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Math 314 A1 Fall 2012 Homework 1 Solutions

Due Thursday Sep. 20

5:30pm (Assignment box CAB 3rd floor)

Problem 1. (3 pts) Suppose we need to prove uniform boundedness of a function f (x) – that is|f (x)| some M > 0 for all x - by contradiction, what should the starting assumption be? Explain informal logic. (Hint: You should first write uniform boundedness into formal expression using ∀ and ∃)

Solution. Uniform boundedness means

∃M > 0 ∀x∈R, |f (x)|M. (1)

Note that this means there is one M  that works for all x. If we write ∀x ∈ R ∃M > 0, then any realfunction can be called “bounded”, since for any given x, we can always take M = |f (x)|+ 1.

Therefore to start a proof by contradiction, we need to “assume f (x) is not uniformly bounded”,which means

∀M > 0 ∃x∈R, |f (x)|> M. (2)

Note that in particular, “not uniformly bounded” does not mean “∃x∈R ∀M > 0, |f (x)|> M ” whichis the same as saying there is x where |f (x)|=∞.

Problem 2. (2 pts) Prove that x > 0 x2> 0 is false.

Proof. As x > 0 x2> 0 means x > 0 x2> 0 and x2> 0 x > 0, or equivalently

∀x > 0, x2> 0 and ∀x such that x2> 0, x > 0, (3)

we only need one particular x which makes one of the two statement false. Take x =−1 makes the 2ndstatement false. Thus ends the proof.

Problem 3. (2 pts) Construct the truth table for (A and B) or B and determine its relation to B.Justify your answer.Solution. We have

A B A and B (A and B) or B

T T T T  

T F F F  

F T F T  

F F F F  

. (4)

It is clear that (A and B) or B are equivalent to B since they have exactly the same truth table. Inother words, as logic statements, [(A and B) or B] = B.

Problem 4. (3 pts) Let A , B , X   be sets. Prove X \(A∩B) = (X \A)∪ (X \B).

Proof. We need to show

X \(A∩B)⊆ (X \A)∪ (X \B) (5)

and

(X \A)∪ (X \B)⊆X \(A∩B). (6)

• X \(A ∩ B) ⊆ (X \A) ∪ (X \B). We need to show that every x ∈ X \(A ∩ B) is an element of (X \A)∪ (X \B), that is either x∈X \A or x∈X \B.

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Let x∈X \(A∩B) be an arbitrary element. There are only two possibilities for x (true for allx, not just those in X \(A∩B)):

1. x A. Since x∈X \(A∩B), we have x∈X  therefore x∈X \A⊆ (X \A)∪ (X \B);

2. x ∈ A. Since x ∈ X \(A ∩ B), we have x A ∩ B which together with x ∈ A gives x B

because by definition

x∈A and x∈B x∈A∩B. (7)

Now combine x B and x∈X  we reach x∈X \B ⊆ (X \A)∪ (X \B).

Summarizing, we have X \(A∩B)⊆ (X \A)∪ (X \B).

• (X \A)∪ (X \B)⊆X \(A∩B). Let x∈ (X \A)∪ (X \B) be an arbitrary element. Then there aretwo possibilities: either x∈X \A or x∈X \B.

1. If  x ∈ X \A, then we have x ∈ X  and x A. Since A ∩ B ⊆ A we have x A ∩ B. Sox∈X \(A∩B).

2. If  x ∈ X \B, then we have x ∈ X  and x B. Since A ∩ B ⊆ B we have x A ∩ B. Sox∈X \(A∩B).

Summarizing, we have (X \A)∪ (X \B)⊆X \(A∩B).

The proof ends.

Remark 1. A common mistake is to argue: x A∩B x A or x B, so .... But note that this is a special case

of what we need to prove (think of a huge X that contains everything). So unless the statementx A∩B x A

or x B is proved separately, the whole argument is circular.

