### Basic Topology

Written by Men-Gen Tsai email: b89902089@ntu.edu.tw 1. Prove that the empty set is a subset of every set.

Proof: For any element x of the empty set, x is also an element of every set since x does not exist. Hence, the empty set is a subset of every set.

2. A complex number z is said to be algebraic if there are integers a_{0}, ..., a_{n},
not all zero, such that

a_{0}z^{n}+ a_{1}z^{n−1}+ ... + a_{n−1}z + a_{n} = 0.

Prove that the set of all algebraic numbers is countable. Hint: For every positive integer N there are only finitely many equations with

n + |a_{0}| + |a_{1}| + ... + |a_{n}| = N.

Proof: For every positive integer N there are only finitely many equa- tions with

n + |a_{0}| + |a_{1}| + ... + |a_{n}| = N.

(since 1 ≤ n ≤ N and 0 ≤ |a_{0}| ≤ N ). We collect those equations as
C_{N}. Hence^{S}C_{N} is countable. For each algebraic number, we can form
an equation and this equation lies in C_{M} for some M and thus the set
of all algebraic numbers is countable.

3. Prove that there exist real numbers which are not algebraic.

Proof: If not, R^{1} = { all algebraic numbers } is countable, a contra-
diction.

4. Is the set of all irrational real numbers countable?

Solution: If R − Q is countable, then R^{1} = (R − Q)^{S}Q is countable,
a contradiction. Thus R − Q is uncountable.

5. Construct a bounded set of real numbers with exactly three limit points.

Solution: Put

A = {1/n : n ∈ N }^{[}{1 + 1/n : n ∈ N }^{[}{2 + 1/n : n ∈ N }.

A is bounded by 3, and A contains three limit points - 0, 1, 2.

6. Let E^{0} be the set of all limit points of a set E. Prove that S^{0} is
closed. Prove that E and E have the same limit points. (Recall that
E = E^{S}E^{0}.) Do E and E^{0} always have the same limit points?

Proof: For any point p of X − E^{0}, that is, p is not a limit point E,
there exists a neighborhood of p such that q is not in E with q 6= p for
every q in that neighborhood.

Hence, p is an interior point of X − E^{0}, that is, X − E^{0} is open, that
is, E^{0} is closed.

Next, if p is a limit point of E, then p is also a limit point of E since
E = E^{S}E^{0}. If p is a limit point of E, then every neighborhood Nr(p)
of p contains a point q 6= p such that q ∈ E. If q ∈ E, we completed
the proof. So we suppose that q ∈ E − E = E^{0}− E. Then q is a limit
point of E. Hence,

N_{r}^{0}(q)

where r^{0} = ^{1}_{2} min(r−d(p, q), d(p, q)) is a neighborhood of q and contains
a point x 6= q such that x ∈ E. Note that N_{r}^{0}(q) contains in N_{r}(p)−{p}.

That is, x 6= p and x is in Nr(p). Hence, q also a limit point of E. Hence, E and E have the same limit points.

Last, the answer of the final sub-problem is no. Put E = {1/n : n ∈ N },

and E^{0} = {0} and (E^{0})^{0} = φ.

7. Let A_{1}, A_{2}, A_{3}, ... be subsets of a metric space. (a) If B_{n} =^{S}^{n}_{i=1}A_{i},
prove that Bn =^{S}^{n}_{i=1}Ai, for n = 1, 2, 3, ... (b) If B =^{S}^{∞}_{i=1}, prove that
B ⊃^{S}^{∞}_{i=1}Ai. Show, by an example, that this inclusion can be proper.

Proof of (a): (Method 1) B_{n} is the smallest closed subset of X that
contains B_{n}. Note that ^{S}A_{i} is a closed subset of X that contains B_{n},
thus

Bn⊃

n

[

i=1

Ai.

If p ∈ B_{n}− B_{n}, then every neighborhood of p contained a point q 6= p
such that q ∈ B_{n}. If p is not in A_{i} for all i, then there exists some
neighborhood N_{r}_{i}(p) of p such that (N_{r}_{i}(p) − p)^{T}A_{i} = φ for all i. Take
r = min{r_{1}, r_{2}, ..., r_{n}}, and we have N_{r}(p)^{T}B_{n} = φ, a contradiction.

Thus p ∈ A_{i} for some i. Hence
B_{n}⊂

n

[

i=1

A_{i}.
that is,

B_{n}=

n

[

i=1

A_{i}.

(Method 2) Since ^{S}^{n}_{i=1}A_{i} is closed and B_{n} = ^{S}^{n}_{i=1}A_{i} ⊂ ^{S}^{n}_{i=1}A_{i},
Bn⊂^{S}^{n}_{i=1}Ai.

Proof of (b): Since B is closed and B ⊃ B ⊃ A_{i}, B ⊃ A_{i} for all i.

Hence B ⊃ ^{S}A_{i}.

Note: My example is A_{i} = (1/i, ∞) for all i. Thus, A_{i} = [1/i, ∞), and
B = (0, ∞), B = [0, ∞). Note that 0 is not in A_{i} for all i. Thus this
inclusion can be proper.

8. Is every point of every open set E ⊂ R^{2} a limit point of E? Answer
the same question for closed sets in R^{2}.

Solution: For the first part of this problem, the answer is yes.

(Reason): For every point p of E, p is an interior point of E. That
is, there is a neighborhood N_{r}(p) of p such that N_{r}(p) is a subset of
E. Then for every real r^{0}, we can choose a point q such that d(p, q) =
1/2 min(r, r^{0}). Note that q 6= p, q ∈ N_{r}^{0}(p), and q ∈ N_{r}(p). Hence, every
neighborhood Nr^{0}(p) contains a point q 6= p such that q ∈ Nr(p) ⊂ E,
that is, p is a limit points of E.

