Math 317 HW #8 Solutions
1. Exercise 3.4.8. A set E is totally disconnected if, given any two points x, y ∈ E, there exist
separated sets A and B with x ∈ A, y ∈ B, and E = A ∪ B.
(a) Show that Q is totally disconnected.
Proof. Let x, y ∈ Q with x < y. By Corollary 1.4.4, there exists an irrational number a
such that x < a < y. Now, define
A := (−∞, a) ∩ Q and B := (a, +∞) ∩ Q.
Then certainly A ⊂ (−∞, a], so A ∩ B = ∅. Likewise, B ⊂ [a, +∞), so B ∩ A = ∅.
Finally, x ∈ A, y ∈ B, and A ∪ B = Q.
Since our choice of x, y ∈ Q with x < y was arbitrary, we see that such sets can be
constructed for all pairs of rational numbers, so Q is totally disconnected.
(b) Is the set of irrational numbers totally disconnected?
Answer. Yes. If x, y ∈ I with x < y, then, by Theorem 1.4.3, there exists a ∈ Q such
that x < a < y. Define
A := (−∞, a) ∩ I and B := (a, +∞) ∩ I.
Then, just as in part (a), we will have that A ∩ B = ∅, B ∩ A = ∅, x ∈ A, y ∈ B, and
A ∪ B = I. Therefore, since the choice of x and y was arbitrary, we can conclude that I
is totally disconnected.
2. Exercise 3.4.10. Let {r1 , r2 , r3 , . . .}
be an enumeration of the rational numbers, and for each
S∞
n
n ∈ N set n = 1/2 . Define O = n=1 Vn (rn ), and let F = Oc .
(a) Argue that F is a closed, nonempty set consisting only of irrational numbers.
Proof. Since O is a union of open intervals (and, thus, a union of open sets), we know
that O is open by Theorem 3.2.3. In turn, Theorem 3.2.13 ensures that F = Oc is closed.
Moreover, since every rational number is the center of one of the intervals which make
up O, we know that Q ⊆ O. Hence, every element of F must be irrational.
To see that F is nonempty, note that, for each n, the length of the interval
Vn (rn ) = (rn − n , rn + n )
is 2n = 2 21n =
up O is
1
.
2n−1
Therefore, the total length of all the intervals Vn (rn ) that make
∞
X
n=1
1
2n−1
=
∞
X
1
= 2.
2n
n=0
Hence, the length of O is no bigger than 2. Since R has infinite length and F = Oc , we
see that F must also have infinite length and so, in particular, must be nonempty.
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(b) Does F contain any nonempty open intervals? Is F totally disconnnected?
Answer. No, F does not contain any nonempty open intervals. Let x, y ∈ R with x < y.
Then, since Q is dense, there exists a rational number rn between x and y. Thus, since
rn ∈ O, we see that (x, y) ∩ O 6= ∅, so it cannot be the case that (x, y) ⊆ F . Since our
choice of x and y was arbitrary, we see that F contains no nonempty open intervals.
F is totally disconnected since it is a subset of I, which we showed in Problem 1(b) is
totally disconnected.
3. Exercise 3.5.4.
(a) Starting with n = 1, inductively construct a nested sequence of closed intervals I1 ⊇
I2 ⊇ I3 ⊇ · · · satisfying In ⊆ Gn . Give special attention to the issue of the endpoints of
each In .
Construction. Since G1 is dense, it must be nonempty, so there exists x1 ∈ G1 . Now,
since G1 is open, there exists 1 > 0 such that V1 (x1 ) ⊆ G1 . In other words,
(x1 − 1 , x1 + 1 ) ⊆ G1 .
Now, notice that the closed interval
[x1 − 1 /2, x1 + 1 /2] ⊆ (x1 − 1 , x1 + 1 ) ⊆ G1 .
Hence, if we define I1 := [x1 − 1 /2, x1 + 1 /2], then we have that I1 ⊆ G1 .
