MATH 325 University of California Irvine Algebra Equations Problems

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< hw1.pdf .. Math 325 Analysis, Section 010 Fall 2021 Homework 1 Due: Monday, September 20th by 11:59 PM ET • Collaboration with other students is highly encouraged, but you must w own solutions independently. • Please make sure your submission is well-written and legible. Typed accepted. . You can use any result proved in the course text, in class, or on a previo question provided you clearly mention the result you are using. Chapter 0 Exercises Problem 1 (5 points). Let A, B, C be sets. Prove the following set relation properties: (i) (Transitivity of set inclusion) If ACB and BCC, then ACC (ii) (Reflexivity of set equality) A = A (iii) (Transitivity of set equality) If A = B and B=C, then A=( Problem 2 (5 points). For each function, determine if it is (i) injective and (ii) surjective. Justify your answer with a proof. 1/3 (a) f:(0,1) + (0,0) where f(x) (b) g: R0,0) where g(x) = x2 Problem 3 (5 points). Show that Z (the set of all integers) is countable. Problem 4 (5 points). Prove Proposition 0.3.16: Consider f : A + B. Let C, D be subsets of A. Then, f(CUD) = f(C) Uf(D) f(CAD) Cf( CD B and sets C, D such that f(COD) D FC). Additionally, find a f Chapter 1 Exercises Problem 5 (5 points). Let E = (a,0) := {x ER: x > a} where a € R. Compute sup E
and inf E if they exist, or prove that E is unbounded above/below if they do not exist.
Justify your answer by proof
(Note: do not use the extended reals for this problem)
Math 325 Analysis, Section 010
Fall 2021
Problem 6 (10 points). Suppose A, B are non-empty sets that are both bounded above
and below, and furthermore that ACB. Prove that
inf B x=y
=> x= + 5y ER
X² is surjective.
then
ges=x² is NOT
injective,
then goo= x²
Chapter 0
Problem 1
(i) Conditions: A CB and BCC
XE
a. assumption
b. We have ACB, this means
if XE A > XE B
given (a) we
have
XEB
e. We have BCC, this means
XE B » XEC
given (b), we have
XEC
d. we have
XEO » Xtc
Therefore, Aço
Set is the subset of it self.
a.
b
b.
AÇA, this means
f XEA
Xt А
9 given (b), we have
A SA
he have
ACA and AZA,
C.
this is equivalent to
А – А
m3
If zB countable, then we can list them as a bejection
with the natural number Set N
Aim: create a
function that bijection between Z and N
The function
can look like this
Z
2
N
O
–
->
f. z>N
O fon fizeco, co> > Necven nueter
fl&s= 22
for Ze8-00, 0) > W odd number
flz= 27+
4
3
> 4
– N M & h
2
2.
Now we pione f. is bijective,
for ZETo, o fiz= 22.
suppose fiz,= fiz) 28,=222 Z=22
this f is injective,
for YEN, y = f1zx= 28 » z = € Z
this fis surjective
then fis bijective.
y
f. Zelo, o) > NE odd number
f(2)= 22 +1
suppose fizi) = f(Za) -> 22,+1=2Z2+1 =>2,= 22
fis injective
suppose y = f(2)=> z= g / Ez
thus fis surjective
then fB bijective
Therefore. Integer set zis countable
Problem 4
B it exists y oficUD)
then, it has an XECUD so that fix=y
– XEC
on XED,
= f(x) E f(C) f(x) & flo)
= f(x) + f(c) u f(B)
= fc cude fic, ufio
Let yt fra) u fio), then y = f(c) on y Efio)
there exists an
x so that fix) = y
hence fix) e f (c) uftos
since
XE CUD
this leads to f(c) Ufo CF (CUD)
thus fic UD) = f(c) ufio
(11) Let xef End), then there’s
yo and such that
fly) = x, Then
YEC, so x= f(y) Efic).
similarly, because
YED, x-fly, & f(D), then for
XE fron flos
thus fcc no) c fccon fo)
Function
Let ‘X=fa, b,c} ,x={1,24
flas= 1 f16=2 fcc)= 1
Cafa, by D=fb.ch
fccnp) = f(b) = 2
f(0) = { 1,29
f(0) = f + 1, 24
froon f(s) = {1, 2}
ficap) & ficnfio)
m3
If zB countable, then we can list them as a bejection
with the natural number Set N
Aim: create a
function that bijection between Z and N
The function
can look like this
Z
2
N
O
–
->
f. z>N
O fon fizeco, co> > Necven nueter
fl&s= 22
for Ze8-00, 0) > W odd number
flz= 27+
4
3
> 4
– N M & h
2
2.
