Download 1 Selected Homework Solutions

1
Selected Homework Solutions
Mathematics 46001
A. Bathi Kasturiarachi
September 2006
1.1
Selected Solutions to HW #1
HW #1: (1.1) 5, 7, 8, 10; (1.2): 1, 2 ; (1.4): 11; (1.5): 3
(1.1): #10 For each of the following subsets of R R, determine whether it is equal
to the Cartesian product of two subsets of R.
(a) f(x; y)jx is an integerg = Z
(b) f(x; y)j0 < y 6 1g = R
R.
(0; 1]:
(c) f(x; y)jy > xg. This cannot be written as a Cartesian product of two
subsets of R. Suppose X = f(x; y)jy > xg = A B for some A; B
subsets of R. The points (0; 1) 2 X and (1; 2) 2 X. Since 1 2 A
and 1 2 B, we conclude that (1; 1) 2 A B. This is not true since
(1; 1) 2
= X.
(d) f(x; y)jx is not an integer and y is an integerg = Z
2
Z.
2
(e) f(x; y)jx + y < 1g. This cannot be written as a Cartesian product
of two subsets of R. For instance the points (0:8; :1) and (0:1; 0:8) are
both in the set, since 0:82 + 0:12 = 0:65 < 1. However, if the given
set could be written as a Cartesian product, then (0:8; 0:8) should
also be in the set. But it is not, since 0:82 + 0:82 = 1:28 > 1. Noticed
we used 0:8 since this is a number greater than p12 .
(1.5): #3 Let A = A1
A2
and B = B1
B2
.
(a) Show that if Bi Ai for all i, then B A.
For purposes
of the entire
[
[problem, we will de…ne the following. Let
X =
Bj and Y =
Aj . Then a a !-tuple can be de…ned as
j2Z+
j2Z+
follows.
x : Z+
X, with bi
x : Z+
Y , with ai
! X where x = (b1 ; b2 ;
2 Bi for each i and
! Y where x = (a1 ; a2 ;
2 Ai for each i.
) is an in…nite sequence of elements of
) is an in…nite sequence of elements of
Now let x be an element of B, and treated as a !-tuple of elements
of X, xi = x(i) 2 Bi Ai by hypothesis for all i. Therefore, we can
treated x as a !-tuple of elements of Y , xi = x(i) 2 Ai for all i. This
implies x be an element of A. Hence, B A.
1
(a) Show that the converse of (a) holds if B is non empty.
If B = ? then B
A trivially. From this it will not follow that
Bi Ai for all i. For example, if B1 = ?, B2 = fb1 ; b2 g, A1 = fa1 g,
and A2 = fb1 g then
B = B1
B2 = ?, A = A1
A2 = f(a1 ; b1 )g.
In this example, clearly B
A, but B2 * A2 . Therefore, we will
assume that B 6= ?. Then there exists
(b1 ; b2 ;
) 2 B such that bi 2 Bi for all i.
Fix an index j 2 Z+ . Let b0j 2 Bj be an arbitrary element of Bj .
Then
(b1 ; b2 ;
; bj
0
1 ; bj ; bj+1 ;
)2B
Since b0j was arbitrary, we have Bj
desired result.
A (by hypothesis) =) b0j 2 Aj .
Aj . Varying j we obtain the
(b) Show that if A is non empty, then each Ai is non empty. Does the
converse hold?
Suppose A is non empty. Then there exists
(a1 ; a2 ;
) 2 A such that ai 2 Ai for all i.
Therefore, Ai 6= ?. Conversely, suppose that for each i, Ai 6= ?.
Choose ai 2 Ai and construct an !-tuple (a1 ; a2 ; ) which will
belong to A by de…nition, so A is non empty.
(c) What is the relationship between the set A [ B and the Cartesian
product of the sets Ai [ Bi ? i.e.
What is the relationship between A [ B and
1
Y
i=1
(Ai [ Bi ) ?
