# Problem Set #1 1 Set Theory

```Dr. Marques Sophie
Office 519
Algebra 1
Spring Semester 2015
[email protected]
Problem Set #1
1
Set Theory
1.1
Basics
Exercise 1 :
Consider the subsets of R defined as follows : An is the interval s0, 1{nr for all n P N.
Show that :
1. X8
A “ H;
n“1 n
2. Y8
“s0, 1r.
n“1
Hint : In 1., use the Archimedean Property of the integers in the real number system :
For every x P R, there exists an integer n P N such that n ą x.
Solution :
We use the Arichimedean property of the integers N in R : for any x P R, Dn P N such
that n ě x. If x P X8
A then x P An , @n, therefore 0 ă x ă 1{n, @n, and then
n“1 n
1{x ą 1{p1{nq “ n for all n P N, in conflicts with the Archimedean Principle.
Conclusion : there can’t be any such x P R and X8
A “ H.
n“1 n
Exercise 2 :
In the space X “ R consider the intervals A “ r2, 5r and B “s1, `8r. Describe the
sets :
1. Ac ;
2. Bc ;
3. AzB ;
4. BzA ;
Solution :
1. Ac “s &acute; 8, 2rYr5, `8r ;
2. Bc “s &acute; 8, 1s ;
3. AzB “ H ;
4. BzA “ r5, `8rYs1, 2r ;
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Exercise 3 :
Prove the following statements from the definitions :
1.
AzB “ A X Bc
2.
AzpB X Cq “ pAzBq Y pAzCq
Solution :
Note that x P Bc &ocirc; x R B.
Now,
1.
x P A X Bc &ocirc; x P A AND x P Bc &ocirc; x P A AND x R B
But AzB “ tx P R : x P A AND x R Bu. So, AzB “ A X Bc .
2.
AzpB X Cq “
“
“
“
A X pB X Cqc pby the previous questionq
A X pBc Y Cc qpDe Morgan law
pA Y Bc q X pA Y Cc q (distributive law for ”Y”q
pAzBq X pAzCq
You can also reprove Morgan law and distributive law if you want.
1.2
Mappings
Exercise 4 :
Let S2 “ tx P R3 : x21 ` x22 ` x23 “ 1u and J : S2 &Ntilde; S2 the inversion map Jpx1 , x2 , x3 q “
p&acute;x1 , &acute;x2 , &acute;x3 q.
1. Is J injective ? surjective ? Bijective ? Is there a formula for J&acute;1 (if the
inverse exists) ?
2. Let ppx1 , x2 , x3 q “ px2 , x3 q, a projection map p : R3 &Ntilde; R2 . Let q “ p|S2 “
restriction of p to S2 Ď R3 . Find Rangepqq. Is q one-to-one ?
Solution :
1. x ‰ y P R3 &ntilde; Di such that xi ‰ yi &ntilde; &acute;xi ‰ &acute;yi &ntilde; Jpxq “ p&acute;x1 , &acute;x2 , &acute;x3 q ‰
Jpyq “ p&acute;y1 , &acute;y2 , &acute;y3 q. Hence, J is one-to-one. Also surjective (hence bijective and
invertible) because JpJpxq “ Jp&acute;xq “ &acute;p&acute;xq “ x, implies every x P S2 is the J-image
of some point in S2 .
J&acute;1 “ J
(J is its own inverse).
ř
2. Rangepqq “ qpS2 q “ ppS2 q. Points x “ px1 , x2 , x3 q P S2 have i x2i “ 1 ; then project
to qpxq “ px1 , x2 q. Obviously qpxq lies in the closed unit disc D̄ “ tpx, yq P R2 :
2
1, but in fact qpS2 q is all of D,
x2 ` y2 ď 1u in R2 because x21 ` x22 ď x21 ` x22 ` x23 “ b
for if px1 , x2 q has x21 ` x22 ď 1. We may take x3 “ ˘ 1 &acute; px21 ` x22 q to get x P R3
with
(a) x P S2
(b) qpxq “ px1 , x2 q
Therefore, Rangepqq “ D.
The map q : S2 &Ntilde; D is not one-to-one, indeed for any x “ px1 , x2 , x3 q P S2 with
x3 ‰ 0 and y “ px1 , x2 , &acute;x3 q then x ‰ y and y21 ` y22 ` y23 “ 1 so that y P S2 and
qpxq “ qpxq.
Exercise 4 :
For φ : X &Ntilde; Y and A, B Ď X, show that
1.
φpA Y Bq “ φpAq Y φpBq
2. Give one counter example to show that we do not have always the
equality
φpA X Bq “ φpAq X φpBq
Solution :
1. We prove the double inclusion.
First prove pĎq :
y P φpA Y Bq &ntilde; Dx P A Y B such that φpxq “ y &ntilde; Dx P A such that φpxq “ y
or Dx P B such that φpxq “ y &ntilde; y P φpAq or y P φpBq &ntilde; y P φpAq Y φpBq.
