1. 1.14
For n=1: Let δ = ε. This works trivially. Induction step: Suppose it’s true for
n:
∀y∀ε > 0∃δ > 0|u − y| < δ → |un − y n | < ε
Let’s show it for n + 1: Let y and ε > 0 be given. Then
|un+1 − y n+1 | ≤ |u − y||un + un−1 y + . . . + y n |
≤ |u − y|(|u|n + |u|n−1 |y| + . . . |y|n )
≤ |u − y|(n max{|u|, |y|}n )
ε
Fix δ so that δ < 1 and δ < n(|y|+1)
n . If |u − y| < δ, then
|un − y n | ≤ δ · n · max{|u|, |y|}n ≤ δ · n · (|y| + 1)n <
ε
n · (|y| + 1)n = ε
n(|y| + 1)n
2. 1.15
Need to show that for any x > 0 there is a unique y > 0 with y n = x.
Let y = sup(S) where S := {s ∈ R | sn ≤ x}
Claim 2.1. y n = x
Proof. We’ll use the ε principle and show |y n − x| < ε for every positive ε. Use 14
to get a δ so that |u − y| < δ → |un − y n | < ε.
Then y, being sup(S) is within δ of an element of S and an element of S C . So,
∃z ∈ S so |z − y| < δ. Thus |un − y n | < ε, showing that y n < ε + un ≤ ε + x (the
last inequality being because u ∈ S.
Similarly, ∃z ∈
/ S so |z − y| < δ. Thus |un − y n | < ε, showing that y n > un −ε >
x − ε. Putting these two inequalities together |y n − x| < ε. As ε > 0 was arbitrary,
|y n − x| = 0.
We now need to show uniqueness. Suppose 0 < a < b. We will show that
0 < ak < bk for any k. We show this by induction. The base case, when k = 1 is
trivial (the hypothesis and conclusion are the same). Suppose it is true for k, that
is 0 < a < b implies 0 < ak < bk . We now show it for k + 1: Using the properties
of ordered fields: 0 · a < ak · a < bk · a < bk · b, so 0 < ak+1 < bk+1 , which is what
we needed.
3. 1.16
a) I add a bit of notation to make the following more streamlined: x0 = N . By
xk−1
Definition, for k ≥ 0, xk is the greatest integer ≤ 10k (x − (N + x101 + . . . 10
k−1 )). So,
xk−1
x1
xk ≤ 10k (x − (N +
+ . . . k−1 )) < xk + 1
10
10
, so:
x1
xk−1
xk
0 ≤ 10k (x − (N +
+ . . . k−1 + k )) < 1
10
10
10
, so:
x1
xk−1
xk
0 ≤ 10k+1 (x − (N +
+ . . . k−1 + k )) < 10
10
10
10
xk−1
xk
As xk+1 is the least integer ≤ 10k+1 (x − (N + x101 + . . . 10
k−1 + 10k )) < 10, we
see that xk+1 is between 0 and 9. As this works for any k ≥ 0, we have all xi are
between 0 and 9 for i ≥ 1.
1
2
b) Suppose toward a contradiction that k is the least so that xl = 9 for all l ≥ k.
Then for any l ≥ k
x1
x1
9
9
xk−1 + 1
xk−1
N+
>x>N+
+ ...
+ . . . k−1 + k + . . . + l
10
10k−1
10
10
10
10
Subtracting from both sides:
1
x1
9
9
xk−1
> x − (N +
+ . . . k−1 ) > k + . . . + l
10k−1
10
10
10
10
, or:
1
1
x1
xk−1
0 > x − (N +
+ . . . k−1 + k−1 ) > − l+1
10
10
10
10
. Or, considering absolute values,
x1
1
1
xk−1
|x − (N +
+ . . . k−1 + k−1 )| < l+1
10
10
10
10
xk−1
for any l. For any positive ε, there is some l so that 101l+1 , so |x − (N + x101 + . . . 10
k−1 +
xk−1
x1
1
ε for any positive ε, so |x − (N + 10 + . . . 10k−1 + 10k−1 )| = 0. But this contradicts
the definition of xk−1 (it wasn’t the greatest integer it could have been). This
contradiction shows that there cannot be a final sequence of 9’s.
c) Consider the set S = {N, N + x101 , . . .}.
Claim 3.1. N + 1 is an upper bound for S.
Proof. Let’s consider an element of S:
N+
x1
xk
9
9
10k − 1
+ ... + k ≤ N +
+ ... + k ≤ N +
≤N +1
10
10
10
10
10k
Let us consider the decimal expansion of x, the least upper bound for S: M.y1 y2 . . ..
As N ≤ x < N + 1 (the second inequality is due to the fact that one of the xk
is < 9): M = N .
x2
, . . .}, so 10(x − N ) is
x − N is the least upper bound for the set {0, x101 , x101 + 100
x2
the least upper bound for the set {x1 , x1 + 10 , . . .}. So, x1 ≤ 10(x − N ) ≤ x1 + 1,
showing that x1 = y1 . Continuing as such, we see that xi = yi for all i.
d) Repeat everything with 10 replaced by a general base.
4. 1.30
a) Define f (x+) = inf{f (y) | y > x} and f (x−) = sup{f (y) | y < x}, and the
jump of f at x to be f (x+) − f (x−).
Claim 4.1. f is continuous at x if and only if the jump of f at x is 0.
Proof. →: We’ll show that the jump is < ε for each positive ε. Let ε > 0 be fixed.
By continuity, there is a δ so that |x − y| < δ → |f (x) − f (y)| < 2ε .
Then f (x+) ≤ f (x + 3δ ) ≤ f (x) + 2ε , and similarly f (x−) ≥ f (x − 3δ ) ≥ f (x) − 2ε .
So f (x+) − f (x−) ≤ ε. As ε > 0 was arbitrary, f (x+) − f (x−) = 0.
←: Suppose the jump of f at x is 0, and let ε > 0 be given.
As the jump is 0, f (x+) = f (x−) = c. Let d ∈ {f (y) | y > x} be so d < c + ε,
and let e ∈ {f (y) | y < x} be so e > c − ε (these exist by the definition of f (x+)
as an inf and f (x−) as a sup). Let z > x be so f (z) = d and w < x be so
that f (w) = e. Finally, let δ = min{x − w, z − x}. If |y − x| < δ, then either
1
)|
10k−1
<
3
w < y < x, thus f (w) < f (y) < f (x) or x ≤ y < z, thus f (x) ≤ f (y) < f (z). So,
f (w) < f (y) < f (z), showing |f (y) − c| < ε. It remains to see that c = f (x).
If it were true that c < f (x), then by the definition of f (x+) as an inf, there would
have to be a z > x so that f (z) < f (x) which would contradict f being monotone.
Similarly, we can contradict the possibility that c > f (x). Thus c = f (x), and our
δ above witnesses continuity of f .
Note that if x < y, then f (x−) ≤ f (x) ≤ f (x+) ≤ f (y−) ≤ f (y) ≤ f (y+). So,
if there were n jumps of size at least q, say at x1 , . . . , xn , then f (a) ≤ f (x1 −) ≤
f (x1 +) − q ≤ f (x2 −) − q ≤ f (x2 +) − 2q . . . ≤ f (xn +) − nq ≤ f (b). It follows
(a)
that there can be at most f (b)−f
many jumps of size ≥ q. Thus the set of points
q
S
with non-zero jump q>0,q∈Q {jumps of size at least q}. This is a countable union
of countable sets, so is countable. Thus there are countably many discontinuities
of f in the interval [a, b].
b)Yes! Consider f on the intervals [n, n + 1] for each n ∈ Z. On each interval,
there are only countably many discontinuities of f by part a. Since there are only
countably many such intervals (they are indexed by Z, which is countable), the set
of discontinuous points is a countable union of countable sets, so is countable.
