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Math 310: Solutions to Homework 4
- Page 117 problem 12.
Suppose A ⊆ Z. Write the following statement entirely in symbols using the quan-
tifiers ∀ and ∃. Write out the negative of this statement in symbols.
There is a greatest number in the set A.
In preparation, first note: An element m is the greatest element in a set of num-
bers A iff it is greater or equal to all the elements of the set. (We can’t use strictly
greater than, because m itself is an element of the set and certainly m 6 > m!)
Statement, in symbols: ∃m ∈ A, ∀a ∈ A, m ≥ a.
Negation of statement: ∀m ∈ A, ∃a ∈ A, m < a.
Give an example of a set A for which this statement is true. Give another ex-
ample for which it is false.
The statement is true for any set A that is finite or bounded above. For example,
A = { 1 , 2 , 3 } has a greatest element (m = 3), and so does the set
Z
− = {− 1. − 2 , − 3 , − 4 , ...} (here m = − 1 ).
The statement is false for any subset of Z which is not bounded above, such as
A = Z, A = Z
, or A = { 2 k|k ∈ Z}.
- Use the formal definition of the limit of a sequence to show that lim
n→∞
n
2
PROOF:
By the formal definition of the limit of a sequence, lim
n→∞
n
2
= 0 if and only if:
∀ > 0 , ∃N ∈ Z
, ∀n ≥ N, |
n
2
Let be any positive real number, and let N be some integer greater than
1 √
Then for all n ≥ N >
1 √
we have that
n >
Multiplying both sides of this inequality by
> 0 and dividing by n > 0 , we
obtain:
n
Since both sides of this inequality are positive, squaring both sides preserves the
direction of the inequality:
n
2
i.e. |
n
2
QED.
- Use the (negation of the) formal definition of limit of a sequence to prove that
the sequence bn = n
2 does not have limit equal to 0 (that is, bn is NOT what the
book calls a null sequence).
PROOF:
By the formal definition of the limit of a sequence, lim
n→∞
n
2 = 0 iff
∀ > 0 , ∃N ∈ Z
, ∀n ≥ N, |n
2 | < ,
and therefore lim
n→∞
n
2
6 = 0 iff
∃ > 0 , ∀N ∈ Z
, ∃n ≥ N, |n
2 |>.
First note that for all n ≥ 1 we have n
2 ≥ (n)(1) ≥ 1.
Let = 1. Then |n
2 | = n
2 ≥ 1 = for all n ∈ Z
. This shows that lim
n→∞
n
2 6 = 0. QED.
- Page 118 problem 16 parts ii, iii and v.
Determine which of the following functions f i
: R → R are injective, which are
surjective and which are bijective. Write down an inverse of each of the bijec-
tions.
(ii) f 2 (x) = x
3 .
- INJECTIVE: Suppose f 2 (x 1 ) = f 2 (x 2 ), i.e. x
3
1
= x
3
2
. Taking cube root of both
sides, we get x 1
= x 2
. This shows that f 2
is injective.
- SURJECTIVE: For all y ∈ R, there exists x =
3
y which is also a real number
and which satisfies f (x) = f (
3
y) = (
3
y)
3 = y. Hence the function is surjective.
- Since f 2 is both surjective and injective, it is BIJECTIVE. Consequently, it
has an inverse, f
− 1
2
: R → R, f
− 1
2
(x) =
3
x.
(iii) f 3
(x) = x
3 − x.
- NOT Injective: There exist x 1
= 0 ∈ R and x 2
= 1 ∈ R such that x 1
= x 2
but
f 3 (x 1 ) = f 3 (x 2 ) = 0. This shows that f 3 is not injective.
- SURJECTIVE: Since f 3 (x) = x
3 −x is a continuos function on R, with limx→−∞ f 3 (x) =
−∞ and lim x→+∞
f 3
(x) = +∞, by the Intermediate Value Theorem, Im(f 3
) = R.
That is, f 3 is surjective.
- Since f 2 is not injective, it is not bijective and, consequently, it has no inverse.
(v) f 5
(x) = e
x .
- INJECTIVE: If f 5
(x 1
) = f 5
(x 2
), then e
x 1 = e
x 2
. Applying natural log to both
sides of the equality, x 1
= x 2
. This shows that f 5
is injective.
- NOT Surjective: There exists − 1 ∈ R such that − 1 6 = e
x ∀x ∈ R. That is, f 5
is
not surjective. (In fact, Imf 5
= R
≥ ⊂ R)
- Since f 5 is not surjective, it is not bijective and, consequently, it has no inverse.
Tip: We need to prove a set inclusion, namely that
f (A 1
f (A 2
) (subject to the
hypothesis that A 1 ⊆ A 2 .) So we will start with an arbitrary element y in
f (A 1 ),
and show that y ∈
f (A 2
PROOF: Let y ∈
f (A 1 ). By definition of
f (A 1 ), this means that there exists some
x ∈ A 1
such that y = f (x). But A 1
⊆ A
2
, so x ∈ A 1
⇒ x ∈ A 2
. This shows that
y = f (x) for x ∈ A 2 (y is the image through f of an element in A 2 ), which is
equivalent to saying y ∈
f (A 2
). This proves the desired set inclusion.
