (B)). This shows that B ⊆ f (f −1 (B)). The
converse inclusion was proved above, so equality holds.
Now consider f : N → N defined by f (x) = 2x. Let E be the set of pos-
itive even integers and O the set of odd integers. Then N = f −1 (N), but
f (f −1 (N)) = f (N) = E 6= N. Hence equality need not hold if f is not surjec-
tive.
1.3.3 Unions. Choose any point
S
y ∈ f Ai .
i∈I
Then y = f (x) for some x ∈ ∪Ai . Hence x ∈ Ai for some i, so y ∈ f (Ai ) for
that i, so S
y ∈ f (Ai ).
i∈I
, 2 Solutions c 2018 Christopher Heil
Conversely, if S
y ∈ f (Ai ),
i∈I
Then y ∈ f (Ai ) for some i, so there is some point x ∈ Ai such that y = f (x).
Since x ∈ ∪Ai , it follows that
S
y ∈ f Ai .
i∈I
Intersections. Choose any point
T
y ∈ f Ai .
i∈I
Then y = f (x) for some x ∈ ∩Ai . This point x belongs to every Ai , so
y ∈ f (Ai ) for every i. Therefore
T S
f Ai ⊆ f (Ai ).
i∈I i∈I
On the other hand, suppose that f is injective and choose any point
S
y ∈ f (Ai ).
i∈I
Then y ∈ f (Ai ) for every i, so for each i there exists some point xi ∈ Ai such
that f (xi ) = y. However, since f is injective there is a unique point x such
that f (x) = y. Therefore we must have xi = x for every i. Hence y = f (x)
and x = xi ∈ Ai , so y ∈ f (Ai ) for every i. Therefore, for injective functions
we have
S T
f (Ai ) ⊆ f Ai .
i∈I i∈I
However, if f is not injective then the preceding inclusion need not hold.
For example, define f : R → R by f (x) = |x|. Then
f [−2, 1] ∩ [−1, 2] = f [−1, 1] = [0, 1],
while
f [−2, 1] ∩ f [−1, 2] = [0, 2] ∩ [0, 2] = [0, 2].
Complements. Choose any point y ∈ f (X)\f (A). Then y = f (x) where
x ∈ X, but y 6= f (z) for any z ∈ A. If we had x ∈ A, then we would have
y = f (x) with x ∈ A, which is a contradiction. Therefore x ∈
/ A, so we have
y = f (x) where x ∈ AC . Thus y ∈ f (AC ). Therefore
f (X)\f (A) ⊆ f (AC ).