5. Multiplication is well-defined
and (bn ) ∼ (b0n ). We need to show that (an bn ) ∼ (a0n b0n ).
Suppose (an ) ∼
We use the fact that every Cauchy sequence is bounded. Let M ∈ Q be so that
(an ), (bn ), (a0n ), (b0n ) are sequences in [−M, M ]. Let ε > 0 be given. We need to
show that there is some N ∈ N so that n > N implies that |an bn − a0n b0n | < ε.
ε
ε
Let N be large enough that n > N implies |an − a0n | < 4M
and |bn − b0n | < 4M
.
0
0 0
0
0
0 0
Then for n > N , |an bn − an bn | = |an bn − an bn + an bn − an bn | ≤ |an bn − an bn | +
ε
ε
|an b0n − a0n b0n | = |an ||bn − b0n | + |b0n ||an − a0n | < M · 4M
+ M · 4M
< ε.
(a0n )
6. First written problem
We need to show that the following two conditions are equivalent:
(1) ∀ε > 0∃N (ε) ∈ N(n > N (ε) → an > bn − ε)
(2) (an ) ∼ (bn ) OR ∃N ∈ N(n > N → an > bn )
Let’s first show 1 → 2. For each ε > 0, let K(ε) > N (ε) be large enough that
n, m > K implies |an − am | < 4ε and |bn − bm | < 4ε . We proceed based on cases:
Case 1: For each ε > 0, There is no n > K(ε) so that an ≥ bn + ε. In this case,
n > K(ε) (thus n > N (ε)) implies that −ε < an − bn < ε, so |an − bn | < ε. As this
works for every ε > 0, we have that (an ) ∼ (bn ), so condition 2 holds.
Case 2: For some ε > 0, there is an n > K(ε) so that an ≥ bn + ε. In this case,
we will show that for any m > K(ε), am > bm . For m > K(ε):
ε
ε
ε
am − bm = (am − an ) + (an − bn ) + (bn − bm ) ≥ − + ε − ≥ > 0
4
4
2
So condition 2 holds.
Now let’s show 2 → 1. Suppose the first case of condition 2 holds. Then
∀ε > 0∃N ∈ N so that n > N implies |an − bn | < ε. So, in particular, n > N
implies an > bn − ε.
If the other case holds, for some N ∈ N, n > N implies that an > bn . In
particular, it implies an > bn − ε for any positive ε.
4
7. Second written problem
We need to show that [(an )] < [(bn )] if and only if ∃ε > 0∃N ∈ N so n > N
implies bn > an + ε.
Let’s first show the rightward implication. Assume [(an )] < [(bn )] and suppose
towards a contradiction that the condition on the right fails. That means for any
ε > 0 and any N ∈ N, there is some k > N with ak ≥ bk −ε. By the problem above,
after some M , an < bn . We now will derive a contradiction by proving (an ) ∼ (bn ).
Let ε > 0 be given. Let N > M large enough that n, m > N implies |an − am | < 4ε
and |bn − bm | < 4ε . Let k > N be so that ak ≥ bk − 2ε . Then for any m > N we
have that
ε ε ε
0 < bm − am = (bm − bk ) + (bk − ak ) + (ak − am ) < + + = ε
4 2 4
Now let’s show the leftward implication. By the above problem, it’s clear that
[(an )] ≤ [(bn )]. We need only show that (an ) 6∼ (bn ). Suppose towards a contradiction that (an ) ∼ (bn ). Let ε and N be so that n > N implies bn > an + ε. Let
M be so that n > M implies |an − bn | < ε. But then for any k > N and k > M ,
bn − an > ε and bn − an < ε. This is a contradiction.
8. 2.17
Suppose f : M → N has the property that it sends convergent sequences to
convergent sequences. We need to show that it sends limits to limits.
Claim 8.1. If (pn ) is a sequence in M converging to p, then (f (pn )) is a sequence
in N which converges to f (p)
Proof. Consider the sequence (p1 , p, p2 , p, p3 , p, . . .) in M . This is still a convergent
sequence with limit p. Thus, by the hypothesis, the sequence (f (p1 ), f (p), f (p2 ), f (p) . . .)
is a convergent sequence in N . The subsequence (f (p), f (p), f (p), . . .) must converge to the same limit, thus the limit is f (p). But then the other subsequence
(f (p1 ), f (p2 ), . . .) must converge to f (p).
By the sequence convergence condition for continuity, we see that f is continuous.
9. 2.18
a) Let δ = ε.
b) It is a bijection by definition. Check (the definition) that the inverse of an
isometry is an isometry, so by part a, they are both continuous.
c) Suppose f : [0, 2] → [0, 1] were an isometry. Then d[0,1] (f (0), f (2)) = 2. But
there are no two points in [0, 1] of distance 2! This is a contradiction, so there can
be no isometry between these two spaces.
10. 2.19
Check the definitions: The identity map: M → M is always an isometry. If
f : M → N is an isometry, so is f −1 : N → M . If f : M → N and g : N → K are
isometries, then gf : M → K is also an isometry.
11. 2.21
Draw pictures and play
5
12. 2.22
No, R and Q are not even bijective as they have different cardinalities.
13. 2.23
Suppose f : Q → Z were a homeomorphism. Let f (q) = 0, and let ε = 1. Which
δ > 0 will ensure that |q − r| < δ → |f (q) − f (r) < 1|? Well, f (q) − f (r) can only
be < 1 if it is 0, i.e., if q = r. Thus no δ > 0 will suffice. So, f is not continuous at
q. This is a contradiction.
Alternatively, use Corollary 11: in Z, every set is open. If there were a homeomorphism with Q, then every subset of Q would also be open. But this is not
true!
14. 2.26
a) Suppose (xn ) converges to the point x in R. We’ll see that (|xn |) converges
to the point |x| in R. Let ε > 0 be given. There exists an N ∈ N so that n >
N → |xn − x| ≤ ε. But |xn | − |x| ≤ |xn − x| and |x| − |xn | ≤ |xn − x| (both by the
triangle inequality). So, ||xn | − |x|| < ε, so this same N shows that (|xn |) converges
to |x|.
b) Consider the sequence (−1, 1, −1, 1, . . .). The absolute values converge, but
the sequence does not.
15. 2.30
a) Let (pn ) be a sequence which is monotone and bounded. Suppose it increases
(decreases is symmetric). Let p be the least upper bound for the set S = {pn | n ∈
N}. We need to show that pn converges to p. Let ε > 0 be given. Since p is the
least upper bound for S, there is a pN in S so that p − ε < pN ≤ p. But then,
since the sequence is increasing, n > N → pn > pN , so p − ε < pn ≤ p (the second
inequality because pn ∈ S and p is an upper bound for S). Thus for any n > N ,
|p − pn | < ε.
b)We used the least upper bound property to prove the monotone sequence
condition (the conclusion of part a). Now let’s do the reverse. Suppose S is a
bounded set. We’ll define a sequence pn of points in S so that pn converges to the
least upper bound of S.
Let p1 be any point in S. Having define pk , we will pick pk+1 as follows: Let
n be least so that there exists any element of S which is > pk + 2−n . Let pk+1
be such an element. If there is no such n, then pk is in S but there is no greater
element of S, so pk is the least upper bound. So, we’re either done, or we have a
sequence (pn ) which is increasing. Since it’s in S, it’s also bounded. So, by the
monotone sequence condition, (pn ) converges, say to the point p. We’ll show that
p is the least upper bound for S.
Suppose p were not an upper bound for S. Then there is a point a ∈ S which is
bigger than p. Let l be so that 2−l < a−p. As (pn ) converges to p, there is a k so that
p − 2−l−1 < pk < p. But then there is a point (namely a) of S so that a > pk + 2−l ,
so we chose pk+1 to be such a point. So pk+1 > pk + 2−l > p − 2−l−1 + 2−l > p.