What is the relationship between the set A \ B and the Cartesian
product of the sets Ai \ Bi ? i.e.
What is the relationship between A \ B and
Proposition 1. Suppose A = A1
non empty. Then
A[B
1
Y
i=1
A2
and B = B1
1
Y
i=1
(Ai \ Bi ) ?
B2
are both
(Ai [ Bi ) with equality if and only if Ai = Bi for all i.
2
0
Proof : First let K = Y [ X = @
[
j2Z+
1
0
Aj A [ @
[
j2Z+
Then we can de…ne an !-tuple x by
x : Z+ ! K where x = (x1 ; x2 ;
K, with xi 2 Ai [ Bi for each i.
1
Bj A =
[
j2Z+
(Aj [ Bj ).
) is an in…nite sequence of elements of
Now let x be an element of A[B, then x 2 A or x 2 B. That is, xi = x(i) 2 Ai
for all i or xi = x(i) 2 Bi for all i. Therefore,
xi = x(i) 2 Ai [ Bi for all i 2 Z+ ,
establishing A [ B
1
Y
i=1
(Ai [ Bi ).
Conversely, if we take an !-tuple x = (x1 ; x2 ;
)2
all i
1
Y
i=1
(Ai [ Bi ), then for
xi 2 Ai [ Bi =) xi 2 Ai or xi 2 Bi .
However, from the above we cannot conclude that x 2
1
Y
i=1
Ai or x 2
1
Y
Bi
i=1
(unless Ai = Bi - some component can belong to an Am and not to any
Bi while another component can belong to a Bn and not to any Ai )
Suppose, for some i 2 Z+ , Ai 6= Bi :Then there exists a 2 Ai such that
a2
= Bi . Construct the following !-tuple.
y= (y1 ; y2 ; ) = (b1 ; b2 ;
for all j 6= i and yi = a.
; bi
1 ; a; bi+1 ;
) where each yj = bj 2 Bj
It is clear that, y 2
= A and y 2
= B, so y 2
= A [ B. This leaves us with only
one case to consider, namely, Ai = Bi for all i 2 Z+ . In this case,
1
Y
i=1
(Ai [ Bi ) =
1
Y
i=1
Ai = A = B = A [ B.
Proposition 2. Suppose Ai \ Bi 6= ? for each i 2 Z+ . Then
A\B =
1
Y
i=1
(Ai \ Bi ).
Proof : Suppose A \ B = ?. Then all pairings Ai \ Bj must also be empty.
1
Y
Hence
(Ai \ Bi ) = ? and we have equality. Therefore, let us assume
i=1
that A \ B 6= ?. Let x 2 A \ B. Then x 2 A and x 2 B. This means
3
that x = (x1 ; x2 ;
) with xi 2 Ai and xi 2 Bi =) xi 2 Ai \ Bi for each
1
1
Y
Y
i. This implies that, x 2
(Ai \ Bi ). Therefore, A \ B
(Ai \ Bi ).
i=1
i=1
To prove the converse, we will assume that Ai \ Bi 6= ? for each i 2 Z+ .
1
Y
Then by part (c),
(Ai \ Bi ) is non empty.
i=1
So select x 2
1
Y
i=1
xi
(Ai \ Bi ). Then
2
Ai \ Bi =) xi 2 Ai and xi 2 Bi for all i
1
1
Y
Y
=) x 2
Ai and x 2
Bi =) x 2 A \ B.
i=1
Therefore,
1
Y
i=1
1.2
(Ai \ Bi )
i=1
A \ B.
Selected Solutions to HW #2
HW #2: (1.6) 2, 5; (1.7): 1, 4
(1.7): #1 Show that Q is countably in…nite.
First note that Q = Q [ f0g [ Q+ , where the unions are disjoint and
Q = fx 2 Q j x > 0g, Q+ = fx 2 Q j x < 0g. First show that Q+ is
countable.