For the last part of this problem, the answer is no. Consider A =
{(0, 0)}. A^{0} = φ and thus (0, 0) is not a limit point of E.

9. Let E^{o} denote the set of all interior points of a set E.

(a) Prove that E^{o} is always open.

(b) Prove that E is open if and only if E^{o} = E.

(c) If G is contained in E and G is open, prove that G is contained in
E^{o}.

(d) Prove that the complement of E^{o} is the closure of the complement
of E.

(e) Do E and E always have the same interiors?

(f) Do E and E^{o} always have the same closures?

Proof of (a): If E is non-empty, take p ∈ E^{o}. We need to show that
p ∈ (E^{o})^{o}. Since p ∈ E^{o}, there is a neighborhood N_{r} of p such that
N_{r} is contained in E. For each q ∈ N_{r}, note that N_{s}(q) is contained in
N_{r}(p), where s = min{d(p, q), r − d(p, q)}. Hence q is also an interior
point of E, that is, N_{r} is contained in E^{o}. Hence E^{o} is always open.

Proof of (b): (⇒) It is clear that E^{o} is contained in E. Since E is
open, every point of E is an interior point of E, that is, E is contained
in E^{o}. Therefore E^{o} = E.

(⇐) Since every point of E is an interior point of E (E^{o}(E) = E), E
is open.

Proof of (c): If p ∈ G, p is an interior point of G since G is open. Note
that E contains G, and thus p is also an interior point of E. Hence
p ∈ E^{o}. Therefore G is contained in E^{o}. (Thus E^{o} is the biggest open
set contained in E. Similarly, E is the smallest closed set containing
E.)

Proof of (d): Suppose p ∈ X − E^{o}. If p ∈ X − E, then p ∈ X − E
clearly. If p ∈ E, then N is not contained in E for any neighborhood
N of p. Thus N contains an point q ∈ X − E. Note that q 6= p, that
is, p is a limit point of X − E. Hence X − E^{o} is contained in X − E.

Next, suppose p ∈ X − E. If p ∈ X − E, then p ∈ X − E^{o} clearly. If
p ∈ E, then every neighborhood of p contains a point q 6= p such that
q ∈ X − E. Hence p is not an interior point of E. Hence X − E is
contained in X − E^{o}. Therefore X − E^{o} = X − E.

Solution of (e): No. Take X = R^{1} and E = Q. Thus E^{o} = φ and
E^{o} = (R^{1})^{o} = R^{1} 6= φ.

Solution of (f ): No. Take X = R^{1} and E = Q. Thus E = R^{1}, and
E^{o} = φ = φ.

10. Let X be an infinite set. For p ∈ X and q ∈ X, define d(p, q) =

1 (if p 6= q) 0 (if p = q)

Prove that this is a metric. Which subsets of the resulting metric space are open? Which are closed? Which are compact?

Proof: (a) d(p, q) = 1 > 0 if p 6= q; d(p, p) = 0. (b) d(p, q) = d(q, p) since p = q implies q = p and p 6= q implies q 6= p. (c) d(p, q) ≤ d(p, r) + d(r, q) for any r ∈ X if p = q. If p 6= q, then either r = p or

r = q, that is, r 6= p or r 6= q. Thus, d(p, q) = 1 ≤ d(p, r) + d(r, q). By (a)-(c) we know that d is a metric.

Every subset of X is open and closed. We claim that for any p ∈ X,
p is not a limit point. Since d(p, q) = 1 > 1/2 if q 6= p, there exists an
neighborhood N_{1/2}(p) of p contains no points of q 6= p such that q ∈ X.

Hence every subset of X contains no limit points and thus it is closed.

Since X − S is closed for every subset S of X, S = X − (X − S) is open. Hence every subset of X is open.

Every finite subset of X is compact. Let S = {p_{1}, ..., p_{n}} be finite.

Consider an open cover {G_{α}} of S. Since S is covered by G_{α}, p_{i} is
covered by G_{α}_{i}, thus {G_{α}_{1}, ..., G_{α}_{n}} is finite subcover of S. Hence S is
compact. Next, suppose S is infinite. Consider an open cover {G_{p}} of
S, where

G_{p} = N^{1}

2(p)

for every p ∈ S. Note that q is not in G_{p} if q 6= p. If S is compact,
then S can be covered by finite subcover, say

G_{p}_{1}, ..., G_{p}_{n}.

Then there exists q such that q 6= p_{i} for all i since S is infinite, a con-
tradiction. Hence only every finite subset of X is compact.

11. For x ∈ R^{1} and y ∈ R^{1}, define

d_{1}(x, y) = (x, y)^{2},
d_{2}(x, y) = ^{q}|x − y|,
d_{3}(x, y) = |x^{2}− y^{2}|,
d4(x, y) = |x − 2y|,
d5(x, y) = |x − y|

1 + |x − y|.

Determine, for each of these, whether it is a metric or not.

Solution: (1) d_{1}(x, y) is not a metric. Since d_{1}(0, 2) = 4, d_{1}(0, 1) = 1,
and d_{1}(1, 2) = 1, d_{1}(0, 2) > d_{1}(0, 1) + d_{1}(1, 2). Thus d_{1}(x, y) is not a
metric.