This is a contradiction, since the sequence (pn ) is increasing, so it will never (after
k + 1) get closer to p than pk+1 is.
6
Suppose p were not the least upper bound for S: Say y < p is an upper bound
for S. But since (pn ) converges to p, there is a k so that |pk − p| < p − y. But then
pk > y showing that y isn’t an upper bound for S.
16. 2.31
a) Suppose there are infinitely many elements of the sequence which are ≥ x1 .
If not, there are infinitely many elements of the sequence which are ≤ x1 , and
a symmetric approach will work. We’ll attempt to define a sequence which is
increasing. If we fail, we will build a sequence which is decreasing.
Let p1 = x1 . Pick p2 to be xk2 where k2 > 1 is the least with the property that
xk2 ≥ p1 and infinitely many xl are ≥ xk2 . Then pick p3 to be xk3 where k3 > k2
is the least with the property that xk3 ≥ p2 and infinitely many xl are ≥ xk3 . If
this process continues infinitely, this builds an increasing sequence.
Suppose it fails: At some step, there is no xkn so that xkn ≥ xkn−1 and there
are infinitely many xl ≥ xkn . What’s this get us? For any xkn ≥ xkn−1 , there are
only finitely many xl with xl > xkn . Note that for any xl for l > kn−1 there must
be an xj with j > l so that xkn < xj < xl , since only finitely many are larger.
Let’s now build a decreasing sequence. Choose q1 to be any xl with l > kn and
xl > xkn . Then there must be an xj with j > l and xkn < xj < xl . Let q2 be this
xj . Repeating as such, we build a decreasind sequence.
b)Any bounded sequence has a monotone and bounded subsequence (by part a).
By 2.30a, this is convergent.
17. 1.39
a) If f is uniformly continuous, that means for every ε > 0 there exists a δ > 0
which simultaneously works for every point x (i.e., δ doesn’t even depend on x).
This is certainly good enough to see that for every x and every ε > 0 there is a
δ > 0 which works. The other way, f (x) := sin( x1 ) has the property that in any
interval (0, δ) there is an a and a b so that f (a) = −1 and f (b) = 1. Thus if any
single δ was claimed to work for ε = 1, it would fail for the point x = δ.
b) Yes! For any ε, let δ = 2ε . (Fill in the details)
1
c) No! For any δ, you can always find an x so that (x + δ)2 − x2 > 1 (say x = 2δ
will work). This shows that for ε = 1, no δ can work.
18. 2.1
a) (0, 1) is open in R:
Given p ∈ (0, 1), let ε = min{p, 1 − p}. Then Bε(p) = (p − ε, p + ε) ⊇ (0, 1). Since
p ∈ (0, 1), ε is > 0, so this shows that (0, 1) is open.
b) (0, 1) is not open in R2 , where (0, 1) represents the associated interval on the
x-axis.
Let p = 21 ∈ (0, 1). Note that in R2 , p is the point with x-coordinate 21 and
y-coordinate 0. Let ε > 0 be given. Then the point with x-coordinate 12 and
y-coordinate 2ε is within the ball Bε (p) but is not in the set (0, 1).
19. 2.2
Answer: For the intervals [a, b] where both a and b are irrational numbers.
Claim 19.1. If either a or b is rational, then [a, b] ∩ Q is not open in Q.
7
Proof. Consider the point p which is either a or b (whichever is rational). Then
there is no ε so that Bε (p) ∩ Q ⊆ [a, b].
Claim 19.2. If both a and b are irrational, then [a, b] ∩ Q is clopen in Q.
Proof. Openness: Let p be a point in [a, b] ∩ Q. By assumption, p cannot be a or
b. Let ε = min{p − a, b − p}. Then Bε (p) ∩ Q ⊆ [a, b] ∩ Q.
Closedness: We need to show that if (pn )n∈N is a sequence in [a, b] ∩ Q which
converges to p ∈ Q, then p ∈ [a, b] ∩ Q. But [a, b] is closed in R, so p ∈ [a, b] and
p ∈ Q, so p ∈ [a, b] ∩ Q.
20. 2.3
We need to show that S = {p} is closed in M for any p ∈ M . Let (pn )n∈N be
a sequence in S. Then since S contains only the point p, the sequence is so that
pn = p for each n ∈ N. But then the limit of (pn ) is the point p, which is in S.
Thus S is closed.
Since the finite union of closed sets is closed, we see that any finite set is closed.
21. 2.5
C
S is open if and only if S is closed. This means all of the limit points of S C
are in S C . But this just means that all of the limit points of S C are not in S, or
no points of S are limit points of its complement.
22. 2.11
Let f be the function sending an open set U to the closed set U C . Let g be the
function sending a closed set K to the open set K C . Note that both f g and gf are
the identity function, so they are both bijections.
23. 2.12
a) First suppose M is a metric space with the discrete metric. We will show that
every set is open, which (by taking complements) shows that every set is closed as
well. To see that every set is open, since every set is a union of single points, we
need to only see that every set of the form S = {p} is open. This is true by taking
ε = 21 . Bε (p) = {p} ⊆ S. Thus S is open, and every set is open. As there is a
bijection between the topologies of homeomorphic metric spaces, this shows that
every subset of any homeomorph of a discrete space is open.
b)Recall that f : M → N is continuous is f pre (U ) is open for every open U ⊆ N .
But if every subset of M is open, this this condition must hold.
c)Suppose (pn ) converges to the point p. As the set S = {p} is open, there is
some ε so that {p} ⊆ Bε (p) ⊆ {p}, so Bε (p) = {p}. But then there is some N ∈ N
so that n > N → pn ∈ Bε (p) = {p}, so from some point onwards the sequence is
constant. Thus convergent implies constant from some point onwards. Conversely,
if a sequence is constant from some point onwards, it is certainly convergent.
8
24. 2.14
Recall dist(p, S) = inf{d(p, s) | s ∈ S}.
a)
Claim 24.1. For every p ∈ M and S ⊆ M , p ∈ lim(S) if and only if dist(p, S) = 0
Proof. →: If p is a limit point of S, then for every ε > 0, there is a point s ∈ S
with d(p, s) < ε. Thus dist(p, S) < ε. By the ε-principle, dist(p, S) ≤ 0. But 0 is a
lower bound for {d(p, s) | s ∈ S}, so dist(p, S) ≥ 0, thus dist(p, S) = 0
←: If dist(p, S) = 0, we’ll show that p is a limit point of S. We know it
suffices to show that for every ε > 0, Bε (p) ∩ S 6= ∅. Let ε > 0 be given. Then
dist(p, S) = 0 < ε means that ε is not a lower bound for {d(p, s) | s ∈ S}. Thus
there is a point s ∈ S so that d(p, s) < ε. But then s ∈ Bε (p) ∩ S, showing that the
intersection is non-empty, and thus p is a limit point of S.
25. 2.34
f : M → N is open if for each open set U ⊆ M , the image set f (U ) ⊆ N is open
in N .
a) Let f : R → Z be the rounding function sending x to the nearest integer. By
problem 2.12 (or an easy variant of it), every subset of Z is open, so f is open. It
is not continuous, however. In particular, it is not continuous at the point 21 ∈ R.
b)Yes, we showed that f gives a bijection between the open sets of M and
the open sets of N . In particular, f −1 is continuous and since f is a bijection,
f (U ) = (f −1 )pre (U ) is open.
c)Yes. Since f is a bijection, (f −1 )pre (U ) = f (U ), so f −1 is continuous (since f
is open). Thus f and f −1 are continuous bijections, so f is a homeomorphism.
d)No. Consider the map f (x) = x(x − 1)(x + 1), and let U = (0, 1) Then
f (U ) = (0, √13 ], which is not open.
e)Yes! We just need to show that f is injective (1-to-1). Suppose not. Then
we have an a < b with f (a) = f (b). Then f attains a maximum and minimum on
the interval [a, b] (we’ll discuss this in detail when talking about the compactness
theorem. Or: See Theorem 23 in Ch. 1). Let U = (a, b). We’ll see that f (U ) is not
open. If both of these are f (a), then f is constant in the interval and f (U ) = {f (a)}
is not open. If at least one of the max/min is not f (a), then we see that f (U ) is
either a half-open or a closed interval, so f (U ) is not open.
f)Consider the map that send the point with polar coordinates (1, θ) to the
point with polar coordinates (1, 2θ). This loops the circle around itself twice. This
function is surjective, open, and continuous, but it is not injective (both (1, 0) and
(1, π) are sent to (1, 0) = (1, 2π)
26. 2.6
T
T
a) If S ⊆ T , then cl(S) = closed K⊇S K ⊆ closed K⊇T K = cl(S), where the
⊆ above holds since the intersection is over a smaller collection of sets K. OR
Consider a limit point of S. Since S is a subset of T , it’s also a limit point of T .
