Solutions to Homework #4
1. (Lemma 7.6 from class) Let {xn } be a sequence in a metric space
(X, d), and let x be some element of X. Prove that the following conditions
are equivalent:
(i) some subsequence of {xn } converges to x
(ii) for every ε > 0 there are infinitely many n for which xn ∈ Nε (x).
When proving the implication (ii)⇒(i) make it clear how you use that xn ∈
Nε (x) for infinitely many n (and not just for some n).
Solution: “(i)⇒ (ii)” Suppose that some subsequence {xnk } converges
to x. Then for any ε > 0 there exists M ∈ N such that xnk ∈ Nε (x)
for all k ≥ M . In particular, xn ∈ Nε (x) for infinitely many n (namely
n = nM , nM +1 , . . .)
“(ii)⇒ (i)” Now assume that (ii) holds. Then we can construct a sequence
of integers n1 < n2 < . . . s.t. xnk ∈ N1/k (x) for each k. The sequence is
constructed inductively as follows – once n1 , . . . , nk−1 are chosen, using the
fact that xn ∈ N1/k (x) for infinitely many n, we can always choose one of
those n’s s.t. n > nk−1 ; this n will be our nk ).
Since xnk ∈ N1/k (x), we have d(xnk , x) < k1 . Hence d(xnk , x) → 0 as
k → ∞, and therefore {xnk } converges to x.
2. Let (X, d) be a metric space and Y a subset of X.
(a) Suppose that X is sequentially compact and Y is a closed subset of
X. Prove that Y is also sequentially compact.
(b) Now assume that Y is sequentially compact. Prove that Y is closed
in X (we are not assuming anything about X).
Note: Since sequential compactness is equivalent to compactness for metric spaces, the results of parts (a) and (b) of this problem are equivalent to
Theorems 6.1 and 6.2 from class (=Theorems 2.35 and 2.34 in Rudin), respectively. The point of this exercise is to prove the results working directly
with the definition of sequential compactness.
Solution: (a) Take any sequence {yn } in Y . Then {yn } is also a sequence
in X, so (since X is sequentially compact), it must have a subsequence {ynk }
which converges to some x ∈ X. Then x ∈ Y , and since Y is closed, we
have Y = Y by Theorem 5.1(1), so x ∈ Y . Thus, the subsequence {ynk }
converges in Y .
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We showed that any sequence in Y has a subsequence which converges in
Y , so by definition Y is sequentially compact.
(b) By Theorem 5.1(1), Y is closed in X ⇐⇒ Y = Y . Since Y ⊆ Y for
any Y , we just need to prove the reverse inclusion Y ⊆ Y .
So take any x ∈ Y . By Proposition 7.4, there is a sequence {yn } in Y
which converges to x.
Since Y is sequentially compact, the sequence {yn } must have a subsequence {ynk } which converges to some y ∈ Y . On the other hand, since
the entire sequence {yn } converges to x ∈ X, any of its subsequences must
converge to the same x. Since the limit of a sequence (if exists) is unique,
we must have x = y, so x ∈ Y . Thus we showed that Y ⊆ Y .
3. Let X = C[a, b], considered as a metric space with uniform metric
dunif (as defined in Problem 4 in HW#2). Prove that the set B1 (0), the
closed ball of radius 1 centered at 0 in X, is not sequentially compact (and
hence not compact).
Solution: By Problem 7 in HW#2 there exists a sequence of functions
{fn } in X such that dunif (fn , fm ) = 1 for all n 6= m. Moreover, an explicit
construction of such sequence given in online solutions shows that one can
choose fn ∈ B1 (0). We claim that the sequence {fn } has no convergent
subsequences.
Indeed, suppose that {fnk } is a convergent subsequence. Then {fnk } is
also Cauchy, so (by the definition of Cauchy with ε = 1), there exists M ∈ R
such that dunif (fnk , fnl ) < 1 for all k, l ≥ M . This is clearly a contradiction
since dunif (fnk , fnl ) = 1 whenever k 6= l.
4. Recall the concept of the completion of a metric space introduced in
Lecture 8. In the statement of the problem we will use the notion of isometric
metric spaces. Metric spaces (X, d) and (X 0 , d0 ) are called isometric (to each
other) if there is a bijection f : X → X 0 such that d(x, y) = d0 (f (x), f (y))
for all x, y ∈ X.