(2) d_{2}(x, y) is a metric. (a) d(p, q) > 0 if p 6= q; d(p, p) = 0. (b)
d(p, q) = ^{q}|p − q| =^{q}|q − p| = d(q, p). (c) |p − q| ≤ |p − r| + |r − q|,

q|p − q| ≤^{q}|p − r| + |r − q| ≤^{q}|p − r| +^{q}|r − q|. That is, d(p, q) ≤
d(p, r) + d(r, q). (3) d_{3}(x, y) is not a metric since d_{3}(1, −1) = 0.

(4) d4(x, y) is not a metric since d4(1, 1) = 1 6= 0.

(5) d5(x, y) is a metric since |x − y| is a metric.

Claim: d(x, y) is a metric, then

d^{0}(x, y) = d(x, y)
1 + d(x, y)
is also a metric.

Proof of Claim: (a) d^{0}(p, q) > 0 if p 6= q; d(p, p) = 0. (b) d^{0}(p, q) =
d^{0}(q, p). (c) Let x = d(p, q), y = d(p, r), and z = d(r, q). Then x ≤ y+z.

d^{0}(p, q) ≤ d^{0}(p, r) + d^{0}(r, q)

⇔ x

1 + x ≤ y

1 + y + z 1 + z

⇔ x(1 + y)(1 + z) ≤ y(1 + z)(1 + x) + z(1 + x)(1 + y)

⇔ x + xy + xz + xyz ≤ (y + xy + yz + xyz) + (z + xz + yz + xyz)

⇔ x ≤ y + z + 2yz + xyz

⇐ x ≤ y + z

Thus, d^{0} is also a metric.

12. Let K ⊂ R^{1} consist of 0 and the numbers 1/n, for n = 1, 2, 3, ....

Prove that K is compact directly from the definition (without using the Heine-Borel theorem).

Proof: Suppose that {O_{α}} is an arbitrary open covering of K. Let
E ∈ {O_{α}} consists 0. Since E is open and 0 ∈ E, 0 is an interior point
of E. Thus there is a neighborhood N = N_{r}(0) of 0 such that N ⊂ E.

Thus N contains

1

[1/r] + 1, 1

[1/r] + 2, ...

Next, we take finitely many open sets En ∈ {Oα} such that 1/n ∈ En

for n = 1, 2, ..., [1/r]. Hence {E, E_{1}, ..., E_{[1/r]} is a finite subcover of K.

Therefore, K is compact.

Note: The unique limit point of K is 0. Suppose p 6= 0 is a limit point
of K. Clearly, 0 < p < 1. (p cannot be 1). Thus there exists n ∈ Z^{+}
such that

1

n + 1 < p < 1 n.

Hence N_{r}(p) where r = min{_{n}^{1} − p, p − _{n+1}^{1} } contains no points of K,
a contradiction.

13. Construct a compact set of real numbers whose limit points form a countable set.

Solution: Let K be consist of 0 and the numbers 1/n for n = 1, 2, 3, ...

Let xK = {xk : k ∈ K} and x + K = {x + k : k ∈ K} for x ∈ R^{1}. I
take

S_{n} = (1 − 1

2^{n}) + K
2^{n+1},

S =

∞

[

n=1

S_{n}^{[}{1}.

Claim: S is compact and the set of all limit points of S is K^{S}{1}.

Clearly, S lies in [0, 1], that is, S is bounded in R^{1}. Note that S_{n} ⊂
[1 − _{2}^{1}n, 1 − _{2}n+1^{1} ]. By Exercise 12 and its note, I have that all limit
points of S^{T}[0, 1) is

0,1 2, ..., 1

2^{n}, ...

Clearly, 1 is also a limit point of S. Therefore, the set of all limit points
of S is K^{S}{1}. Note that K^{S}{1} ⊂ S, that is, K is compact. I com-
pleted the proof of my claim.

14. Give an example of an open cover of the segment (0, 1) which has no finite subcover.

Solution: Take {O_{n}} = {(1/n, 1)} for n = 1, 2, 3, .... The following is
my proof. For every x ∈ (0, 1),

x ∈ ( 1

[1/x] + 1, 0) ∈ {O_{n}}

Hence {O_{n}} is an open covering of (0, 1). Suppose there exists a finite
subcovering

( 1

n_{1}, 1), ..., ( 1
n_{k}, 1)

where n_{1} < n_{2} < ... < n_{k}, respectively. Clearly _{2n}^{1}

p ∈ (0, 1) is not in any elements of that subcover, a contradiction.

Note: By the above we know that (0, 1) is not compact.

15. Show that Theorem 2.36 and its Corollary become false (in R^{1}, for ex-
ample) if the word ”compact” is replaced by ”closed” or by ”bounded.”

Theorem 2.36: If {K_{α}} is a collection of compact subsets of a metric
space X such that the intersection of every finite subcollection of {K_{α}}
is nonempty, then ^{T}K_{α} is nonempty.

Corollary: If {K_{n}} is a sequence of nonempty compact sets such that
K_{n} contains K_{n+1} (n = 1, 2, 3, ...), then ^{T}K_{n} is not empty.

Solution: For closed: [n, ∞). For bounded: (−1/n, 1/n) − {0}.

16. Regard Q, the set of all rational numbers, as a metric space, with
d(p, q) = |p − q|. Let E be the set of all p ∈ Q such that 2 < p^{2} < 3.

Show that E is closed and bounded in Q, but that E is not compact.

Is E open in Q?

Proof: Let S = (√ 2,√

3)^{S}(−√
3, −√

2). Then E = {p ∈ Q : p ∈ S}.