C C
b) Either use
S part a and the
S fact that int(S) = (cl(S )) OR:
int(S) = open U ⊆S U ⊆ open U ⊆T U = int(T ), where the ⊆ is because the
union is over a larger collection of sets.
9
27. 2.81
a)We’ll use the ε-δ definition of continuity to show that dM : M × M → R is
continuous with respect to the sum-metric on M × M . Note that the function dM
is simply the distance on M . That is, d takes in a pair of points from M : (p, q)
and outputs the distance between the two points in M : dM (p, q).
Fix (p, q) ∈ M × M and an ε > 0. We need to show that there is a δ > 0 so that
dsum ((p, q), (p0 , q 0 )) < δ → dR (dM (p, q), dM (p0 , q 0 )) < ε. We’ll see that δ = ε works.
Without loss of generality, dM (p, q) > dM (p0 , q 0 ) (if it’s not clear why, read the rest
and fill in the other case). Note that dsum ((p, q), (p0 , q 0 )) = dM (p, p0 ) + dM (q, q 0 )
and dR (dM (p, q), dM (p0 , q 0 )) = |dM (p, q) − dM (p0 , q 0 )| = dM (p, q) − dM (p0 , q 0 ) ≤
dM (p, p0 ) + dM (p0 , q 0 ) + dM (q 0 , q) − dM (p0 , q 0 ) ≤ dM (p, p0 ) + dM (q, q 0 ) < δ. Thus if
δ = ε, we see that the conclusion is dR (dM (p, q), dM (p0 , q 0 )) < ε.
b)Yes! We showed (in class) that a function from M × N to any metric space is
continuous in the sum-metric if and only if it is continuous in the Euclidian metric
if and only if it is continuous in the max-matric.
28. 2.83
a)The difficult condition in each case is the triangle inequality, which is what I’ll
explain for each of the three below.
dE : We need to show that for any p = (p1 , p2 ), q = (q1 , q2 ), and r = (r1 , r2 ),
dE (p, q) ≤ dE (p, r) + dE (r, q). Let’s consider what (dE (p, q))2 is (just so we don’t
have to think in terms of square roots yet): it’s d(p1 , q1 )2 + d(p2 , q2 )2 , i.e. it’s just
|(d(p1 , q1 ), d(p2 , q2 ))|2 ≤ |(d(p1 , r1 ) + d(r1 , q1 ), d(p2 , r2 ) + d(r2 , q2 ))|
Let x be the vector (d(p1 , r1 ), d(p2 , r2 )) and let y be the vector (d(r1 , q1 ), d(r2 , q2 )).
Then we need to bound hx + y, x + yi:
hx + y, x + yi = hx, xi + 2hx, yi + hy, yi ≤ |x|2 + 2|x||y| + |y|2 = (|x| + |y|)2
where the inequality is due to the Cauchy-Schwartz inequality.
What this means is that (dE (p, q))2 ≤ (dE (p, r)+dE (r, q))2 . Taking square roots
of both sides yields the triangle inequality.
dsum :dsum (p, q) = dsum ((p1 , p2 ), (q1 , q2 )) = d(p1 , q1 ) + d(p2 , q2 ) ≤ d(p1 , r1 ) +
d(r1 , q1 ) + d(p2 , r2 ) + d(r2 , q2 ) = (d(p1 , r1 ) + d(p2 , r2 )) + (d(r1 , q1 ) + d(r2 , q2 )) =
dsum (p, r) + dsum (r, q)
dmax : dmax (p, q) = dmax ((p1 , p2 ), (q1 , q2 )) = max{d(p1 , q1 ), d(p2 , q2 )} ≤ max{d(p1 , r1 )+
d(r1 , q1 ), d(p2 , r2 )+d(r2 , q2 )} ≤ max{d(p1 , r1 ), d(p2 , r2 )}+max{d(r1 , q1 ), d(r2 , q2 )} =
dmax (p, r) + dmax (r, q)
b) I am currently of the opinion that this part of the problem is false as stated.
Extra points for anyone who can prove that.
29. 2.84
a)The difficult part is to show ρ(p, q) satisfies the triangle inequality: Assuming
d(p, q) ≥ d(p, r) and d(p, q) ≥ d(r, q):
ρ(p, q) =
d(p, r) + d(r, q)
d(p, r)
d(r, q)
d(p, r)
d(r, q)
d(p, q)
≤
=
+
≤
+
= ρ(p, r)+ρ(r, q)
1 + d(p, q)
1 + d(p, q)
1 + d(p, q) 1 + d(p, q)
1 + d(p, r) 1 + d(r, q)
10
If d(p, q) < d(p, r), then since x 7→
ρ(p, q)
x
1+x
is an increasing function:
d(p, q)
d(p, r)
≤
= ρ(p, r) ≤ ρ(p, r) + ρ(r, q)
1 + d(p, q)
1 + d(p, r)
Similarly, if d(p, q) < d(r, q).
b) The identity map is certainly a bijection. We need to see that it is continuous
in both directions. One direction: Given ε > 0 and p ∈ M , we need to find δ > 0
x
is
so that d(p, q) < δ → ρ(p, q) < ε. Let δ = ε. Then since the function x 7→ 1+x
ε
an increasing function, ρ(p, q) < 1+ε < ε.
The other direction: Given ε > 0 and p ∈ M , we need to find δ > 0 so that
ρ(p, q) < δ → d(p, q) < ε.
d(p,q)
< δ, so d(p, q)(1 − δ) < δ, and
If δ < min{ 2ε , 12 } and ρ(p, q) < δ, then 1+d(p,q)
d(p, q) <
δ
1−δ
≤
ε
2
1− 12
≤ ε.
30. 2.92
a)
Claim 30.1. S is clopen if and only if ∂S is empty.
Proof. →: Since S is closed, cl(S) = S. Since S is open int(S) = S. Thus ∂S =
cl(S) r int(S) = S r S = ∅.
←: ∂Scl(S) r int(S) is empty. Thus cl(S) = int(S). But int(S) ⊆ S ⊆ cl(S), so
S = cl(S) = int(S). Thus S is both closed (being equal to its closure) and open
(being equal to its interior).
b) ∂S C = cl(S C ) ∩ cl((S C )C ) = cl(S C ) ∩ cl(S) = cl(S) ∩ cl(S C ) = ∂S.
c)
Claim 30.2. ∂S is closed for any set S.
Proof. ∂S = cl(S) r int(S) = cl(S) ∩ (int(S))C . But this is the intersection of two
closed sets, so is closed.
Then since the closure of a closed set is itself, ∂∂S = cl(∂S) r int(∂S) = ∂S r
int(∂S) ⊆ ∂S. d)
Claim 30.3. If A ⊆ B, then int(A) ⊆ int(B).
