Math 3210-3
Exam 1
Solutions
Name:
You may use your dictionary of definitions. Please use a pencil and keep your proofs neat
and organized. Make sure you use complete “sentences,” and remember you need to give
justifications for each step in your proofs. Each problem is worth one point. Write on this
front page which problems you want graded. Notice that you have some choice as to which
problems you will do.
Grade Problems:
Do one of the following:
1. State the Density of Q in R Theorem.
If x, y ∈ R with x < y, then there exists r ∈ Q such that x < r < y.
2. State the Heine-Borel Theorem.
A set S ⊂ R is compact if and only if it is closed and bounded.
Do two of the following:
3. Let R be the relation on the real numbers given by xRy iff x − y is rational. Prove R is an equivalence
relation. Describe Eπ .
Proof: To show that R is an equivalence relation, we need to show it is reflexive, symmetric and
transitive. So let x ∈ R. Then x − x = 0 which is rational, hence xRx, so R is reflexive. Now assume
that xRy. Then x − y = r ∈ Q, so y − x = −(x − y) = −r which is rational, so yRx. So we can
conclude that R is symmetric. Finally, assume xRy and yRz. Then x − y = r and y − z = s where
r, s ∈ Q, so x − z = x − y + y − z = (x − y) + (y − z) = r + s which is rational since Q is closed under
addition. Hence R is transitive. Therefore R is an equvalence relation.
Eπ = {x ∈ R : xRπ} = {x ∈ R : π − x ∈ Q} = {π}.
1
1
1
n
1
+
+
··· +
=
for all n ∈ N.
1·2 2·3 3·4
n(n + 1)
n+1
1
1
1
1
n
Proof: (By Induction) Let P (n) be the statement
+
+
···+
=
. Notice
1·2 2·3 3·4
n(n + 1)
n+1
1
1
that P (1) is the statement 1+1
= 1·2
= 12 which is true. Now assume that P (k) is true for some k ∈ N.
1
1
1
k
1
+
+
···+
=
. We need to show that P (k + 1) is true. Consider the
Then
1·2 2·3 3·4
k(k + 1)
k+1
following:
4. Prove that
1
1
1
1
+
+
··· +
1·2 2·3 3·4
k(k + 1)
1
1
1
1
1
+
+
··· +
+
1·2 2·3 3·4
k(k + 1) (k + 1)(k + 2)
=
=
=
=
=
=
k
so we have
k+1
k
1
+
k + 1 (k + 1)(k + 2)
k(k + 2) + 1
(k + 1)(k + 2)
k 2 + 2k + 1
(k + 1)(k + 2)
(k + 1)(k + 1)
(k + 1)(k + 2)
k+1
k+2
Hence P (k + 1) is true. Therefore we conclude that P (n) is true for all n ∈ N.
Do one of the following:
1
.
inf A
Proof: Let n = inf A. Then by definition, n ≤ a for all a ∈ A. This implies that n1 ≥ a1 for all a ∈ A.
Since S = { a1 : a ∈ A} we see that n1 ≥ s for all s ∈ S. Hence n1 is an upper bound for S. To show
1
that it is the least upper bound for S, let m < n1 . Then m
> n. Since n is the greatest lower bound
1
of A, there is some a ∈ A such that m > a. But this implies that m < a1 ∈ S. Thus m is not a bound
for the set S. Therefore n1 is the least upper bound for S, and we conclude that sup S = inf1 A .
5. Let A be a bounded set of positive real numbers. Let S = { a1 : a ∈ A}. Prove that sup S =
6. For any set S ⊆ R let S denote the intersection of all the closed sets containing S. Prove that S = cl S.
Proof: Notice that S = ∩Ci where Ci are closed sets and S ⊆ Ci for all i. We know by a theorem
from class that the intersection of closed sets is closed, so S is closed.
We also proved in class that cl S is a closed set, so cl S = Cj for some j. Thus S ⊆ cl S.
Conversely, let x ∈ cl S. Then x ∈ S or x ∈ S 0 . If x ∈ S then x ∈ S. If x ∈ S 0 r S then every
neighborhood of x intersects S. Thus x ∈ Ci for all i. Hence x ∈ S.
Therefore S = cl S.
Do one of the following:
7. Suppose that f : A → B is injective. Let C1 and C2 be subsets of A. Prove f (C1 ∩C2 ) = f (C1 )∩f (C2 ).
Proof: Let y ∈ f (C1 ∩ C2 ). Then there exists x ∈ C1 ∩ C2 such that f (x) = y. Thus x ∈ C1 and
x ∈ C2 , which implies that f (x) = y ∈ f (C1 ) and f (x) = y ∈ f (C2 ). Hence y ∈ f (C1 ) ∩ f (C2 ).
Conversely suppose y ∈ f (C1 ) ∩ f (C2 ). Then there is some x1 ∈ C1 and x2 ∈ C2 such that f (x1 ) =
y = f (x2 ). Since f is injective, this implies that x1 = x2 . Let’s let x = x1 = x2 . Then x ∈ C1 ∩ C2
and f (x) = y. Thus f (x) = y ∈ f (C1 ∩ C2 ).
Therefore f (C1 ∩ C2 ) = f (C1 ) ∩ f (C2 ).
8. Prove R is uncountable.
Proof: It suffices to show that (0, 1) is uncountable. Suppose (0, 1) is countable. Then we can list
off the elements in the set as (0, 1) = {x1 , x2 , x3 , . . .}. Each xi has a decimal expresstion of the form
xi = 0.ai1 ai2 ai3 · · · where aij ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. So we can write
x1
=
0.a11 a12 a13 a14 · · ·
x2
=
0.a21 a22 a23 a24 · · ·
x3
=
0.a31 a32 a33 a34 · · ·
x4
=
..
.
0.a41 a42 a43 a44 · · ·
Now we will construct y ∈ (0, 1) which is not equal to any of the xi values. Let y = 0.b1 b2 b3 b4 · · · where
2 if aii 6= 2
bi =
. Clearly y ∈ (0, 1), but by construction, y 6= xi for any i. This contradicts that
3 if aii = 2
(0, 1) is countable. Therefore R is uncountable.