Name: ___ANSWERS___________________________
Math 101
Take Home Midterm 2
August 5th 2005
Problem 1. (10 points)
If A and B are subsets of some universal set X, and B’ as usual means the complement of B (in X), then prove
that A  B'   if and only if A  B .
Solution: Prove the equivalence of the two statements, A  B'   and A  B , by proving two
implications. First suppose A  B and show that this implies that A  B'   , then next suppose that
A  B'   and show this likewise implies A  B .
So, first we suppose A  B . Then suppose that there is an element x  A  B' then by definition x  A
and x  B' . However since A  B , then since x  A then x  B . So x  B and x  B' , which is
impossible, so there are no elements in A  B' , i.e. A  B'   .
Next, suppose A  B'   . Then this means that if x  A , then x  B' , otherwise x  A  B' , which is
supposedly the null set. However, x  B' is equivalent to x  B (double negation), so that if x  A , then it
must be the case that x  B , i.e. A  B .
With these two implications proven, then the original statement of equivalence follows.
Problem 2. (12 points)
Call a subset B of a set A cofinite if the complement of B in A (i.e. A  B ) is a finite set. If B and C are both
cofinite subsets of A then prove that B  C is also a cofinite subset of A.
Solution: Assume that B and C are both cofinite subsets of A. Thus A  B and A  C are both finite
subsets. Now since these are both finite sets, then their union ( A  B)  ( A  C ) is also finite. Next note
that A  B  A  B' (assuming the existence of a universal set containing all three sets A, B and C), and that
A  C  A  C' . These two facts imply that ( A  B)  ( A  C ) equals ( A  B' )  ( A  C ' ) which by a set
distribution law is the same as A  ( B'C ' ) . Now DeMorgan’s Law holds that B'C '  ( B  C )' so that
A  ( B'C ' ) equals A  ( B  C )' which is the same as A  ( B  C ) . So ( A  B)  ( A  C ) equals
A  ( B  C ) , and since ( A  B)  ( A  C ) is finite, then so is A  ( B  C ) , which by definition means that
B  C is a cofinite subset of A.
Problem 3. (4 points)
Find the problem in this problem. There are three incorrect statements in this problem.
(i)
(ii)
(iii)
(iv)
(v)
22  4
3 7  20
13  8  5
10  2  4
10  2  8
Solution: The problem in the problem is the paradox. Statements (ii) and (iv) are incorrect. However, this
means that only two statements are incorrect, so in fact the statement “There are three incorrect statements in
this problem” is itself incorrect (as there are only two incorrect statements). But then this makes a total of
three incorrect statements, so “There are three incorrect statements in this problem” is now correct, but then
that means there are only two incorrect statements, the arithmetic ones, but that makes “There are three
incorrect statements in this problem” incorrect and so there a total of three incorrect statements again, so
“There are three incorrect statements in this problem” is now correct, so now there are only two... three...
two... a paradox (not much different from “This statement is false”).
Problem 4. (8 points)
Determine whether each of the following statements is true or false (where as usual Z+ is the set of positive
integers not including 0, and R is the real numbers). Be sure to explain your answers:

]

]
(i) x  R [ y  Z  ( yx 2  y)  ( yx 3  y)
(ii) x  R [ y  R ( yx 2  y)  ( yx 3  y)
Solution: In statement (i), the statement supposes the existence of a real number, x, such that for any positive
integer, y, it is the case that both yx2 = y and yx 3  y . Such an x does exist, namely x = -1. Thus statement
(i) is true, such an x exists.
However, in statement (ii) there is no such element x, as now the statements yx2 = y and yx 3  y have to
hold for all real numbers y, not just positive integers, and when y = 0, there is no such x, as 0x3 = 0 for all
x  R . Thus statement (ii) is false.
Problem 5. (6 points)
Suppose proposition P(x,y) is the statement "x = 2y" (so, for example, P(3, 2) is false, and P(6, 3) is true).