Clearly, E is bounded in Q. Since Q is dense in R, every limit point of Q is in Q. (I regard Q as a metric space). Hence, E is closed in Q.

To prove that E is not compact, we form a open covering of E as follows:

{G_{α}} = {N_{r}(p) : p ∈ E and (p − r, p + r) ⊂ S}

Surely, {G_{α}} is a open covering of E. If E is compact, then there are
finitely many indices α_{1}, ..., α_{n} such that

E ⊂ G_{α}_{1}^{[}...^{[}G_{α}_{n}.

For every Gαi = Nri(pi), take p = max1≤i≤npi. Thus, p is the nearest point to √

3. But N_{r}(p) lies in E, thus [p + r,√

3) cannot be covered since Q is dense in R, a contradiction. Hence E is not compact.

Finally, the answer is yes. Take any p ∈ Q, then there exists a neighborhood N (p) of p contained in E. (Take r small enough where Nr(p) = N (p), and Q is dense in R.) Thus every point in N (p) is also in Q. Hence E is also open.

17. Let E be the set of all x ∈ [0, 1] whose decimal expansion contains only the digits 4 and 7. Is E countable? Is E dense in [0, 1]? Is E compact?

Is E perfect?

Solution:

E =^{n}

∞

X

n=1

a_{n}

10^{n} : a_{n} = 4 or a_{n} = 7^{o}

Claim: E is uncountable.

Proof of Claim: If not, we list E as follows:

x_{1} = 0.a_{11}a_{12}...a_{1n}...

x2 = 0.a21a22...a2n...

... ...

x_{k} = 0.a_{k1}a_{k2}...a_{kn}...

... ...

(Prevent ending with all digits 9) Let x = 0.x_{1}x_{2}...x_{n}... where

x_{n}=

4 if a_{nn} = 7
7 if a_{nn} = 4

By my construction, x /∈ E, a contradiction. Thus E is uncountable.

Claim: E is not dense in [0, 1].

Proof of Claim: Note that E^{T}(0.47, 0.74) = φ. Hence E is not dense
in [0, 1].

Claim: E is compact.

Proof of Claim: Clearly, E is bounded. For every limit point p of E, I show that p ∈ E. If not, write the decimal expansion of p as follows

p = 0.p_{1}p_{2}...p_{n}...

Since p /∈ E, there exists the smallest k such that p_{k} 6= 4 and p_{k} 6= 7.

When p_{k} = 0, 1, 2, 3, select the smallest l such that p_{l} = 7 if possible.

(If l does not exist, then p < 0.4. Thus there is a neighborhood of p such that contains no points of E, a contradiction.) Thus

0.p_{1}...p_{l−1}4p_{l+1}...p_{k−1}7 < p < 0.p_{1}...p_{k−1}4.

Thus there is a neighborhood of p such that contains no points of E, a contradiction.

When p_{k} = 5, 6,

0.p_{1}...p_{k−1}47 < p < 0.p_{1}...p_{k−1}74.

Thus there is a neighborhood of p such that contains no points of E, a contradiction.

When p_{k} = 8, 9, it is similar. Hence E is closed. Therefore E is
compact.

Claim: E is perfect.

Proof of Claim: Take any p ∈ E, and I claim that p is a limit point
of E. Write p = 0.p_{1}p_{2}...p_{n}... Let

x_{k} = 0.y_{1}y_{2}...y_{n}...

where

y_{n} =

p_{k} if k 6= n
4 if p_{n} = 7
7 if p_{n} = 4

Thus, |x_{k}− p| → 0 as k → ∞. Also, x_{k} 6= p for all k. Hence p is a limit
point of E. Therefore E is perfect.

18. Is there a nonempty perfect set in R^{1} which contains no rational num-
ber?

Solution: Yes. The following claim will show the reason.

Claim: Given a measure zero set S, we have a perfect set P contains no elements in S.

Proof of Claim: (due to SYLee). Since S has measure zero, there
exists a collection of open intervals {I_{n}} such that

S ⊂^{[}I_{n} and ^{X}|I_{n}| < 1.

Consider E = R^{1} −^{S}I_{n}. E is nonempty since E has positive mea-
sure. Thus E is uncountable and E is closed. Therefore there exists
a nonempty perfect set P contained in E by Exercise 28. P ^{T}S = φ.

Thus P is our required perfect set.

19. (a) If A and B are disjoint closed sets in some metric space X, prove that they are separated.

(b) Prove the same for disjoint open sets.

(c) Fix p ∈ X, δ > 0, define A to be the set of all q ∈ X for which d(p, q) < δ, define B similarly, with > in place of <. Prove that A and B are separated.

(d) Prove that every connected metric space with at least two points is uncountable. Hint: Use (c).

Proof of (a): Recall the definition of separated: A and B are sep-
arated if A^{T}B and A^{T}B are empty. Since A and B are closed sets,
A = A and B = B. Hence A^{T}B = A^{T}B = A^{T}B = φ. Hence A and
B are separated.

Proof of (b): Suppose A^{T}B is not empty. Thus there exists p such
that p ∈ A and p ∈ B. For p ∈ A, there exists a neighborhood N_{r}(p) of
p contained in A since A is open. For p ∈ B = B^{S}B^{0}, if p ∈ B, then
p ∈ A^{T}B. Note that A and B are disjoint, and it’s a contradiction.

If p ∈ B^{0}, then p is a limit point of B. Thus every neighborhood of p
contains a point q 6= p such that q ∈ B. Take an neighborhood N_{r}(p)of
p containing a point q 6= p such that q ∈ B. Note that N_{r}(p) ⊂ A,
thus q ∈ A. With A and B are disjoint, we get a contradiction. Hence
A^{T}(B) is empty.