Let (X, d) be a metric space. Let Ω(X) be the set of all Cauchy sequences
{xn }n∈N with xn ∈ X for each n. Define the relation ∼ on Ω by setting
{xn } ∼ {yn } ⇐⇒ lim d(xn , yn ) = 0.
n→∞
(a) Prove that ∼ is an equivalence relation.
b = Ω/ ∼, the set of equivalence classes with respect to ∼. The
Now let X
equivalence class of a sequence {xn } will be denoted by [xn ]. For instance,
[ n1 ] = [ n12 ] since the sequences xn = n1 and yn = n12 are equivalent. Given
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b the equivalence class of the
an element x ∈ X, we will denote by [x] ∈ X
constant sequence all of whose elements are equal to x.
b ×X
b → R≥0 by setting
Now define the function D : X
D([xn ], [yn ]) = lim d(xn , yn )
n→∞
(∗ ∗ ∗)
(b) Prove that the limit on the right-hand side of (***) always exists
and that the function D is well-defined (that is, if [xn ] = [x0n ] and
[yn ] = [yn0 ], then limn→∞ d(xn , yn ) = limn→∞ d(x0n , yn0 )).
b D) is a metric space
(c) Prove that (X,
b given by ι(x) = [x] (that is, ι sends each
(d) Consider the map ι : X → X
x to the equivalence class of the corresponding constant sequence).
Prove that ι is injective and D(ι(x), ι(y)) = d(x, y) for all x, y ∈ X.
This implies that (X, d) is isometric to the metric space (ι(X), D) (so
b D)).
identifying X with ι(X), we can think of X as a subset of (X,
Solution: (a) Reflexivity: {xn } ∼ {xn } since limn→∞ d(xn , xn ) = 0.
Symmetry ({xn } ∼ {yn } ⇒ {yn } ∼ {xn }) follows from the fact that d is
symmetric (d(x, y) = d(y, x)).
Finally, transitivity follows from the triangle inequality: if {xn } ∼ {yn }
and {yn } ∼ {zn }, then limn→∞ d(xn , yn ) = 0 and limn→∞ d(yn , zn ) = 0.
Since 0 ≤ d(xn , zn ) ≤ d(xn , yn ) + d(yn , zn ), by the squeeze theorem we have
limn→∞ d(xn , yn ) = 0, so {xn } ∼ {zn }.
(b) First we show that for any Cauchy sequences {xn } and {yn }, the limit
limn→∞ d(xn , yn ) exists. Since R is complete, it suffices to show that the
sequence {d(xn , yn )} is Cauchy.
Fix ε > 0. Since {xn } and {yn } are Cauchy, there exists M1 , M2 ∈ N such
that d(xn , xm ) < 2ε for all n, m ≥ M1 and d(yn , ym ) < 2ε for all n, m ≥ M2 .
Let M = max{M1 , M2 }. Then for all n, m ≥ M we have
d(xn , yn ) ≤ d(xn , xm ) + d(xm , ym ) + d(ym , yn ) < d(xm , ym ) + ε,
and similarly, d(xm , ym ) < d(xn , yn ) + ε. Hence |d(xn , yn ) − d(xm , ym )| < ε
for all n, m ≥ M , and therefore {d(xn , yn )} is Cauchy.
Now we prove independence on the equivalence class.
{xn } ∼ {x0n } and {yn } ∼ {yn0 }. Then
Suppose that
d(x0n , yn0 ) ≤ d(x0n , xn ) + d(xn , yn ) + d(yn , yn0 ).
(∗ ∗ ∗)
Since limn→∞ d(x0n , xn ) = limn→∞ d(yn , yn0 ) = 0, taking limits on both
sides of (***), we conclude that limn→∞ d(x0n , yn0 ) ≤ limn→∞ d(xn , yn ). By
the same argument, the reverse inequality also holds: limn→∞ d(xn , yn ) ≤
limn→∞ d(x0n , yn0 ), so limn→∞ d(x0n , yn0 ) = limn→∞ d(xn , yn ).
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(c) The first two conditions in the definition of a metric space are clear,
so we only need to check triangle inequality. Let {xn }, {yn } and {zn } be
any Cauchy sequences. We know that d(xn , zn ) ≤ d(xn , yn ) + d(yn , zn ) for
each n, and each of the sequences {d(xn , zn )}, {d(xn , zn )} and {d(yn , zn )}
converges, so passing to the limit in the above inequality, we get
D([xn ], [zn ]) = lim d(xn , zn ) ≤ lim d(xn , yn ) + lim d(yn , zn )
n→∞
n→∞
n→∞
= D([xn ], [yn ]) + D([yn ], [zn ]).
(d) By definition D(ι(x), ι(y)) is the limit of the constant sequence d(x, y),
so D(ι(x), ι(y)) = d(x, y). Since d(x, y) 6= 0 for x 6= y, we also conclude that
ι(x) 6= ι(y) for x 6= y, so ι is injective.