Determine whether the following statement is true or false (be sure to explain your answer):
w, x, y, z(w  2x)  ( y  2z)  P(w, y)  P( x, z) (assuming w, x, y, z are all real numbers)
Solution: To show that the statement is true we would need to show that ( w  2 x)  ( y  2 z ) implies
that P( w, y )  P( x, z ) for all possible combinations of real numbers w, x, y, z. So, assume that
( w  2 x)  ( y  2 z ) , then P( w, y ) is equivalent to the statement “w = 2y” and since w = 2x then this is
equivalent to 2x = 2y. Now since y = 2z, then this is equivalent to 2x = 2 (2z) = 4z, or that x = 2z, which is
the same as proposition P( x, z ) . So if ( w  2 x)  ( y  2 z ) , then P( w, y ) is indeed equivalent to P( x, z ) , so
the statement (w  2x)  ( y  2z)  P(w, y)  P( x, z) is true for all real numbers w, x, y, z.
Problem 6. (12 points)
Given a set X with two closure operators K1 and K2 , say that K1 is smaller than K2 if the set of all closed
sets produced by operator K1 is properly contained in the set of all closed sets produced by operator K2 .
Suppose X = Z.
(a) Describe the smallest closure operator K such the set M = x  Z | k  Z, x  2k  1 is a closed set
under K (i.e. K(M) = M). “Describe” means describe what K(A) equals for any subset A  Z .
(b) Describe the smallest closure operator K such that the set of the closed sets produced by K contains all
sets of the form m, m  1 where m is any integer.
Solution:
In part (a) we know that the list of closed sets of K contains at the very least  , M and Z. In fact this list is
the shortest possible list of closed sets for a closure operator. This list of three sets includes  and Z, and
the union or intersection of any combination of these three yields a set already on the list. So the closure
operator defined by this set , M , Z is the smallest possible one. Now note that set M is simply the odd
integers. Next note that the closure of any set, A, is simply the smallest possible closed set that contains A.
Then the action of K is very easy to describe, as there are only three possible outcomes. If A =  , then
K  =  . If A is contained in M (i.e. it’s a subset consisting only of odd integers), then KA = A. Otherwise,
if A contains any even integers, then KA = Z.
For part (b), we know that the set of closed sets includes everything of the form m, m  1. However, since
the list of closed sets of any closure operator includes all possible intersections, then the list must include all
singleton element sets as well, as for any integer m, then both m, m  1 and m  1, m are closed sets, and
their intersection is just {m}. Next you might think that if every set of the form is closed, then since unions
of closed sets are also closed, then any subset A of Z must be closed as well, but not quite! Remember that
the rules governing closed sets simply dictates that any finite union of closed sets must be closed. Thus to
make the smallest list of closed sets possible (i.e. to find the “smallest closure operator”), then we simply
need it to be the case that any finite subset A of Z is closed.
So... the description of the smallest closure operator such that all sets of the form m, m  1 are closed is
fairly straightforward. As usual K  =  , next if A is a finite set then KA = A, otherwise KA = Z (i.e.
whenever A is an infinite set). Note this is not quite the discrete topology, which would just make KA = A
regardless of the “size” of A. The discrete topology is in some sense “larger” than this finite-discrete
topology as the list of closed sets for the discrete topology includes all possible subsets of Z, whereas this
one includes only the finite subsets of Z.
Problem 7. (16 points)
Let An  1,2,3,  , n . I claim that by induction that the power set of the positive integers, (Z  ) , is
countable. The following is a sketch of my proof:
I. Induction on n shows that ( An ) has 2 n  2 elements.
II. A countable union of finite sets is itself countable.
III. Therefore (Z  ) is countable.
Critique this argument by answering the following questions:
Are the premises (I) and (II) true statements on their own? For each statement fix whatever is not right and
give a brief proof of the correct version. Does (III) follow from the premises as you have fixed them? Why
or why not? If (III) doesn't follow, then what have your corrected statements (I) and (II) actually proved?
Solution: Premise I is incorrect as stated, and should read “Induction on n shows that ( An ) has 2 n
elements.” Premise II is correct as stated.