Now, since G2 is dense, there exists an element x2 ∈ G2 such that
x1 − 1 /2 < x2 < x1 + 1 /2.
In turn, since G2 and (x1 − 1 /2, x1 + 1 /2) are open, the set G2 ∩ (x1 − 1 /2, x1 + 1 /2)
is open, so there exists 2 > 0 such that V2 (x2 ) ⊆ G2 ∩ (x1 − 1 /2, x1 + 1 /2). Define
I2 := [x2 − 2 /2, x2 + 2 /2].
Then, by construction, I2 ⊆ V2 (x2 ) ⊆ G2 and I2 ⊆ (x1 − 1 /2, x1 + 1 /2) ⊆ I1 .
Now we iterate this process. In general, if we have Ik = [xk −k /2, xk +k /2], Gk+1 being
dense implies there exists xk+1 ∈ Gk+1 such that xk+1 ∈ (xk − k /2, xk + k /2) ⊆ Ik .
In turn, Gk+1 and (xk − k /2, xk + k /2) being open implies there exists k+1 such that
Vk+1 (xk+1 ) ⊆ Gk+1 ∩ (xk − k /2, xk + k /2). Then Ik+1 := [xk+1 − k+1 /2, xk+1 + k+1 /2]
is contained in both Gk+1 and Ik .
Thus, this gives us a nested sequence of intervals I1 ⊇ I2 ⊇ I3 ⊇ · · · with In ⊆ Gn for
all n.
(b) Now, use Theorem 3.3.5 or the Nested Interval Property to finish the proof of Theorem
3.5.2.
Proof. By the Nested Interval Property, there exists
x∈
∞
\
n=1
2
In .
Since In ⊆ Gn for all n, we know that
∞
\
In ⊆
n=1
∞
\
Gn ,
n=1
so x is in the intersection of the Gn as well.
4. Exercise 4.2.3. Use Corollary 4.2.5 to show that each of the following limits does not exist.
(a) limx→0 |x|/x
Proof. Let f (x) = |x|/x. Consider the sequences (xn ) and (yn ) given by
xn =
1
n
and yn = −
1
n
for all n.
Then, by construction, xn 6= 0 and yn 6= 0 for all n, but (xn ) → 0 and (yn ) → 0. Note
that
|1/n|
lim f (xn ) = lim f (1/n) = lim
= lim 1 = 1,
1/n
but
lim f (yn ) = lim f (−1/n) = lim
| − 1/n|
= lim(−1) = −1.
−1/n
Therefore, by Corollary 4.2.5, limx→0 f (x) does not exist.
(b) limx→1 g(x) where g is Dirichlet’s function from Section 4.1.
Proof. Consider the sequences (xn ) and (yn ) given by
√
2
1
xn = 1 +
and yn = 1 +
n
n
for all n.
By construction, xn 6= 1 and yn 6= 1 for all n, but (xn ) → 1 and (yn ) → 1. Now,
lim g(xn ) = lim g(1 + 1/n) = lim 1 = 1
since 1 + 1/n is rational for all n. However, by HW #2 Problem 5, yn is irrational for
every n, so
lim g(yn ) = lim 0 = 0.
Hence, by Corollary 4.2.5, limx→1 g(x) does not exist.
5. Exercise 4.2.8. Assume f (x) ≥ g(x) for all x in some set A on which f and g are defined.
Show that for any limit point c of A we must have
lim f (x) ≥ lim g(x).
x→c
x→c
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Proof. Assume limx→c f (x) = L1 and limx→c g(x) = L2 exist. Suppose (xn ) is a sequence
contained in A such that xn 6= c for all n and (xn ) → c (Theorem 3.2.5 implies that at least
one such sequence exists since c is a limit point of A). Then the forward direction of Theorem
4.2.3 implies that
lim f (xn ) = L1 and lim g(xn ) = L2 .
n→∞
n→∞
Since f (xn ) ≥ g(xn ) for all n, the Order Limit Theorem for sequences (Theorem 2.3.4) implies
that L1 ≥ L2 , as desired.
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