MELENDREZ, Romeo Dane MATH 55.
1
OLMON, Christian Cedrick July 12, 2020
VITUG, Mikko Problem Set 1
1
1. Let ∆ = {x ∈ R : x ≥ 1}. Determine ∪α∈∆ α, α + α
. Explain your answer:
Solution:
Claim: ∪α∈∆ (α, α + α1 ) = (1, +∞)
Proof:
Part 1 : Let x ∈ ∪α∈∆ (α, α + α1 ). Then 1 ≤ k < x < k + k1 for some k ∈ ∆. This
means that x ∈ (1, k + k1 ), but since (1, k + k1 ) ⊆ (1, +∞) then x ∈ (1, +∞). So
∪α∈∆ (α, α + α1 ) ⊆ (1, +∞).
Part 2 : Let y be an arbitrary number such that y ∈ (1, +∞). We have two cases:
Case 1: y ∈ (1, 2)
If y ∈ (1, 2), then it’s easy to see that (1, 2) ⊆ ∪α∈∆ α, α + α1 .
So y ∈
∪α∈∆ α, α + α1
Case 2: y ∈ [2, +∞)
1 1
Consider y(x) = x + 2x . So for any b ∈ R − {0}, we have limx→b x + 2x =
1 1
b + 2b = y(b). So y = x + 2x is continuous for all x ≥ 1. This also means that y is
dy
differentiable for x > 1, and since dx = 1− 2x12 > 0 for all x > 1, y must be strictly
increasing. This means that y can take up any value starting from y = 1 + 1/2
(the initial value when x = 1) and all the numbers higher than 32 .
1
From here, notice that x < x+ 2x < x+ x1 for all x ≥ 1, so this means x < y <x+ x1 .
Therefore y ∈ x, x + x1 . However, 1
notice that x, x + x ⊆ ∪α∈∆ α, α + α , this
1
1
means that y ∈ ∪α∈∆ α, α + α .
In either case, we have y ∈ ∪α∈∆ α, α + α1 . So (1, +∞) ⊆ ∪α∈∆ α, α + α1
1
Combining the results from Part 1 and Part 2, we get that ∪ α∈∆ α, α + α ⊆ (1, +∞)
and (1, +∞) ⊆ ∪α∈∆ α, α + α1 . Therefore ∪α∈∆ α, α + α1 = (1, +∞).
2. Let A, B be non-empty sets such that f : A → B and S, T ⊆ A. Show that f (S ∩ T ) ⊆
f (S) ∩ f (T ). Is it possible to have f (S ∩ T ) ⊂ f (S) ∩ f (T )? Justify your answer.
Solution:
Let y ∈ f (S ∩ T ). By definition, we know that f (S ∩ T ) = {f (k) : k ∈ S ∩ T }, so
∃x ∈ S ∩ T such that y = f (x). Since x ∈ S ∩ T , it follows that x ∈ S and x ∈ T .
Also by definition, we have f (S) = {f (s) : s ∈ S} and f (T ) = {f (t) : t ∈ T }, which
means f (x) ∈ f (S) and f (x) ∈ f (T ), or equivalently f (x) ∈ f (S) ∩ f (T ). Finally,
since f (x) = y, then y ∈ f (S) ∩ f (T ).
∴ f (S ∩ T ) ⊆ f (S) ∩ f (T )
Now, answering the question, the answer is yes, it is possible. Consider f : R → R
such that f (x) = x2 . Let S = {1, 3, 5} and T = {2, 4, 6}, then f (S ∩ T ) = ∅ and
f (S) ∩ f (T ) = {1, 4, 9, 16, 25, 36} which in this case, f (S ∩ T ) ⊂ f (S) ∩ f (T )
2x
3. Let f : R − {1} → R − {2} with f (x) = x−1
for each x ∈ R − {1}. Show that f is a
one-to-one correspondence.
1 of 2
MELENDREZ, Romeo Dane MATH 55.1
OLMON, Christian Cedrick July 12, 2020
VITUG, Mikko Problem Set 1
Solution:
Suppose f (a) = f (b) for some arbitrary a, b ∈ R − {1}. Then we have:
2a 2b
=
a−1 b−1
2a (a − 1)(b − 1) 2b (a − 1)(b − 1)
· = ·
a−1 2 b−1 2
a(b − 1) = b(a − 1)
ab − a = ab − b
−a = −b
a=b
Since a and b were arbitrary, this holds ∀a, b ∈ R − {1}. Thus f is a one-to-one
correspondence.
4. Let A, B, C, D be sets that satisfy the following:
A ∪ B ⊆ C ∪ D, A ∩ B = ∅, C ⊆ A.
Show that B ⊆ D.
Solution:
Let x ∈ B. Then the following statements are true:
x∈B ⇒x∈A∪B
⇒ (x ∈ C ∪ D) ∧ (x ∈ / A) (since A ∪ B ⊆ C ∪ D and
A ∩ B = ∅ respectively)
x∈
/A⇒x∈ / C (since C ⊆ A)
⇒ x ∈ D (since x ∈ C ∪ D)
x∈B⇒x∈D
∴B⊆D
5. Let f : A → B, g : B → A such that f ◦ g = iB , the identity function on B. Is it
possible that f −1 and g −1 do not exist? Justify your answer.
Solution:
The question only asks if it’s possible for f −1 and g −1 to not exist, so providing an
example will be enough to solve the problem.
Consider A = R and B = Z, so f : R → Z and g : Z → R such that f (x) = bxc
and g(x) = x. This follows (f ◦ g)(x) = x. However notice that f is not one-to-one
(counter example is b1.5c = b1.2c = 1), and g is not onto (Since g can only have integer
outputs, all non-integer elements in the codomain R will have no corresponding value
in the domain). In this case f −1 and g −1 don’t exist. Therefore the answer is yes, it is
possible.
2 of 2