Let f : Z+ Z+ ! Q+ be de…ned by, f ((m; n)) = m
n . Let x 2 Q+ . Then
x= m
Z+ . Therefore,
n for some (m; n) 2 Z+
f ((m; n)) = x, so f is surjective.
Since Z+ Z+ is countably in…nite, there exists a surjection g : Z+ !
Z+ Z+ (by Theorem 7.1). The function f g is a surjection since it the
composition of two surjections, and is given by,
f
g : Z+ ! Q+ .
Now appeal to Theorem 7.1 again to conclude that Q+ is countable.
Next we want to show that Q is countable.
Let h : Z+ Z+ ! Q be de…ned by, h((m; n)) = nm . Let x 2 Q . Then
x = nm for some (m; n) 2 Z+ Z+ . Therefore,
h((m; n)) = x, so h is surjective.
Like before the composition h g is a surjection, and is given by,
4
h g : Z+ ! Q .
By Theorem 7.1 we conclude that Q is countable. Finally, Q = Q [ f0g [
Q+ is the disjoint union of three countable sets, so is countable.
1.3
Selected Solutions to HW #3
HW #3: (2.13) 1, 4; (2.16): 3, 4, 6
(2.13): #1 Let X be a topological space; let A be a subset of X. Suppose that for
each x 2 A there is an open set U containing x such that U
A. Show
that A is open in X.
Let x 2 A. Then there
A (by
[ exists an open set Ux such that x 2 Ux
hypothesis). Let U =
Ux . Notice that U is open since X is a topological
x2A
space. We claim that [
U = A. For each x 2 A there exists an open set Ux
such that x 2 Ux
Ux , so A
U . The converse is easy, because each
Ux
A =) U =
[
x2A
Ux
A. Therefore, A is open.
x2A
(2.13): #4
(a). If fT g is a family of topologies on X, show that
[
X. Is
T a topology on X?
\
\
T is a topology on
T :
\
(i) ? 2
T , since ? 2 T for all .
\
X2
T , since X 2 T for all .
\
(ii) Let fU g be a non empty collection of open sets in
T . Then fU g 2
[
T for all . Then
U 2 T for all , since each T is a topology. Hence
[
U 2
\
T .
n
(iii) Let fUi gi=1 be a …nite non empty collection of open sets in
fUi g 2 T for all
. Then
n
\
i=1
Ui 2 T for all
5
\
T . Then
, since each T is a topology.
n
\
\
Ui 2
T .
i=1
\
Therefore,
T is a topology.
[
In general,
T is not a topology, unless we agree to include all arbitrary
Hence
unions and …nite intersections of open sets into the collection. Consider the
following example. Let X = fa; b; cg. Let T1 = f?; X; fag; fa; cgg and T2 =
f?; X; fbg; fb; cgg. It is easy to show that these are topologies. However,
fa; cg \ fb; cg = fcg 2
= T1 [ T2 , thus it is not a topology.
(b). If fT g is a family of topologies on X, show that there is a unique smallest
topology on X containing all the collections T , and a unique largest
topology contained in all T .
is
The unique smallest topology that contains all of the topologies T of X,
[
T together with all arbitrary unions and …nite intersections. Call this
collection F . One can easily check that F is a topology. Certainly, F contains
all the topologies T . Suppose G is another topology that contains all the
topologies T and is strictly coarser than F . This means that there is an open
set U in G such that U 2
= F . It cannot be the case that U 2 T for some . If so,
U 2 F . Therefore, U must be an open set that is constructed by using arbitrary
unions and (possibly) …nite intersection of open sets from di¤erent T and T
topologies. But in this case, U 2 F by the de…nition of F . Therefore, there
cannot be a smaller topology other than F that contains all of the topologies
T .
The unique
largest topology that is contained in all of the topologies T
\
of X, is
T . We have already shown that this is a topology. It is easy to
show that this topology is the largest unique topology contained in all of the
topologies T . This is left as an exercise.