Proof of the corrected version for Premise I: “Induction on n shows that ( An ) has 2 n elements:”
Base case: Let n = 1, then A1 = {1}. Then ( A1 ) = {  ,{1}} has 21 = 2 elements as required.
Induction step: Suppose that ( An ) has 2 n elements for some integer n, then consider the set An+1
= An  {n  1} . Now the sets in ( An 1 ) can be divided into two distinct groups, (1) those that include the
number n+1 and (2) those that don’t. The subsets in the second group (2) are just equivalent to the subsets in
( An ) , i.e. there are 2 n of them. Next, consider any set in group (1). These are all the sets that include the
number n+1. If you remove n+1 from any of them then what remains is a subset of An , i.e. a member of
( An ) , of which there are 2 n possible sets. Note that given any two subsets A and B in group (1), then if A*
and B* are the same sets with the number n+1 removed, then A* and B* must be different, thus there really
are 2 n possible sets in group (1). As ( An 1 ) is composed of sets from group (1) and sets from group (2),
and as each group of sets has exactly 2 n members, then ( An 1 ) must have 2 n 1 elements, which proves the
induction step, and so by the principle of mathematical induction ( An ) has 2 n  2 elements for any
positive integer n.
Proof of Premise II: A countable union of finite sets is itself countable:
Suppose we are given a countable union of finite sets,  Ai where each Ai is a finite set. Then the set of
iZ
elements in this union is countable, as we can simply list them out in order, listing the elements of
A1.followed by those of A2 , then those of A3 etc., which is possible as each Ai has a finite number of
elements. Thus if A1 has m elements listed in some order as {a11, a12, ..., a1m}, A2 has n elements, {a21, a22,
..., a2n}, etc., then the elements in  Ai can be listed in order as {a11, a12, ..., a1m, a21, a22, ..., a2n,, ... }. Note
iZ
that some elements might show up more than once if they show up in different Ai‘s, but that’s fine as any
subset of a countable set must also be countable.
Next, does Premise III follow from (corrected) Premise I and II? Obviously not, as (corrected) Premise I and
Premise II are true, and Premise III is not true. What does follow from (corrected) I and II is a bit subtle.
Note that any finite subset B of Z+ is a subset of An for some positive integer n (if B is finite, then let n be the
largest number in B, and then B  An ). Thus B  ( An ) . Then any finite subset of Z+ is an element in
 ( An ) . Now since  ( An ) is a countable union of finite sets (finite as Premise I shows that ( An )
nZ
nZ
n
has 2 elements), then Premise II shows that this must be a countable set, so the set of finite subsets of Z+ is
countable, even though the full set of subsets (finite and otherwise) of Z+ (i.e. (Z  ) ) is itself uncountable.
Problem 8. (12 points)
Suppose that P is the set of all primes (i.e. P = {2, 3, 5, 7, ...}) Is (P) countable? Give a convincing proof
one way or the other (i.e. that it is countable, or that it is not countable).
Solution: There are a couple of approaches to this problem. One proof involves recognizing a bijection
between the set of primes and Z+. This bijection is simple. Just let pi stand for the ith prime where i  Z  .
Thus p1 = 2, p2 = 3, p3 = 5, p4 = 7, etc. Next for any subset A  Z  n let B be the corresponding subset of P
where each integer i in A is replaced by pi, the prime corresponding to integer i. This gives a one-to-one
correspondence between subsets of Z+ and subsets of P, and since (Z  ) is uncountable, then it must be
the case that (P) is uncountable as well.
Another approach is to assume that (P) is countable and derive a contradiction. So suppose it is countable,
and let Ai for i  Z  be a list of all of the subsets in (P) . Next construct a new subset B of P using the
following recipe – (1) if prime pi is in subset Ai then don’t include it in set B, (2) if prime pi is not in subset
Ai then put it in set B. Then by construction this new subset B is different from any subset Ai on the list, and
so the list doesn’t have all possible subsets of P, i.e. (P) is not countable.