Now we pione f. is bijective,
for ZETo, o fiz= 22.
suppose fiz,= fiz) 28,=222 Z=22
this f is injective,
for YEN, y = f1zx= 28 » z = € Z
this fis surjective
then fis bijective.
y
Chapter 0
Problem 1
(i) Conditions: A CB and BCC
XE
a. assumption
b. We have ACB, this means
if XE A > XE B
given (a) we
have
XEB
e. We have BCC, this means
XE B » XEC
given (b), we have
XEC
d. we have
XEO » Xtc
Therefore, Aço
Set is the subset of it self.
a.
b
b.
AÇA, this means
f XEA
Xt А
9 given (b), we have
A SA
he have
ACA and AZA,
C.
this is equivalent to
А – А
– Conditions A =B and B=c. .
if A=B, we have
A CB and A ZB
ne have
b. if B=c,
BCC,
B2C
c. given in
we have
A CB, B e C – ACC
A ZB, B 2C » ASC
d. ACC and A ZC is equivalent
to A=c
em z
)
fi 10,1)>(0,00
If x, y E (0,1), and & fx) fy) if y E (0,0), y=fx
fox= Ž= flys-
*
y
=> x=
–
Not
If y=0.01
X=100
=> x=y
in 10,!).
then fix= Ž is injectine
then fot is not sujective
g
R 1o, p)
‘If fyz= fx
If yt. 10.), and y- gix)= x²
y = x > x=y
=> x= + 5y ER
X² is surjective.
then
ges=x² is NOT
injective,
then goo= x²
Chapter 0
Problem 1
(i) Conditions: A CB and BCC
XE
a. assumption
b. We have ACB, this means
if XE A > XE B
given (a) we
have
XEB
e. We have BCC, this means
XE B » XEC
given (b), we have
XEC
d. we have
XEO » Xtc
Therefore, Aço
Set is the subset of it self.
a.
b
b.
AÇA, this means
f XEA
Xt А
9 given (b), we have
A SA
he have
ACA and AZA,
C.
this is equivalent to
А – А
– Conditions A =B and B=c. .
if A=B, we have
A CB and A ZB
ne have
b. if B=c,
BCC,
B2C
c. given in
we have
A CB, B e C – ACC
A ZB, B 2C » ASC
d. ACC and A ZC is equivalent
to A=c
em z
)
fi 10,1)>(0,00
If x, y E (0,1), and & fx) fy) if y E (0,0), y=fx
fox= Ž= flys-
*
y
=> x=
–
Not
If y=0.01
X=100
=> x=y
in 10,!).
then fix= Ž is injectine
then fot is not sujective
g
R 1o, p)
‘If fyz= fx
If yt. 10.), and y- gix)= x²
y = x > x=y
=> x= + 5y ER
X² is surjective.
then
ges=x² is NOT
injective,
then goo= x²
m3
If zB countable, then we can list them as a bejection
with the natural number Set N
Aim: create a
function that bijection between Z and N
The function
can look like this
Z
2
N
O
–
->
f. z>N
O fon fizeco, co> > Necven nueter
fl&s= 22
for Ze8-00, 0) > W odd number
flz= 27+
4
3
> 4
– N M & h
2
2.
Now we pione f. is bijective,
for ZETo, o fiz= 22.
suppose fiz,= fiz) 28,=222 Z=22
this f is injective,
for YEN, y = f1zx= 28 » z = € Z
this fis surjective
then fis bijective.
y
f. Zelo, o) > NE odd number
f(2)= 22 +1
suppose fizi) = f(Za) -> 22,+1=2Z2+1 =>2,= 22
fis injective
suppose y = f(2)=> z= g / Ez
thus fis surjective
then fB bijective
Therefore. Integer set zis countable
Problem 4
B it exists y oficUD)
then, it has an XECUD so that fix=y
– XEC
on XED,
= f(x) E f(C) f(x) & flo)
= f(x) + f(c) u f(B)
= fc cude fic, ufio
Let yt fra) u fio), then y = f(c) on y Efio)
there exists an
x so that fix) = y
hence fix) e f (c) uftos
since
XE CUD
this leads to f(c) Ufo CF (CUD)
thus fic UD) = f(c) ufio
(11) Let xef End), then there’s
yo and such that
fly) = x, Then
YEC, so x= f(y) Efic).
similarly, because
YED, x-fly, & f(D), then for
XE fron flos
thus fcc no) c fccon fo)
Function
Let ‘X=fa, b,c} ,x={1,24
flas= 1 f16=2 fcc)= 1
Cafa, by D=fb.ch
fccnp) = f(b) = 2
f(0) = { 1,29
f(0) = f + 1, 24
froon f(s) = {1, 2}
ficap) & ficnfio)