5. A metric space (X, d) is called ultrametric if for any x, y, z ∈ X the
following inequality holds:
d(x, z) ≤ max{d(x, y), d(y, z)}.
(Note that this inequality is much stronger than the triangle inequality).
If X is any set and d is the discrete metric on X (that is, d(x, y) = 1 if
x 6= y and d(x, y) = 0 if x = y), then clearly (X, d) is ultrametric. A more
interesting example of an ultrametric space is given in the next problem.
Prove that properties (i) and (ii) below hold in any ultrametric space
(X, d) (note that both properties are counter-intuitive since they are very
far from being true in R).
(i) Take any x ∈ X, ε > 0 and take any y ∈ Nε (x). Then Nε (y) =
Nε (x). This means that if we take an open ball of fixed radius around
some point x, then for any other point y from that open ball, the
open ball of the same radius, but now centered at y, coincides with
the original ball. In other words, any point of an open ball happens
to be its center.
(ii) Prove that a sequence {xn } in X is Cauchy ⇐⇒ for any ε > 0
there exists M ∈ N such that d(xn+1 , xn ) < ε for all n ≥ M . Note:
The forward implication holds in any metric space.
Solution: (i) Let d = d(x, y). Since y ∈ Nε (x), we have d < ε. Take
any z ∈ Nε (y). Then d(y, z) < ε, so d(z, x) ≤ max{d(y, z), d(x, y)} =
max{d(y, z), d} < ε, so z ∈ Nε (x). Thus we showed that Nε (y) ⊆ Nε (x).
The reverse inclusion Nε (x) ⊆ Nε (y) is proved similarly.
(ii) The forward implication is clear. Assume now that for any ε > 0
there exists M ∈ N such that d(xn+1 , xn ) < ε for all n ≥ M . Then
for all n ≥ M we also have d(xn+2 , xn+1 ) < ε whence d(xn+2 , xn ) ≤
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max{d(xn+1 , xn ), d(xn+2 , xn+1 )} < ε. Repeating the same trick several
times, we deduce that d(xn+p , xn ) < ε for all n ≥ M and p ≥ 0; equivalently, d(xm , xn ) < ε for all n, m ≥ M . Therefore, the sequence {xn } is
Cauchy.
6. Let p be a fixed prime number. Define the function | · |p : Q → R≥0 as
follows: given a nonzero x ∈ Q, we can write x = pa dc for some a, c, d ∈ Z
where c and d are not divisible by p . Define |x|p = p−a (note that the
above representation is not unique, but it is easy to see that a is uniquely
determined by x). For instance,
 1
if p = 3


 9
9
4
if p = 2
=
20 5
if p = 5


p

1 for any other p.
Also define |0|p = 0. Now define the function dp : Q×Q → R≥0 by dp (x, y) =
|y − x|p .
(a) Prove that (Q, dp ) is an ultrametric space. (Note: the completion of
this metric space is usually denoted by Qp is called p-adic numbers).
(b) Describe explicitly the set N1 (0) (the open ball of radius 1 centered
at 0) in (Q, dp ).
(c) Let d be the standard metric on Q (that is, d(x, y) = |y − x| where
| · | is the usual absolute value). Give examples of sequences {xn }
and {yn } in Q such that
(i) xn → 0 in (Q, dp ) but {xn } is unbounded as a sequence in (Q, d)
(ii) yn → 0 in (Q, d) but {yn } is unbounded as a sequence in (Q, dp )
Solution: (a) Take any x, y, z ∈ Qp , and let u = y − x, v = z − y, so that
u + v = z − x. Thus, we need to prove that
|u + v|p ≤ max{|u|p , |v|p } for all u, v ∈ Qp .
The inequality is clear if u = 0, v = 0 or u + v = 0, so from now on we will
assume that u, v, u + v 6= 0.
Write u = pa · dc and v = pb · fe where p - c, d, e, f . By symmetry, without
loss of generality we can assume that a ≥ b. Then
u + v = pb · (pa−b
e
pa−b cf + de
c
+ ) = pb ·
.
d f
df
Since p - d and p - f , we have p - df . We do not know whether pa−b cf + de
is divisible by p, but in any case we can write pa−b cf + de = pg · h for some
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h
g, h ∈ Z with g ≥ 0 and p - h. Then u + v = pb+g df
, so
|u + v|p =
1
pb+g
≤
1
= |v|p ≤ max{|u|p , |v|p }.
pb
(b) N1 (0) consists of 0 and all rational numbers
and p | m.
(c) The sequences xn = pn and yn =
1
pn
m
n
such that gcd(m, n) = 1
have the required properties.