Similarly, A^{T}B is also empty. Thus A and B are separated.

Proof of (c): Suppose A^{T}B is not empty. Thus there exists x such

that x ∈ A and x ∈ B. Since x ∈ A, d(p, x) < δ. x ∈ B = B^{S}B^{0}, thus
if x ∈ B, then d(p, x) > δ, a contradiction. The only possible is x is a
limit point of B. Hence we take a neighborhood N_{r}(x) of x contains y
with y ∈ B where r = ^{δ−d(x,p)}_{2} . Clearly, d(y, p) > δ. But,

d(y, p) ≤ d(y, x) + d(x, p)

< r + d(x, p)

= δ − d(x, p)

2 + d(x, p)

= δ + d(x, p) 2

< δ + δ 2 = δ.

A contradiction. Hence A^{T}B is empty. Similarly, A^{T}B is also empty.

Thus A and B are separated.

Note: Take care of δ > 0. Think a while and you can prove the next sub-exercise.

Proof of (d): Let X be a connected metric space. Take p ∈ X, q ∈ X with p 6= q, thus d(p, q) > 0 is fixed. Let

A = {x ∈ X : d(x, p) < δ}; B = {x ∈ X : d(x, p) > δ}.

Take δ = δ_{t} = td(p, q) where t ∈ (0, 1). Thus 0 < δ < d(p, q). p ∈ A
since d(p, p) = 0 < δ, and q ∈ B since d(p, q) > δ. Thus A and B are
non-empty.

By (c), A and B are separated. If X = A^{S}B, then X is not connected,
a contradiction. Thus there exists y_{t}∈ X such that y /∈ A^{S}B. Let

E = E_{t}= {x ∈ X : d(x, p) = δ_{t}} 3 y_{t}.

For any real t ∈ (0, 1), E_{t} is non-empty. Next, E_{t} and E_{s} are disjoint
if t 6= s (since a metric is well-defined). Thus X contains a uncount-

able set {y_{t} : t ∈ (0, 1)} since (0, 1) is uncountable. Therefore, X is
uncountable.

Note: It is a good exercise. If that metric space contains only one point, then it must be separated.

Similar Exercise Given by SYLee: (a) Let A = {x : d(p, x) < r}

and B = {x : d(p, x) > r} for some p in a metric space X. Show that A, B are separated.

(b) Show that a connected metric space with at least two points must be uncountable. [Hint: Use (a)]

Proof of (a): By definition of separated sets, we want to show A^{T}B =
φ, and B^{T}A = φ. In order to do these, it is sufficient to show A^{T}B =
φ. Let x ∈ A^{T}B = φ, then we have:

(1) x ∈ A ⇒ d(x, p) ≤ r(2) x ∈ B ⇒ d(x, p) > r
It is impossible. So, A^{T}B = φ.

Proof of (b): Suppose that C is countable, say C = a, b, x3, .... We
want to show C is disconnected. So, if C is a connected metric space
with at least two points, it must be uncountable. Consider the set
S = {d(a, x_{i}) : x_{i} ∈ C}, and thus let r ∈ R − S and inf S < r < sup S.

And construct A and B as in (a), we have C = A^{S}B, where A and B
are separated. That is C is disconnected.

Another Proof of (b): Let a ∈ C, b ∈ C, consider the continuous function f from C into R defined by f (x) = d(x, a). So, f (C) is con- nected and f (a) = 0, f (b) > 0. That is, f (C) is an interval. Therefore, C is uncountable.

20. Are closures and interiors of connected sets always connected? (Look
at subsets of R^{2}.)

Solution: Closures of connected sets is always connected, but interiors of those is not. The counterexample is

S = N_{1}(2, 0)^{[}N_{1}(−2, 0)^{[}{x − axis} ⊂ R^{2}.

Since S is path-connected, S is connect. But S^{o} = N1(2)^{S}N1(−2) is
disconnected clearly.

Claim: If S is a connected subset of a metric space, then S is con- nected.

Pf of Claim: If not, then S is a union of two nonempty separated set
A and B. Thus A^{T}B = A^{T}B = φ. Note that

S = S − T

= A^{[}B − T

= (A^{[}B)^{\}T^{c}

= (A^{\}T^{c})^{[}(B^{\}T^{c})

where T = S − S. Thus

(A^{\}T^{c})^{\}B^{\}T^{c} ⊂ (A^{\}T^{c})^{\}B^{\}T^{c}

⊂ A^{\}B

= φ.

Hence (A^{T}T^{c})^{T}B^{T}T^{c} = φ. Similarly, A^{T}T^{c}^{T}(B^{T}T^{c}) = φ.

Now we claim that both A^{T}T^{c} and B^{T}T^{c} are nonempty. Suppose
that B^{T}T^{c} = φ. Thus

A^{\}T^{c} = S ⇔ A^{\}(S − S)^{c} = S

⇔ A^{\}(A^{[}B − S)^{c} = S

⇔ A^{\}((A^{[}B)^{\}S^{c})^{c} = S

⇔ A^{\}((A^{c}^{\}B^{c})^{[}S) = S

⇔ (A^{\}S)^{[}(A^{\}A^{c}^{\}B^{c}) = S

⇔ A^{\}S = S.

Thus B is empty, a contradiction. Thus B^{T}T^{c} is nonempty. Similarly,
A^{T}T^{c} nonempty. Therefore S is a union of two nonempty separated
sets, a contradiction. Hence S is connected.