Problem 5. (3 pts) Find infinitely many nonempty sets of natural numbers

N⊃S 1⊃S 2⊃ (8)

such that ∩n=1∞ S n=∅. You need to rigorously justify your claim.

Solution. Take S n= {m∈N: m > n}= {n + 1, n + 2, }.

• First show S n is nonempty. By construction we have n + 1∈S n so it is nonempty.• Next show N⊃S 1⊃S 2⊃ . By definition S 1⊆N. As 1∈N, 1 S 1 we have S 1⊂N. Next we show

S k+1⊂S k for every k∈N. Take any m∈S k+1. By definition of  S k+1 we must have m > k + 1 > k

therefore m ∈ S k. So S k+1 ⊆ S k. Since k + 1 ∈ S k, k + 1 S k+1, we have S k S k+1. ThereforeS k+1⊂S k.

• Finally show ∩n=1∞ S n = ∅. We prove by contradiction. Assume ∩n=1

∞ S n ∅. Then there ism∈∩n=1

∞ S n. However by construction of  S m, m S m. Contradiction.

Problem 6. (4 pts) Let f : X  Y  be a function. Let A, B ⊆X  and S , T ⊆Y . Prove

a) If  A⊆B then f (A)⊆ f (B).

b) If  S ⊆T  then f −1(S )⊆ f −1(T ).

c) Is it true that A⊂B implies f (A)⊂ f (B)? Justify your answer.

d) Is it true that S ⊂T  implies f −1(S )⊂ f −1(T )? Justify your answer.

Proof.

a) To show f (A)⊆ f (B) all we need to do is to show that every y∈ f (A) also belongs to f (B). Takean arbitrary y ∈ f (A). By definition of  f (A) there is x∈A such that y = f (x). Since A ⊆B wehave x∈B. Therefore y ∈ f (B), by definition of the image f (B).

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b) Take an arbitrary x∈ f −1(S ). Then by definition f (x)∈S ⊆ T . So f (x)∈T  x∈ f −1(T ), by

definition of the pre-image f −1.

c) No. Because f  may not be one-to-one. Let f :N N be defined as f (n) = 1 for all n ∈N. LetA = {1, 2}, B = {1, 2, 3}. Then A⊂B but f (A) = {1}= f (B).

d) No. Because f  may not be onto. Let f :R R be f (x) = sin x. Take S ={y∈R:−1 y1} and

T  = {y ∈R:−2 y 2}. Then S ⊂T  but f −1

(S ) =R= f −1

(T ).

Problem 7. (3 pts) Let A⊆X , B ⊆Y  and f : X  Y . Prove that

a) f (f −1(B))⊆B.

b) f −1(f (A))⊇A.

c) If  B ⊆ f (X ), then f (f −1(B)) = B.

Proof.

a) Take an arbitrary y ∈ f (f −1(B)). By definition of the image f (·) there is x ∈ f −1(B) such that

y = f (x). By definition we have x∈ f −1(B) means y = f (x)∈B. Therefore y∈ f (f −1(B)) y∈B

so a) is proved.

b) Take an arbitrary x ∈ A. Then f (x) ∈ f (A). By definition1 of  f −

1 we have x ∈ f −

1(f (A)). SoA⊆ f −1(f (A)).

c) In a) we already proved f (f −1(B)) ⊆ B, so to show f (f −1(B)) = B we need to prove further

B⊆ f (f −1(B)). Take any y∈B. Since B⊆ f (X ) we have y∈ f (X ). Then there is x∈X  such thatf (x) = y. By definition of  f −1, f (x) = y ∈B x∈ f −1(B). This means y = f (x)∈ f (f −1(B)).So y∈B y ∈ f (f −1(B)) which means B ⊆ f (f −1(B)).

1. In the notes I define the preimage as

f −1(S ) {a∈A:∃b∈S  such that f (a) =b} (9)

it is the same asf −1(S ) {a∈A: f (a)∈S }. (10)

Feel free to use the latter one (as definition of preimage).

Due Thursday Sep. 20 3