S
S
Proof. int(A) = open U ⊆A U ⊆ open U ⊆B U = int(B)
So, ∂∂∂(S) = cl(∂∂S) r int(∂∂S) = ∂∂S r int(∂∂S) = (cl(∂S) r int(∂S)) r
int(∂∂S) = cl(∂S)r(int(∂S)∪int(∂∂S)). But ∂∂S ⊆ ∂S (by part c), so int(∂∂S) ⊆
int(∂S). Thus ∂∂∂(S) = cl(∂S) r (int(∂S) ∪ int(∂∂S)) = cl(∂S) r int(∂S) = ∂∂S.
e)
Claim 30.4. cl(S ∪ T ) ⊆ cl(S) ∪ cl(T ) and int(S) ∪ int(T ) ⊆ int(S ∪ T ).
Proof. Suppose (pn )n∈N is a sequence in S ∪T which converges to the point p. Then
either infinitely many of the pn are in S or infinitely many are in T . Thus we can
take a subsequence entirely in S or entirely in T . Thus p is either a limit point of S
or a limit point of T . Thus lim(S ∪T ) ⊆ lim(S)∪lim(T ), so cl(S ∪T ) ⊆ cl(S)∪cl(T ).
Every open subset of S or T is an open subset of S ∪ T . Thus int(S) ∪ int(T ) ⊆
int(S ∪ T ).
11
So, ∂(S ∪ T ) = cl(S ∪ T ) r int(S ∪ T ) ⊆ cl(S) ∪ cl(T ) r int(S ∪ T ) ⊆ cl(S) ∪
cl(T ) r (int(S) ∪ int(T )) ⊆ ∂S ∪ ∂T .
f)Q ⊆ R. The boundary is all of R, and the boundary of that is ∅.
g) Consider S = {0} and T = R both as subsets of R. ∂S = S = {0} and
∂T = ∅. So ∂(S ∪ T ) = ∅ ( {0} = ∂S ∪ ∂T .
31. 2.40
a) Suppose ((pn , f (pn )) is a sequence in M × R which converges in M × R, say
to (p, r). Then the sequence (pn ) converges to p in M and the sequence (f (pn ))
converges to r in R. By continuity of f , f (p) = r. But that means (p, r) is in the
graph of f .
b) Consider the set f M . It’s a compact subset of R. Thus M × f M is compact.
But the graph of f is a closed (by a) subset of M × f M . A closed subset of a
compact set is compact (Theorem 35), so the graph of f is compact.
c) We’ll use the convergent sequence condition for continuity. Suppose (pn )
converges in M to p. We need to show that (f (pn )) converges to f (p) in R. Suppose
towards a contradiction that it doesn’t. Then there is some ε > 0 so that infinitely
many f (pk )’s are not in Bε (f (p)). Look at the subsequence (qk ) consisting of those
(pn ) where d(f (pn ), f (p)) > ε. Then the sequence (qk , f (qk )) is a sequence in
the graph of f . That sequence has a convergent subsequence in the graph of f :
(rl , f (rl )). But (rl ) converges to p, so the limit of this sequence is (p, s) for some
s. On one hand, this s cannot be f (p) since f (rl ) is never within ε of f (p). On
the other hand, this s must be f (p) since (p, s) is in the graph of f . This is the
contradiction we wanted.
d)f (x) = x1 for x 6= 0 and f (0) = 0.
32. 2.41
2
S is compact, since it’s a closed and bounded subset of R3 . R2 is not compact
since it’s not a bounded set. Therefore, the two are not homeomorphic.
33. 2.43
The diameter of each of the Kn ’s is ≥ µ. So, for each n, there is a pair of
points (xn , yn ) in Kn so that d(xn , yn ) > µ − n1 (note by problem 71 below, we
really could take d(xn , yn ) ≥ µ). Take a subsequence (pn , qn ) of the sequence
(xn , yn ) which converges to (p, q) (we just used that K1 × K1 is compact). We
need to check two things. First, the limit (p, q) is in ∩n Kn . This is exactly the
same as the proof that ∩n Kn is non-empty. Namely, after the first k entries, the
sequence is entirely in Kk × Kk . Therefore, since Kk × Kk is closed (since it’s
compact), the limit is in Kk . Since this is true for every k, the limit (p, q) is in
∩n Kn . Second, we need to see that d(p, q) ≥ µ. This is by the triangle inequality
and the ε-principle. We’ll show that for each ε > 0, d(p, q) > µ − ε. Given
ε > 0, let k be large enough that d(pk , p) < 3ε , d(qk , q) < 3ε and ε > k3 . Then
µ − 3ε ≤ µ − k1 ≤ d(pk , qk ) ≤ d(pk , p) + d(p, q) + d(q, qk ) < 3ε + d(p, q) + 3ε . So
µ < ε + d(p, q), or d(p, q) > µ − ε.
12
34. 2.71
First let’s note that A × B is compact. Define a function f : A × B → R by
f (a, b) = d(a, b). Once we show that this function is continuous, we will know that
it attains its minimum on A × B. That is, there will be points a0 ∈ A and b0 ∈ B
so that d(a0 , b0 ) is as small as possible.
So, now let’s show f is continuous. We can use any of the three product metrics on A × B, so let’s choose to use dmax . Given ε > 0, we want to find a
δ > 0 so that dmax ((a, b), (a0 , b0 )) < δ → |f (a, b) − f (a0 , b0 )| < ε. Let δ = 2ε .
Then dmax ((a, b), (a0 , b0 )) < δ means d(a, a0 ) < δ and d(b, b0 ) < δ. Then d(a, b) ≤
d(a, a0 ) + d(a0 , b0 ) + d(b0 , b) < 2δ + d(a0 , b0 ). Thus d(a, b) − d(a0 , b0 ) < 2δ = ε. By
symmetry (same argument flipping a with a0 and b with b0 ), d(a0 , b0 ) − d(a, b) < ε,
so |d(a0 , b0 ) − d(a, b)| < ε.
35. Prelim 5
First of all, we’ll use the fact that f isn’t just continuous but is also uniformly
continuous on [0, 1] × [0, 1]. We’ll show the same about g. Given ε > 0, we want a
δ > 0 so that |x − y| < δ → |g(x) − g(y)| < ε. So, let ε > 0 be given. There is some
δ so that if dmax ((p, q), (p0 , q 0 )) < δ, then |f (p, q) − f (p0 , q 0 )| < ε (this is uniform
continuity of f ). We’ll use this δ and show that if |x − y| < δ, then g(y) > g(x) − ε.
By symmetry, this implies that |g(x) − g(y)| < ε.
Given |x − y| < δ, take the point r ∈ [0, 1] so that f (x, −) attains it maximum
value at r (i.e., f (x, r) = g(x)). Then dmax ((x, r), (y, r)) < δ. So |f (x, r) − f (y, r)| <
ε. In particular, g(y) ≥ f (y, r) > f (x, r) − ε = g(x) − ε. By the last paragraph,
this is all we needed.
36. Prelim 13
a) Suppose towards a contradiction that the sequence does not converge to x.
That means that for some ε > 0 there are infinitely many xn so that d(xn , x) >
ε. Look at the subsequence of all xn with d(xn , x) > ε. There is a convergent
subsequence of this, by compactness of A. But it’s limit cannot be x. This is a
contradiction.
b)(0, 1, 0, 2, 0, 3, 0, 4, 0, . . .)
37. Prelim 14
Let’s show first that there are A and B so that |f (x)| ≤ A + Bx for all positive x.
This is really enough, since then (by considering f (−x)) we know there are A0 , B 0
so that |f (x)| ≤ A0 + B 0 (−x) for all negative x. By taking C to be the max of A
and A0 and D to be the max of B and B 0 , we have |f (x)| ≤ C + D|x| for all x ∈ R.