21. Let A and B be separated subsets of some R^{k}, suppose a ∈ A, b ∈ B,
and define

p(t) = (1 − t)a + tb

for t ∈ R^{1}. Put A_{0} = p^{−1}(A), B_{0} = p^{−1}(B). [Thus t ∈ A_{0} if and only
if p(t) ∈ A.]

(a) Prove that A_{0} and B_{0} are separated subsets of R^{1}.

(b) Prove that there exists t_{0} ∈ (0, 1) such that p(t_{0}) /∈ A^{S}B.

(c) Prove that every convex subset of R^{k} is connected.

Proof of (a): I claim that A_{0}^{T}B_{0} is empty. (B_{0}^{T}A_{0} is similar). If
not, take x ∈ A_{0}^{T}B_{0}. x ∈ A_{0} and x ∈ B_{0}. x ∈ B_{0} or x is a limit
point of B_{0}. x ∈ B_{0} will make x ∈ A_{0}^{T}B_{0}, that is, p(x) ∈ A^{T}B, a
contradiction since A and B are separated.

Claim: x is a limit point of B_{0} ⇒ p(x) is a limit point of B. Take any
neighborhood N_{r} of p(x), and p(t) lies in B for small enough t. More
precisely,

x − r

|b − a| < t < x + r

|b − a|.

Since x is a limit point of B_{0}, and (x − r/|b − a|, x + r/|b − a|) is a
neighborhood N of x, thus N contains a point y 6= x such that y ∈ B_{0},
that is, p(y) ∈ B. Also, p(y) ∈ N_{r}. Therefore, p(x) is a limit point of
B. Hence p(x) ∈ A^{T}B, a contradiction since A and B are separated.

Hence A_{0}^{T}B_{0} is empty, that is, A_{0} and B_{0} are separated subsets of
R^{1}.

Proof of (b): Suppose not. For every t0 ∈ (0, 1), neither p(t0) ∈ A
nor p(t_{0}) ∈ B (since A and B are separated). Also, p(t_{0}) ∈ A^{S}B for
all t_{0} ∈ (0, 1). Hence (0, 1) = A_{0}^{S}B_{0}, a contradiction since (0, 1) is
connected. I completed the proof.

Proof of (c): Let S be a convex subset of R^{k}. If S is not connected,
then S is a union of two nonempty separated sets A and B. By (b),
there exists t_{0} ∈ (0, 1) such that p(t_{0}) /∈ A^{S}B. But S is convex, p(t_{0})
must lie in A^{S}B, a contradiction. Hence S is connected.

22. A metric space is called separable if it contains a countable dense sub-
set. Show that R^{k} is separable. Hint: Consider the set of points which
have only rational coordinates.

Proof: Consider S = the set of points which have only rational coor-
dinates. For any point x = (x_{1}, x_{2}, ..., x_{k}) ∈ R^{k}, we can find a rational
sequence {r_{i}_{j}} → x_{j} for j = 1, ..., k since Q is dense in R^{1}. Thus,

r_{i} = (r_{i}_{1}, r_{i}_{2}, ..., r_{i}_{k}) → x

and r_{i} ∈ S for all i. Hence S is dense in R^{k}. Also, S is countable, that
is, S is a countable dense subset in R^{k}, R^{k} is separable.

23. A collection {V_{α}} of open subsets of X is said to be a base for X if the
following is true: For every x ∈ X and every open set G ⊂ X such that
x ∈ G, we have x ∈ V_{α} ⊂ G for some α. In other words, every open set
in X is the union of a subcollection of {V_{α}}.

Prove that every separable metric space has a countable base. Hint:

Take all neighborhoods with rational radius and center in some count- able dense subset of X.

Proof: Let X be a separable metric space, and S be a countable dense
subset of X. Let a collection {V_{α}} = { all neighborhoods with rational
radius and center in S }. We claim that {V_{α}} is a base for X.

For every x ∈ X and every open set G ⊂ X such that x ∈ G, there
exists a neighborhood N_{r}(p) of p such that N_{r}(p) ⊂ G since x is an
interior point of G. Since S is dense in X, there exists {s_{n}} → x. Take
a rational number r_{n} such that r_{n} < ^{r}_{2}, and {V_{α}} 3 N_{r}_{n}(s_{n}) ⊂ N_{r}(p)
for enough large n. Hence we have x ∈ V_{α} ⊂ G for some α. Hence
{V_{α}} is a base for X.

24. Let X be a metric space in which every infinite subset has a limit
point. Prove that X is separable. Hint: Fix δ > 0, and pick x_{1} ∈
X. Having chosen x_{1}, ..., x_{j} ∈ X, choose x_{j+1}, if possible, so that
d(x_{i}, x_{j+1}) ≥ δ for i = 1, ..., j. Show that this process must stop after
finite number of steps, and that X can therefore be covered by finite
many neighborhoods of radius δ. Take δ = 1/n(n = 1, 2, 3, ...), and
consider the centers of the corresponding neighborhoods.

Proof: Fix δ > 0, and pick x_{1} ∈ X. Having chosen x_{1}, ..., x_{j} ∈ X,
choose x_{j+1}, if possible, so that d(x_{i}, x_{j+1}) ≥ δ for i = 1, ..., j. If
this process cannot stop, then consider the set A = {x_{1}, x_{2}, ..., x_{k}}. If
p is a limit point of A, then a neighborhood N_{δ/3}(p) of p contains a
point q 6= p such that q ∈ A. q = x_{k} for only one k ∈ N . If not,
d(x_{i}, x_{j}) ≤ d(x_{i}, p) + d(x_{j}, p) ≤ δ/3 + δ/3 < δ, and it contradicts the
fact that d(x_{i}, x_{j}) ≥ δ for i 6= j. Hence, this process must stop after
finite number of steps.