The converse is not universally true.
For example, let f : R → R, f (x) = x
2
. Let A 1 = [− 1 , 0] and A 2 = [0, 2]. Then
f (A 1
f ([− 1 , 0]) = [0, 1] and
f (A 2
f ([0, 2]) = [0, 4],
so
f (A 1 ) ⊆
f (A 2 ) holds, but A 1 = [− 1 , 0] 6 ⊆ [0, 2] = A 2.
The converse holds if and only if f is injective
Proof of this was not asked for, but here it is anyway, for your fun and enjoyment:
Proof of ”f injective ⇔ converse holds”
”⇒” (”f injective ⇒ [
f (A 1
f (A 2
) ⇒ A
1
⊆ A
2
]”)
Let x 1
be any element of A 1
. Then its image y 1
= f (x 1
) is an element of
f (A 1
Since
f (A 1
f (A 2
), y 1
f (A 1
) ⇒ y 1
f (A 2
). This implies that y 1
= f (x 2
) for some
x 2 ∈ A 2. But f (x 1 ) = f (x 2 ) (since both equal y 1 ), and, if f is injective, it folows
that x 1
= x 2
. Hence, x 1
∈ A
2
This shows A 1 ⊆ A 2 , and hence it proves that f injective ⇒ converse holds.
”⇐”: For the reverse implication, note that if f is not injective, then there
exists x 1
= x 2
∈ X such that f (x 1
) = f (x 2
). Let y be the element of Y equal to
f (x 1 ) = f (x 2 ). Then
f ({x 1 }) =
f ({x 2 }) = {y} but {x 1 } 6 ⊆ {x 2 }. This shows the contra-
positive of the reverse implication, namely: ”f not injective ⇒ converse not true”.
- Let f : X → Y be a function. Prove that
a) f is injective ⇔
f is injective.
SOLUTION:
There are 2 implications to prove. Often, it is easier to work with the contrapos-
itive of the implications instead.
i) Let’s first show ”f is injective ⇒
f is injective”
PROOF:
Suppose there exists two subsets of X, say A 1
and A 2
, such that
f (A 1
f (A 2
We want to prove that A 1
= A
2
, since this would show that
f is injective.
Since we want to prove a set equality, take an arbitrary x 1
∈ A
1
. We want to show
that x 1 ∈ A 2. If this holds, then it follows that A 1 ⊆ A 2. By the symmetry of the
statement, the same argument but with A 1
and A 2
interchanged would show that
A 2 ⊆ A 1 , and thus we have our desired set equality.
So let x 1 ∈ A 1. Define y 1 = f (x 1 ). By definition of
f (A 1 ), since x 1 ∈ A 1 , y 1 ∈
f (A 1 ).
But
f (A 1
f (A 2
), so y 1
f (A 2
). That is, there exists some x 2
∈ A
2
such that
y 1 = f (x 2 ). But since f is injective, y 1 = f (x 1 ) = f (x 2 ) ⇒ x 1 = x 2. This implies that
x 1
∈ A
2
. QED.
ii) ”
f is injective ⇒ f is injective”, or, equivalently,
”f not injective ⇒
f not injective”
PROOF:
Suppose that f : X → Y is not injective. By definition, this means that there
exists x 1
= x 2
elements of X, such that f (x 1
) = f (x 2
). Let y 1
∈ Y denote the
element f (x 1 ) = f (x 2 ) ∈ Y. Then there exist subsets A 6 = B of X (namely, A = {x 1 },
B = {x 2
}) such that
f (A) =
f (B) = {y 1
}. Hence
f is not injective. QED
From (i) and (ii) we see that f is injective if and only if
f is injective.
Prove: b) f is injective ⇔
←
f is surjective.
SOLUTION:
Again, we need to prove two implications:
ı) ”f is injective ⇒
←
f is surjective”
PROOF:
To show that
←
f : P (Y ) → P (X) is surjective, we need to show that
for any A ∈ P (X), there exists some B ∈ P (Y ) such that A =
←
f (B).
By definition of power sets and
←
f , this means that we want to prove that
for any A ⊆ X, there exists some B ⊆ Y such that A = {x ∈ X|f (x) ∈ B}.
Let A be any subset of X. Take the set B to be B =
f (A) = {f (x)|x ∈ A} ⊆ Y.
Then, by definition of B, A ⊆ {x ∈ X|f (x) ∈ B}. On the other hand, suppose for
some x
′ we have y = f (x
′ ) ∈ B. By definition of B, there exists some x ∈ A with
y = f (x). By the injectivity of f , x
′ = x, so x
′ ∈ A. This shows the reverse set
inclusion: A ⊇ {x ∈ X|f (x) ∈ B}.
Since both set inclusions hold, A = {x ∈ X|f (x) ∈ B}, and hence
←
f is surjective.
QED.