By uniform continuity, there is some δ so that |x − y| ≤ δ implies |f (x) − f (y)| <
1. Let A = |f (0)| + 1 and B = 1δ . Let’s see that this works. Consider all x in the
interval [0, δ]. By the condition, |f (x) − f (0)| < 1, so |f (x)| < |f (0)| + 1 = A <
A + Bx. Consider all x in the interval [δ, 2δ]. By the condition |f (x) − f (δ)| < 1,
so |f (x)| < |f (δ| + 1 < A + 1 = A + Bδ ≤ A + Bx. This should give a convincing
picture; now let’s make it formal.
By induction, let’s show |f (kδ)| ≤ A + k − 1. It’s true for 0 by definition
of A as |f (0)| + 1. Suppose it’s true for k. Then |f ((k + 1)δ) − f (kδ)| < 1, so
|f ((k + 1)δ)| ≤ |f (kδ)| + 1 ≤ A + k − 1 + 1 = A + (k + 1) − 1. Now, given any
13
positive x, there is some k so that kδ ≤ x < (k + 1)δ. Then |f (x) − f (kδ)| < 1, so
|f (x)| < |f (kδ)| + 1 ≤ A + k − 1 + 1 = A + k = A + B · kδ ≤ A + Bx. By the first
paragraph, we needed only to worry about positive x, so we’re done.
38. 2.54
No. Consider two closed circles in R2 which intersect at a point on their bounderies. The interior is two open circles, which is not connected.
39. 2.55
No to both. Consider Q as a subset of R. The closure is R which is connected,
and the interior is ∅, which is connected.
40. 2.58
Let a be any point in U . Consider the set S consisting of all points b so that
there is some path in U from a to b. S is non-empty as a ∈ S (the trivial path
connecting a to a). We’ll show that it’s clopen in U . Given a point b ∈ S, there
is some ε > 0 so that Bε (b) ⊆ U . Then there is a path in U from b to any point c
in Bε (b). By concatenating paths, this gives a path from a to any point c in Bε (b)
(first go from a to b, then go from b to the point c). Thus Bε (b) ⊆ S, showing that
S is open. To see that S is closed in U , we’ll check that U r S is also open. Suppose
b ∈ U r S. Then there is some ε > 0 so that Bε (b) ⊆ U . For any c ∈ Bε (b) ⊆ U ,
there is a path from b to c. Thus there is a path from c to b. Thus, if there were a
path from a to c, there would also be a path from a to b. So there is no path from
a to c. We conclude that Bε (b) ⊆ U r S, so U r S is open. Thus S is a clopen
non-empty subset of U . So S must be all of U , since U is connected. Thus for any
point b in U there is a path from a to b. Finally, given any two points b and c in
U , there is a path from b to a and a path from a to c, so there is a path from b to
c. Thus U is path connected.
41. 2.59
In fact, the annulus is path-connected. Given two points in polar coordinates
(θ, x) and (θ2 , y), there is a path going from (θ, x) to (θ, y) (moving directly, not
changing θ throughout the path). There is also a path from (θ, y) to (θ2 , y) moving
around the annulus keeping y fixed. Concatenating these paths gives us a path
from one point to the other. This shows path-connectedness, thus connectedness.
42. 2.68
Suppose, towards a contradiction, that X = (a, b) and Y = [a, b) are homeomorphic. Then Y r {a} is homeomorphic to X r {c} for some point c ∈ (a, b).
But X r {c} = (a, c) ∪ (c, b), which is disconnected and Y r {a} = (a, b), which is
connected. This is a contradiction, showing that X and Y are not homeomorphic.
14
43. 2.78
a)Let M be a connected metric space, ε > 0 and let a, b be two points in M . We
want to show that there is an ε-chain taking a to b.
Let A = {p ∈ M | there is an ε-chain from a to p}. Then A is open. This is
because if p ∈ A, then Bε (p) ⊆ A. Also, A is closed: If (pn ) is a sequence in A
limiting to p, then for some pk , d(pk , p) < ε. So, there is an ε-chain going from a
to pk , which gives an ε-chain from a to p. Thus A is clopen and a ∈ A, so it is
non-empty. Therefore, A = M , and in particular, b ∈ A.
b)Suppose towards a contradiction that M is not connected. Then there are two
closed non-empty disjoint sets A and B so that M = A ∪ B. Then A and B are also
both compact. Thus they attain a minimum distance to each other. That is, there
is a ∈ A and b ∈ B so that whenever c ∈ A and e ∈ B, d(a, b) < d(c, e). Let ε be
less than this distance. Now let’s see, for our contradiction, that M is not ε-chain
connected. That is, if you start in A, there is no way to get to B via an ε-chain.
This is simply because inductively, d(pi , pi+1 ) < ε, so if pi ∈ A, then pi+1 cannot
be in B. Therefore, pi+1 must also be in A.
c)Yes. For any ε and any a < b ∈ R r Z, you can get from a to b taking steps
of size < ε as follows: Let’s define the pi inductively. p0 = a. Given pn < b, we
choose pn+1 as follows. If d(pn , b) < ε, then pn = b. Otherwise, if pn + 2ε ∈ R r Z,
let pn+1 = pn + 2ε . Otherwise, let pn+1 = pn + 2ε + min{ 4ε , 12 }. In any of the three
cases, pn+1 ∈ R r Z and the induction continues. But the distance from pn to b
goes down by at least 2ε at each step, so eventually the distance is < ε, and then
pn+1 = b.
1
}.
d)No! Let M be the following subset of R2 : {(x, y) ∈ R2 | y > |x|
44. Prelim 15
Solution 1: We need to find what g(1) should be. For any ε > 0, there is some
δ > 0 so that |x − y| < δ implies |h(x) − h(y)| < ε. Let’s look at what this means if
we take x ∈ (1 − δ, 1). Then any point y ∈ (1 − δ, 1) is within δ of x. Thus for any
point y ∈ (1 − δ, 1), h(y) ∈ (h(x) − ε, h(x) + ε). Let’s call the set [f (x) − ε, f (x) + ε]
by the name Sε .
Let’s use this over and over again to zoom in on the right value. For each n, let’s
do that above for ε = n1 . Then for some δ, whenever y ∈ (1 − δ, 1) then h(y) ∈ Sε
and the diameter of Sε = 2ε = n2 . Then there is a single number in ∩n S n1 . We
define g(1) = this single number.
Let’s check that g is continuous: Given ε > 0, there is an N so that N1 < 2ε .
But then we saw that there is some δ so that for every y ∈ (1 − δ, 1), g(y) ∈ S N1 .
By construction, g(1) is also in S N1 , but the diameter of S N1 is N2 ≤ ε. Thus
|g(y) − g(1)| < ε whenever y ∈ (1 − δ, 1).
To see that there is only one choice of g(1) which makes g continuous: Suppose
there were two such functions g and g 0 . By continuity of both, (g(1− n1 ))n converges
to g(1) and (g 0 (1 − n1 ))n converges to g 0 (1). But these two sequences are the same,
so g(1) = g 0 (1) and g = g 0 .
Solution 2: (Credit to Tejas)
Lemma 44.1. If f : X → Y is uniformly continuous, and (pn ) is a Cauchy
sequence in X, then (f (pn )) is a Cauchy sequence in Y .
15
Proof. Let ε > 0 be given. We need to show that for some N , n, m > N
d(f (pn ), f (pm )) < ε. By uniform continuity, there is a δ > 0 so that d(p, q) < δ
d(f (p), f (q)) < ε. Also, since (pn ) is Cauchy, there is some N so that n, m > N
d(pn , pm ) < δ, so n, m > N → d(f (pn ), f (pm )) < ε.
→
→
→
It follows that any sequence (pn ) in [0, 1) which converges to 1 is Cauchy, so
(f (pn )) is Cauchy (and in R) so convergent. We want to show that if (pn ) and (qn )
are two sequences in [0, 1) which converge to 1, then (f (pn )) and (f (qn )) converge
to the same limit. To see this, consider the sequence (p1 , q1 , p2 , q2 , . . .). This sequence also converges to 1, so is Cauchy, and so (f (p1 ), f (q1 ), f (p2 ), f (q2 ), . . .) is
a convergent sequence. Let L be the limit. Then (f (pn )) is a subsequence, so it
has the same limit L. Also (f (qn )) is a subsequence so it has the same limit L.