Suppose this process stop after k steps, and X is covered by N_{δ}(x_{1}),
N_{δ}(x_{2}), ..., N_{δ}(x_{k}), that is, X can therefore be covered by finite many
neighborhoods of radius δ.

Take δ = 1/n(n = 1, 2, 3, ...), and consider the set A of the centers of

the corresponding neighborhoods.

Fix p ∈ X. Suppose that p is not in A, and every neighborhood
N_{r}(p). Note that N_{r/2}(p) can be covered by finite many neighborhoods
N_{s}(x_{1}), ..., N_{s}(x_{k}) of radius s = 1/n where n = [2/r] + 1 and x_{i} ∈ A
for i = 1, ..., k. Hence, d(x_{1}, p) ≤ d(x_{1}, q) + d(q, p) ≤ r/2 + s < r where
q ∈ N_{r/2}(p)^{T}N_{s}(x_{1}). Therefore, x_{1} ∈ N_{r}(p) and x_{1} 6= p since p is not
in A. Hence, p is a limit point of A if p is not in A, that is, A is a
countable dense subset, that is, X is separable.

25. Prove that every compact metric space K has a countable base, and that K is therefore separable. Hint: For every positive integer n, there are finitely many neighborhood of radius 1/n whose union covers K.

Proof: For every positive integer n, there are finitely many neighbor- hood of radius 1/n whose union covers K (since K is compact). Collect all of them, say {Vα}, and it forms a countable collection. We claim {Vα} is a base.

For every x ∈ X and every open set G ⊂ X, there exists N_{r}(x) such that
N_{r}(x) ⊂ G since x is an interior point of G. Hence x ∈ N_{m}(p) ∈ {V_{α}}
for some p where m = [2/r] + 1. For every y ∈ N_{m}(p), we have

d(y, x) ≤ d(y, p) + d(p, x) < m + m = 2m < r.

Hence N_{m}(p) ⊂ G, that is, V_{α} ⊂ G for some α, and therefore {V_{α}} is a
countable base of K. Next, collect all of the center of V_{α}, say D, and
we claim D is dense in K (D is countable since V_{α} is countable). For all
p ∈ K and any > 0 we can find N_{n}(x_{n}) ∈ {V_{α}} where n = [1/] + 1.

Note that x_{n} ∈ D for all n and d(p, x_{n}) → 0 as n → ∞. Hence D is
dense in K.

26. Let X be a metric space in which every infinite subsets has a limit
point. Prove that X is compact. Hint: By Exercises 23 and 24, X has
a countable base. It follows that every open cover of X has a countable
subcover {G_{n}}, n = 1, 2, 3, .... If no finite subcollection of {G_{n}} covers
X, then the complement F_{n} of G_{1}^{S}...^{S}G_{n} is nonempty for each n,
but ^{T}F_{n} is empty. If E is a set contains a point from each F_{n}, consider
a limit point of E, and obtain a contradiction.

Proof: By Exercises 23 and 24, X has a countable base. It follows that
every open cover of X has a countable subcover {G_{n}}, n = 1, 2, 3, ....

If no finite subcollection of {G_{n}} covers X, then the complement F_{n} of
G1S

...^{S}Gn is nonempty for each n, but ^{T}Fn is empty. If E is a set
contains a point from each Fn, consider a limit point of E.

Note that F_{k} ⊃ F_{k+1} ⊃ ... and F_{n} is closed for all n, thus p lies in F_{k}
for all k. Hence p lies in ^{T}F_{n}, but ^{T}F_{n} is empty, a contradiction.

27. Define a point p in a metric space X to be a condensation point of a set E ⊂ X if every neighborhood of p contains uncountably many points of E.

Suppose E ⊂ R^{k}, E is uncountable, and let P be the set of all conden-
sation points of E. Prove that P is perfect and that at most countably
many points of E are not in P . In other words, show that P^{c}^{T}E is at
most countable. Hint: Let {V_{n}} be a countable base of R^{k}, let W be
the union of those V_{n} for which E^{T}V_{n} is at most countable, and show
that P = W^{c}.

Proof: Let {Vn} be a countable base of R^{k}, let W be the union of
those V_{n} for which E^{T}V_{n} is at most countable, and we will show that
P = W^{c}. Suppose x ∈ P . (x is a condensation point of E). If
x ∈ V_{n} for some n, then E^{T}V_{n} is uncountable since V_{n} is open. Thus
x ∈ W^{c}. (If x ∈ W , then there exists V_{n} such that x ∈ V_{n} and E^{T}V_{n}

is uncountable, a contradiction). Therefore P ⊂ W^{c}.

Conversely, suppose x ∈ W^{c}. x /∈ Vn for any n such that E^{T}Vn is
countable. Take any neighborhood N (x) of x. Take x ∈ V_{n} ⊂ N (x),
and E^{T}V_{n} is uncountable. Thus E^{T}N (x) is also uncountable, x is a
condensation point of E. Thus W^{c} ⊂ P . Therefore P = W^{c}. Note that
W is countable, and thus W ⊂ W ^{T}E = P^{c}^{T}E is at most countable.

To show that P is perfect, it is enough to show that P contains no
isolated point. (since P is closed). If p is an isolated point of P , then
there exists a neighborhood N of p such that N^{T}E = φ. p is not a
condensation point of E, a contradiction. Therefore P is perfect.