(c). For this part check the following.
Unique smallest topology containing the to given topologies is:
f?; X; fag; fbgfa; bg; fb; cgg.
Unique largest topology contained in each of the given topologies is:
f?; X; fagg.
(2.16): #3 Consider the set Y = [ 1; 1] as a subspace of R. Which of the following
sets are open in Y ? Which are open in R? In this example, it will help a
lot if you can draw these sets on a number line.
(a) A = fx j 12 < jxj < 1g. The set A is open in R, since it is the union
of two open intervals in R. Namely,
6
1
1
) [ ( ; 1).
2
2
A = ( 1;
It is also open in Y since A = ( 1;
intersection of Y with an open set in R.
1
2)
[ ( 12 ; 1) \ Y , which shows it is the
b. B = fx j 12 < jxj 6 1g. The set B is not open in R, since neither
interval [ 1; 21 ) nor ( 12 ; 1] is open in R. However it is open in Y
since,
B=
1
1
) [ ( ; 2) \ Y ,
2
2
( 2;
which shows it is the intersection of Y with an open set in R.
c. C = fx j 12 6 jxj < 1g. The set C is not open in R, since neither
interval ( 1; 21 ] nor [ 12 ; 1) is open in R. It is also not open in Y since
we cannot intersect Y with an open set in R, to obtain an interior
end point of an interval.
d. D = fx j 12 6 jxj 6 1g. The set D is not open in R, nor in Y . Same
argument as part (c) holds.
e. E = fx j 21 < jxj < 1 and x1 2
= Z+ g. The set E is not open in R. We
can write E = ( 1; 0) [ f(0; 1) r Kg, where K = f n1 jn 2 Z+ g. Then
E can be expressed as,
E = ( 1; 0) [
(
1
[
n=1
1
1
;
n+1 n
)
,
open in R. It is also open in Y since E = E \ Y =
( which shows
( 1 that E is ))
[
1
1
\Y.
( 1; 0) [
n+1 ; n
n=1
(2.16): #4 Show that the maps
1
:X
Y
! X and
2
:X
Y
! Y are open.
Let W be a non empty open set of X Y . Consider 1 (W ) X and 2 (W )
Y . We must show that 1 (W ) is open in X and 2 (W ) is open in Y .
Choose x 2 1 (W ). Since 1 is surjective, there exists (x; y) 2 X Y such
that 1 (x; y) = x, where (x; y) 2 W .
Since W is open in X Y , there exists an open set U V of X Y , such
that (x; y) 2 U V
X Y , where U is open in X and V is open in Y . By
de…nition 1 (U V ) = U and it follows that x 2 U
1 (W ), hence 1 (W ) is
open in X.
To show 2 (W ) is open in Y , we simply observe that 2 (U V ) = V and
that y 2 V
2 (W ).
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(2.16): #6 Show that the countable collection
f(a; b)
2
(c; d) j a < b and c < d; a; b; c; d 2 Qg
is a basis for R :
We will …rst prove that the collection f(a; b) j a < b and a; b 2 Qg is a basis
for R in the standard topology.
(i) Let x 2 R. Since rationals are dense in the reals, there exists a 2 Q such
that x 1 < a < x and b 2 Q such that x < b < x + 1. Therefore, x 2 (a; b)
where a; b 2 Q.
(ii) Now suppose x 2 (a; b) \ (c; d) where a; b; c; d 2 Q. Since the non empty
intersection of two open intervals is open, we can easily …nd an open interval
(p; q) with p; q 2 Q, such that x 2 (p; q) (a; b) \ (c; d).
Therefore, by de…nition the collection f(a; b) j a < b and a; b 2 Qg is a basis
for R. Finally, we appeal to Theorem 15.1 in the text (page 86) to conclude
that the collection
f(a; b)
(c; d) j a < b and c < d; a; b; c; d 2 Qg
is a basis for the topology on R
R = R2 .
8