Thus we see that there is a value L so that: For every sequence (pn ) in [0, 1) which
converges to 1, (f (pn )) converges to L. Let g(1) = L, and the convergent sequence
condition of continuity guarantees that g is continuous. Uniqueness follows, since
any other function sends 1 to a value which is not L, so the convergent sequence
condition for continuity guarantees that such a funciton is not continuous.
45. If A ⊆ M is covering compact, then A is closed.
We’ll show that if x ∈
/ A then x ∈
/ cl(A). This is equivalent to cl(A) ⊆ A,
or that A is closed. Given x ∈
/ A, define the following collection of open sets
Un = {y ∈ M | d(y, x) > n1 }. The collection of these U = {Un | n ∈ N} is an
open cover of A (since x is the only non-covered point, and x ∈
/ A). Thus there is
a finite subcover of U. That means that there is some UN so that A ⊆ UN . But
that means d(x, a) > N1 for every a ∈ A. Thus x ∈
/ cl(A).
46. 2.46
Let U be an open covering of [a, b]. We want to show that there is a finite subcover. We define the set C = {x ∈ [a, b] | [a, b] is covered by a finite subcover of U}.
If we can show that b ∈ C, then we will know that [a, b] can be covered by a finite
subfamily of U. The first step will be to show that b is the least upper bound for
C. We know that C has a least upper bound.
Suppose towards a contradiction that it is c and c < b. There is a set U ∈ U
which contains c. Since U is open, it contains Bε (c) for some ε > 0. Since c is the
least upper bound for C, there is some d ∈ C which is in Bε (c). So, [a, d] is covered
by some finite subcover V of U. And thus [a, c + 2ε ] is covered by {U } ∪ V. So, c + 2ε
is in C contradicting c being an upper bound.
Thus b is the least upper bound for C. Now we’ll show there is a finite subset
of U which covers [a, b]. The argument is the same as above. There is a set U ∈ U
which contains b. Since U is open, it contains (b − ε, b] for some ε > 0. Since b is
the least upper bound for C, there is some d ∈ C which is in (b − ε, b]. So, [a, d] is
covered by some finite subcover V of U. And thus [a, b] is covered by {U } ∪ V. So,
b is in C, and we are done.
47. 2.47
A is a closed subset of a covering compact set K. Let U be an open covering of
A. Let V be U ∪ {M r A}. Then V is an open cover of K. Thus there is a finite
16
subcover V0 of K. If we exclude the set K r A from the cover V0 , we have a finite
subset of U which covers A. Thus A is covering compact.
48. 2.48
Suppose U is an open cover of f A. Let V be {f pre (U ) | U ∈ U}. Then since f is
continuous, V is a collection of open sets. Since U covered f A, V covers f pre (f A),
which contains A, so V is an open cover of A. Thus there is a finite subset V0 of V
which still covers A. It remains to show that the corresponding finite subset of U
is a finite subcover of f A.
49. 2.49
If A is a closed subset of M , then by 2.47, it is covering compact, and by 2.48
f A is covering compact. By the lemma, f A is closed. So f takes closed subsets of
M to closed subsets of N . Thus f −1 is continuous (the f −1 pre-image of a set A is
just f A).
50. 2.50
Let V be the set of all sets B 2ε (p) where p ∈ N . Let U consist of the pre-images
of sets in V. That is, U = {f pre (B 2ε (p)) | p ∈ N }. By the Lebesgue number lemma,
there is a λ so that for any q ∈ M , there is some U ∈ U so that Bλ (q) ⊆ U . That
means that Bλ (q) ⊆ f pre (B 2ε (p)) for some p ∈ N . But that means that for any two
a, b ∈ Bλ (q), d(f (a), f (b) ≤ d(f (a), p) + d(f (q), p) < 2ε + 2ε = ε. So, take δ = λ, and
we see that for each q, d(x, q) < δ implies d(f (x), f (q)) < ε.
51. 2.51
T If A1 ⊃ A2 ⊃ . . . are non-empty covering compact sets, we
T want to show that
A
is
non-empty.
Suppose,
towards
a
contradiction,
that
n n
n An = ∅. Then we
form the following
cover
of
A
:
U
=
{M
r
A
},
and
U
=
{U
1
k
k
k | k ∈ N}. Since
T
the intersection n An is empty (our assumption), U is an open cover of A1 . Thus
there is a finite subcover of U. But that means A1 ⊆ U1 ∪ U2 ∪ . . . ∪ Uk for some
finite k. So A1 ⊆ Uk . Thus, Ak = ∅, giving us a contradiction.
52. 2.52
Suppose C = {Ci | i ∈ I} (arbitrarily many of these) which have the finite intersection property (the intersection of any finitely
many of them is non-empty).
T
We want to show that the grand
intersection
C∈C C is non-empty. Suppose toT
wards a contradiction that C∈C C = ∅. Then let U = {M r Ci | i ∈ I}. This
U is an open cover of M . Since M is covering compact, there is a finite subcover
U0 = {M r Ci1 , . . . M r Cik }. But U0 does not cover Ci1 ∩ Ci2 ∩ . . . ∩ Cik . And this
intersection is non-empty. This is a contradiction.
53. 2.53
We are given M so that every family of closed sets with the finite intersection
property has a non-empty intersection. Given U = {Ui | i ∈ I} an open covering of
M , we want to show that there is a finite subcover. Suppose towards a contradiction
that there is no finite subcover. Then let C be the family of sets Ci = M r Ui .
If no finite collection of the U ’s covers M , then no finite intersection of the C’s is
17
empty. Thus C has the
T finite intersection property. Thus it has non-empty grand
intersection. Say x ∈ i∈I Ci . Then x is not covered by U. This is a contradiction
to U being a covering of M .
54. 2.94
n
n
a)Q is a countable subset of R . It is dense, since any open set contains a ball
of radius ε for some ε, and any such ball contains an element of Qn .
b)We will construct a dense countable subset as follows. For each n ∈ N,
let Un = {B n1 (p) | p ∈ M }. Since M is compact, there is a finite subcover:
{B n1 (p0 ), . . . B n1 (pk )}. Let An = {p0 , . . . , pk } and let A = ∪n An . This A should be
a countable dense subset of M . We need to show that for any a and any ε > 0,
Bε (a) ∩ A 6= ∅. Let N be any number so that N1 < ε. Then q is in M , which
is covered by {B N1 (p0 ), . . . B N1 (pk )} for some points p0 , . . . , pk which we put in A.
Thus for one of these points in A, pi ∈ Bε (a) ∩ A.
55. 3.7
This is almost verbatim the same proof as on page 143-144. Mostly, it’s changing
b’s to ∞’s.
56. 3.11
a) i) In this case, one of
f (βn )−f (0)
βn
and
f (0)−f (αn )
−αn
is ≤
f (βn )−f (αn
.
βn −αn
Similarly, one
f (βn )−f (αn
.
βn −αn
of the two is ≥
This is because if the slope of the line from (αn , f (αn ))
to (0, f (0) and the slope of the line from (0, f (0)) to (βn , f (βn )) are both > S, then
the slope from (αn , f (αn )) to (βn , f (βn )) is also > S. (You should probably write
out something algebraic to verify this, but in fact the longer slope is a weighted
average of the other two, so they can’t be both bigger or both smaller than it).
(0)
(αn )
Thus since each of f (βnβ)−f
and f (0)−f
is approaching 0 as n → ∞, it’s also
−αn
n
(αn
true that f (ββnn)−f
must approach 0 as it is sandwiched between the two.