Hence φpA Y Bq Ď φpAq Y φpBq.
First prove pĚq :
If y P φpAq Y φpBq &ocirc; y P φpAq or y P φpBq &ocirc; Dx P A such that y “ φpxq or
Dx P B such that y “ φpxq &ntilde; x P A Y B such that y “ φpxq &ntilde; y P φpA Y Bq.
Hence φpAq Y φpBq Ď φpA Y Bq.
The two sets are equal.
2. Take X “ Y “ R, φ : X &Ntilde; Y such that φpxq “ x2 . If A “s&acute;8, &acute;2s, B “ r2, `8r
then φpAq “ φpBq “ φpAq X φpBq “ r4, `8, but A X B “s &acute; 8, &acute;2s X r2, `8r“
H.
Exercise 5 :
Taking X “ Y “ R, let f : R &Ntilde; R be the map f pxq “ x2 . Compute the following
inverse image sets φ&acute;1 pSq for :
1. S “ r&acute;1, 1s ;
2. S “ the singleton t4u ;
3. S “ r1, `8r ;
3
4. S “ r&acute;10, &acute;4s.
Solution :
1. S “ r&acute;1, 1s &ntilde; f &acute;1 pAq “ r&acute;1, 1s ;
2. S “ the singleton t4u &ntilde; f &acute;1 pt4uq “ t&acute;2, 2u ;
3. S “ r1, `8r&ntilde; f &acute;1 pAq “s &acute; 8, &acute;1s Y r1, `8r ;
4. S “ r&acute;10, &acute;4s &ntilde; f &acute;1 pr&acute;10, &acute;4sq “ H.
Exercise 6 :
For X “ N ˆ N , Y “ N, define φ, η : X &Ntilde; Y as φpx, yq “ x ` y.
1. Find the inverse of φ&acute;1 p5q of the singleton set t5u.
2. If η : X &Ntilde; Y is the product operation ηpx, yq “ xy, find η&acute;1 p4q.
Solution :
1. φpt5uq “ tpx, yq : x ` y “ 5u “ tp1, 4q, p2, 3q, p3, 2q, p4, 1qu (4 points in N ˆ N)
2. η&acute;1 pt5uq “ tpx, yq P N ˆ N : x &uml; y “ 5u “ tp1, 5q; p5, 1qu (2 points).
1.3
Finite sets
Exercise 7 :
Show that the following infinite sets have the same cardinality by finding
explicit bijections between them :
1. The sets 2Z and 2Z ` 1 of even and odd integers ;
2. The sets N and A “ tn P Z : n ď 50u ;
3. The real line R and the bounded interval s0, 1r“ tx P R : 0 ă x ă 1u.
Solution :
1. Take φpxq “ x ` 1 ;
2. Take φpnq “ 50 &acute; pn &acute; 1q, so φp1q “ 50.
3. One (continuous) map that does it is
φpxq “ 1{x ` 1{px &acute; 1q “ ppx &acute; 1q ` xq{px2 &acute; xq “ p2x &acute; 1q{px2 &acute; xq
Exercise 8 :
Using the Shroeder-Bernstein Theorem, prove that the following sets must have the
same cardinality by producing explicit one-to-one (but not necessarily surjective) maps
f : A &Ntilde; B and g : B &Ntilde; A.
1. The interval r0, 1s, r0, 1r and s0, 1r in the real line ;
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2. The unit disc in the plane A “ tpx, yq P R2 : x2 ` y2 ă 1u and the unit square
B “ tpx, yq P R2 : &acute;1 ď x, y ď 1u.
Solution : Show using Schroeder-Bernstein :
1. — Take f : r0, 1s &Ntilde; r0, 1r, f pxq “ 1{2x and g : r0, 1r&Ntilde; r0, 1s, gpxq “ x (identity
map).
— Take f : r0, 1r&Ntilde;s0, 1r, f pxq “ 1{2x ` 1{2 and g :s0, 1r&Ntilde; r0, 1r, gpxq “ x (identity
map).
Note that ”equal cardinality” A &raquo; B is an RST relation, so r0, 1s “ p0, 1q too.
2. A “ tpx, yq : x2 ` y2 ă 1u, B “ tpx, yq : &acute;1 ď x, y ď 1u ; A Ď B so for f : A &Ntilde; B,
we can take the identity map f pxq “ x. For g : B &Ntilde; A take gpxq “ 1{10x “
p1{10x, 1{10yq, x P B.
1.4
Equivalence relation on sets
Exercise 9 :
For n ą 1 define a ” bpmod nq to mean
b &acute; a is an integer multiple of n
Verify that this is an RST relation on X “ Z.