28. Prove that every closed set in a separable metric space is the union of a (possible empty) perfect set and a set which is at most countable.

(Corollary: Every countable closed set in R^{k}has isolated points.) Hint:

Use Exercise 27.

Proof: Let X be a separable metric space, let E be a closed set on
X. Suppose E is uncountable. (If E is countable, there is nothing
to prove.) Let P be the set of all condensation points of E. Since X
has a countable base, P is perfect, and P^{c}^{T}E is at most countable by
Exercise 27. Since E is closed, P ⊂ E. Also, P^{c}^{T}E = E − P . Hence
E = P ^{S}(E − P ).

For corollary: if there is no isolated point in E, then E is perfect. Thus E is uncountable, a contradiction.

Note: It’s also called Cauchy-Bendixon Theorem.

29. Prove that every open set in R^{1} is the union of an at most countable
collection of disjoint segments. Hint: Use Exercise 22.

Proof: (due to H.L.Royden, Real Analysis) Since O is open, for each x in O, there is a y > x such that (x, y) ⊂ O. Let b = sup{y : (x, y) ⊂ O}.

Let a = inf{z : (z, x) ⊂ O}. Then a < x < b, and I_{x} = (a, b) is an open
interval containing x.

Now I_{x} ⊂ O, for if w ∈ I_{x}, say x < w < b, we have by the definition of
b a number y > w such that (x, y) ⊂ O, and so w ∈ O).

Moreover, b /∈ O, for if b ∈ O, then for some > 0 we have (b−, b+) ⊂ O, whence (x, b + ) ⊂ O, contradicting the definition of b. Similarly, a /∈ O.

Consider the collection of open intervals {I_{x}}, x ∈ O. Since each x ∈ O
is contained in Ix, and each Ix⊂ O, we have O = ^{S}Ix.

Let (a, b) and (c, d) be two intervals in this collection with a point in
common. Then we must have c < b and a < d. Since c /∈ O, it does not
belong to (a, b) and we have c ≤ a. Since a /∈ O and hence not to (c, d),
we have a ≤ c. Thus a = c. Similarly, b = d, and (a, b) = (c, d). Thus
two different intervals in the collection {Ix} must be disjoint. Thus
O is the union of the disjoint collection {I_{x}} of open intervals, and it
remains only to show that this collection is countable. But each open
interval contains a rational number since Q is dense in R. Since we
have a collection of disjoint open intervals, each open interval contains
a different rational number, and the collection can be put in one-to-one
correspondence with a subset of the rationals. Thus it is a countable
collection.

30. Imitate the proof of Theorem 2.43 to obtain the following result:

If R^{k} =^{S}^{∞}_{1} F_{n}, where each F_{n} is a closed subset of R^{k}, then
at least one F_{n} has a nonempty interior.

Equivalent statement: If G_{n} is a dense open subset of R^{k}, for

n = 1, 2, 3, ..., then ^{T}^{∞}_{1} G_{n} is not empty (in fact, it is dense
in R^{k}).

(This is a special case of Baire’s theorem; see Exercise 22, Chap. 3, for the general case.)

Proof: I prove Baire’s theorem directly. Let G_{n} be a dense open
subset of R^{k} for n = 1, 2, 3, .... I need to prove that ^{T}^{∞}_{1} G_{n} intersects
any nonempty open subset of R^{k} is not empty.

Let G_{0} is a nonempty open subset of R^{k}. Since G_{1} is dense and G_{0}
is nonempty, G_{0}^{T}G_{1} 6= φ. Suppose x_{1} ∈ G_{0}^{T}G_{1}. Since G_{0} and G_{1}
are open, G_{0}^{T}G_{1} is also open, that is, there exist a neighborhood V_{1}
such that V_{1} ⊂ G_{0}^{T}G_{1}. Next, since G_{2} is a dense open set and V_{1}
is a nonempty open set, V_{1}^{T}G_{2} 6= φ. Thus, I can find a nonempty
open set V_{2} such that V_{2} ⊂ V_{1}^{T}G_{2}. Suppose I have get n nonempty
open sets V_{1}, V_{2}, ..., V_{n} such that V_{1} ⊂ G_{0}^{T}G_{1} and V_{i+1} ⊂ V_{i}^{T}G_{n+1}
for all i = 1, 2, ..., n − 1. Since Gn+1 is a dense open set and Vn is a
nonempty open set, VnT

Gn+1 is a nonempty open set. Thus I can find
a nonempty open set V_{n+1} such that V_{n+1}⊂ V_{n}^{T}G_{n+1}. By induction, I
can form a sequence of open sets {V_{n}: n ∈ Z^{+}} such that V_{1} ⊂ G_{0}^{T}G_{1}
and V_{i+1} ⊂ V_{i}^{T}G_{i+1} for all n ∈ Z^{+}. Since V_{1} is bounded and V_{1} ⊃
V_{2} ⊃ ... ⊃ V_{n} ⊃ ..., by Theorem 2.39 I know that

∞

\

n=1

V_{n}6= φ.

Since V_{1} ⊂ G_{0}^{T}G_{1} and V_{n+1} ⊂ G_{n+1}, G_{0}^{T}(^{T}^{∞}_{n=1}G_{n}) 6= φ. Proved.

Note: By Baire’s theorem, I’ve proved the equivalent statement. Next,
F_{n} has a empty interior if and only if G_{n} = R^{k}− F_{n} is dense in R^{k}.
Hence we completed all proof.