−αn
ii)A bit of a mess. I would recommend first writing out the case where f 0 (0) = 0
and f (0) = 0 then “shift and tilt” the proof.
iii)By the mean value theorem, Dn = f 0 (cn ) for some cn ∈ (αn , βn ). Then
0
f (cn ) → f 0 (0) as cn → 0 (by continuity of f 0 ).
2
b) Let αn = (2n+1)π
. Then f 0 (αn ) = 1. Then for some small enough hn ,
f (αn +h)−f (αn )
h
≥ 1 − n1 . Let βn = αn + hn . Then Dn limits to 1.
57. 3.13
a) Suppose f (0) = b > 0 (the other case is symmetric). Consider the line l
defined by y = Lx + b.
Claim 57.1. For any x > 0, the point (x, f (x)) is below the line l. i.e. f (x) ≤
Lx + b.
Proof. Suppose towards a contradiction that x > 0 is such that f (x) > Lx + b.
(0)
Then by the MVT, there is some point c ∈ (0, x) such that f 0 (c) = f (x)−f
. But
x
0
0
then f (c) > L contradicting our assumption that f (c) < L for every c.
18
Thus for any x so that Lx + b < x (for any large x this holds), we see that
f (x) − x < 0. Since f (0) − 0 > 0, it follows that f (c) − c = 0 for some c in between
(by the IVT).
b)Consider the function f (x) = x + e−x .
58. 3.14
a) This is clear at any point other than x = 0. At 0, we need to consider the
following limit:
1
lim+
h→0
,
1
e− h
= lim+ h1
h
h→0 e h
which, by L’Hopital’s rule is equivalent to
lim
− h12
h→0+
1
− h12 e h
=0
.
1
Using a similar argument, you can show that limh→0+ pq e− x = 0 for any polynomials p, q. Then by induction, you can see that each of the derivatives is defined
and is 0 at 0.
b)No! If it were, then it would be the constant 0 function, but e is not.
c)First, β is zero outside of (−1, 1), since either e(1 − x) or e(x + 1) is 0 outside
of that interval. On (−1, 1), it’s easy to see that each term is > 0, so the product
is > 0.
An induction on k shows that if f, g are C k , then f ◦ g and f g are C k (the
proof is basically the product rule and chain rule used several times). Thus the
composition and product of smooth functions here is smooth. Plugging in 0, you
get e2 · e−1 e−1 = 1, so β(0) = 1.
d) Plug and chug.
59. 3.26
It should also state the assumption that for any ω1 , ω2 , either ω1 ω2 or ω2 ω1 .
2|
a)Suppose towards a contradiction that limΩ f = L1 and L2 . Let ε be < |L1 −L
.
2
Then there is an ω1 and an ω2 so that ω1 ω → |f (ω) − L1 | < ε and ω2 ω →
|f (ω) − L2 | < ε. Then using either ω1 ω2 or vice-versa, you get that L1 −L2 < 2ε,
which contradicts the choice of ε.
b)The usual stuff: If limf = F and limg = G, then let ωf be good enough to
ensure f (ω) is 2ε -close to F and let ωg be good enough to ensure g(ω) is 2ε -close to
G. Then whichever is bigger is good enough to ensure f + g is is ε-close to F + G.
etc.
c) Read the two definitions. They say exactly the same.
d)
lim R(f + cg) = lim(R(f ) + cR(g)) =by (b) lim R(f ) + c lim R(g)
Ω
Ω
Rb
Ω
. The term on the right is a (f +cg) dx and the term on the left is
by (c).
e)Replace |f (ω) − L| < ε by dM (f (ω), L) < ε.
Ω
Rb
a
f dx +c
Rb
a
g dx
19
60. 3.18
a) Suppose (ak )k∈N is a sequence in Dκ converging to a. We need to show that
osca (f ) = limh→0 sup{f (x) | x ∈ (a − h, a + h)} − inf{f (x) | x ∈ (a − h, a + h)} ≥ κ.
We’ll show that for any h, sup{f (x) | x ∈ (a − h, a + h)} − inf{f (x) | x ∈ (a − h, a +
h)} ≥ κ, thus the limit is also ≥ κ.
Given h, choose any ak so that |a − ak | < h2 . Since oscak (f ) ≥ κ, sup{f (x) |
x ∈ (ak − h2 , a + h2 )} − inf{f (x) | x ∈ (a − h2 , a + h2 )} ≥ κ. Thus there are
points b, c ∈ (a − h2 , a + h2 ) ⊆ (a − h, a + h) so that f (b) − f (c) ≥ κ. Thus
sup{f (x) |Sx ∈ (a − h, a + h)} − inf{f (x) | x ∈ (a − h, a + h)} ≥ κ.
b)D = n∈N D n1 , so D is a countable union of closed sets.
T
c)The complement of D is n∈N (Df rac1n )C , so it’s a countable intersection of
open sets.
61. 3.28
a) If f is Riemann integrable, then limmesh(P )→0 → L. Since the partitions
described in the calculus book are a list of partitions with mesh(P ) → 0, it follows
that the Riemann sums limit to L.
b)Consider χQ .
c)Really (if it were defined), we’d want the integral of χQ to be 0, but the
calculus-book style integral thinks it should be 1. So it’s really no good for us.
62. 3.29
1 → 3: For an interval (ai , bi ), the diameter is just bi − ai .
3 → 2: If ∪Si is a covering of Z and the total diameter is < ε. For each Si , by
definition of diameter sup(Si ) − inf(Si ) = diam(Si ). Thus ∪[sup(Si ), inf(Si )] is a
cover of Z satisfying 2.
2 → 1: If ∪[ai , bi ] covers Z, then ∪(ai , bi ) covers Z r ∪i {ai , bi }. But then
Z r ∪i {ai , bi } is a zero-set and so is ∪i {ai , bi }, since it is countable. Thus the union
is a zero-set and contains Z, so Z is a zero-set.
63. 3.30
First show that [0, 1] cannot be covered by finitely many open intervals (ai , bi )
of total length < 1. Suppose towards a contradiction that it was covered by finitely
many open intervals (ai , bi ) of total length < 1. We can order these so that a1 < 0,
a2 < b1 , a3 < b2 ,. . ., an < bn−1 , bn > 1. This is because some interval must cover
b1 , we just choose 2 to index this interval. Similarly, some interval must cover b2 ,
so we just choose 3 to index
P this interval, etc.
P
Now the total length is i≤n (bi −ai ) = bn −a0 + i≤n−1 bi −ai+1 ≥ bn −a0 ≥ 1.
Thus the total length couldn’t have been < 1. So, we can’t cover [0, 1] with finitely
many intervals of total length < 1.
Now, suppose towards a contradiction that [0, 1] is covered by countably many
open intervals of total length < 1. Then, by compactness of [0, 1] a finite subset
suffices to cover it. But the total length of this finite set of open intervals is < 1
contradicting what we just showed.
64. 3.31
Same as for normal cantor set.
20
65. 3.43
If f and g are antiderivatives of the same function, then f 0 (x) = g 0 (x) for all x,
thus (f − g)0 (x) = 0 for all x, so f − g is a constant function by an easy corollary
of the MVT.
66. 3.47
1
M
1
f (x) ,
1
is a bound for
and since f (x) is never 0, f (x)
is continuous at exactly
1
∈ R.
the points where f is continuous, so by the R-L theorem, f (x)
1
If there is no bound M as in the problem, then f (x) is unbounded, thus nonintegrable.
67. 3.50
This is essentially exactly Corollary 29 in the text. Take a look at that.
68. 3.51
a) Corollary 26
b) Consider f (x) = −1 if x ∈ Q and f (x) = 1 if x ∈ R r Q.
c) Corollary 26
d) Corollary 26
e) Consider f as in (b)
f) Corollary 26
√
g) Corollary 26 (g(x) = x for x ≥ 0 and g(x) = 0 for x < 0 is a continuous
function).