Solution :
1. Reflexive : a „R a. Proof : pa &acute; aq “ 0 &uml; 5 is a multiple of 5 ;
2. Symmetric : a „R b &ntilde; b „R a. Proof : If b &acute; a “ 5k for some k P Z then
a &acute; b “ p&acute;1q &uml; k “ 5 &uml; p&acute;kq is also an integer multiple of 5.
3. Transitive : pa „R bq and pb „R cq &ntilde; pa „R cq. Proof : By hypotheses, Dk, l P Z`
such that b “ a ` 5k, c “ b ` 5l. Then c “ b ` 5l “ pa ` 5kq ` 5l “ a ` 5pk ` lq &ntilde;
c &acute; a “ multiple of 5 &ntilde; c „R a.
2
Integers
2.1
Some algebra on the sets of the integers
Exercise 11 :
Prove that in any unitary commutative ring R :
1. &acute;p&acute;xq “ x where &acute;x is the additive inverse x ;
2. p&acute;1q2 “ 1 where &acute;1 is the additive inverse of 1 ;
3. p&acute;xq &uml; y “ x &uml; p&acute;yq “ &acute;px &uml; yq, for all x, y P R.
Solution :
1. x ` p&acute;xq “ 0 “ p&acute;xq ` x this means exactly that x is the unique inverse of &acute;x,
that is x “ &acute;p&acute;xq.
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2. p&acute;xq&uml;y “ &acute;pxyq since p&acute;xqy`xy “ p&acute;x`xqy “ 0&uml;y “ 0 ; similarly xy`p&acute;xqy “
0. Thus &acute;pxyq “ p&acute;xqy.
The proof that xp&acute;yq “ &acute;pxyq is almost the same, so we omit it.
3. p&acute;1q2 ` p&acute;1q “ p&acute;1qp1 ` p&acute;1qq “ p&acute;1q &uml; 0 “ 0. Backtracking (add `1 to both
sides), we see that p&acute;1q2 “ 1 as claimed.
4. x ą y &ntilde; x &acute; y ą 0. But
x&acute;y “
“
“
“
x ` p&acute;yq “ x ` 0 ` p&acute;yq
x ` rc ` p&acute;cqs ` p&acute;yq
px ` cq ` rp&acute;yq ` p&acute;cqsq “ x ` c ` p&acute;py ` cqq
px ` cq &acute; py ` cq
so x ą y &ocirc; px ` cq &acute; py ` cq ą 0 &ocirc; x ` c ą y ` c.
2.2
Order on Z
Exercise 12 :
Prove that in any unitary commutative ordered ring R, for any x, y P R :
1. x ą y &ntilde; x ` c ą y ` c, for all c P R.
2. x ‰ 0 &ntilde; x2 ą 0.
3. If a ą 0 and b ą 0 then a ą b &ocirc; a2 ą b2 . (Hint : pb2 &acute; a2 q “ pb &acute; aqpb ` aq.
Use Rule of Sings).
Solution :
1. x ą y &ntilde; x &acute; y ą 0. But
x&acute;y “
“
“
“
x ` p&acute;yq “ x ` 0 ` p&acute;yq
x ` rc ` p&acute;cqs ` p&acute;yq
px ` cq ` rp&acute;yq ` p&acute;cqsq “ x ` c ` p&acute;py ` cqq
px ` cq &acute; py ` cq
so x ą y &ocirc; px ` cq &acute; py ` cq ą 0 &ocirc; x ` c ą y ` c.
2. x is either ą 0 or ă 0. If x ą 0, then we also have x2 ą 0. If x ă 0, then
&acute;x “ p&acute;1q &uml; x is ą 0 and p&acute;xq2 ą 0. But p&acute;xq2 “ p&acute;1q2 x2 “ x2 , so x2 ą 0 in
this case too.
3. a2 &acute; b2 “ pa ` bqpa &acute; bq by distributive laws. Since a, b ą 0 are automatically have
a ` b ą 0 and by the rules of signs, pa ` bqpa &acute; bq ą 0 &ocirc; pa &acute; bq ą 0. Thus
a2 ą b2 &ocirc; a2 &acute; b2 ą 0 &ocirc; pa ` bqpa &acute; bq ą 0 &ocirc; a &acute; b ą 0 &ocirc; a ą b, if a and b
are ą 0.
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2.3
Induction
Exercise 13 :
řn
Prove n2 “ (sum of first n odd integers) = k“1 p2k &acute; 1q “ 1 ` 3 ` &uml; &uml; &uml; ` p2n &acute; 1q.
Solution : By induction : certainly true if n “ 1. If true at level n, then at level n ` 1
we have
psumq “ p1`3`&uml; &uml; &uml;`2n&acute;1q`p2pn`1q&acute;1q “ n2 `2n`2&acute;1 “ n2 `2n`1 “ pn`1q2
So (Ppnq true ) &ntilde; (Ppn ` 1q true). Ppnq is true for all n P N.
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