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 a0, ..., an, not all zero, such that
a0zn+ a1zn−1+ ... + an−1z + an = 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 + |a0| + |a1| + ... + |an| = N.
Proof: For every positive integer N there are only finitely many equa- tions with
n + |a0| + |a1| + ... + |an| = N.
(since 1 ≤ n ≤ N and 0 ≤ |a0| ≤ N ). We collect those equations as CN. HenceSCN is countable. For each algebraic number, we can form an equation and this equation lies in CM 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, R1 = { 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 R1 = (R − Q)SQ 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 E0 be the set of all limit points of a set E. Prove that S0 is closed. Prove that E and E have the same limit points. (Recall that E = ESE0.) Do E and E0 always have the same limit points?
Proof: For any point p of X − E0, 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 − E0, that is, X − E0 is open, that is, E0 is closed.
Next, if p is a limit point of E, then p is also a limit point of E since E = ESE0. 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 = E0− E. Then q is a limit point of E. Hence,
Nr0(q)
where r0 = 12 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 Nr0(q) contains in Nr(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 E0 = {0} and (E0)0 = φ.
7. Let A1, A2, A3, ... be subsets of a metric space. (a) If Bn =Sni=1Ai, prove that Bn =Sni=1Ai, for n = 1, 2, 3, ... (b) If B =S∞i=1, prove that B ⊃S∞i=1Ai. Show, by an example, that this inclusion can be proper.
Proof of (a): (Method 1) Bn is the smallest closed subset of X that contains Bn. Note that SAi is a closed subset of X that contains Bn, thus
Bn⊃
n
[
i=1
Ai.
If p ∈ Bn− Bn, then every neighborhood of p contained a point q 6= p such that q ∈ Bn. If p is not in Ai for all i, then there exists some neighborhood Nri(p) of p such that (Nri(p) − p)TAi = φ for all i. Take r = min{r1, r2, ..., rn}, and we have Nr(p)TBn = φ, a contradiction.
Thus p ∈ Ai for some i. Hence Bn⊂
n
[
i=1
Ai. that is,
Bn=
n
[
i=1
Ai.
(Method 2) Since Sni=1Ai is closed and Bn = Sni=1Ai ⊂ Sni=1Ai, Bn⊂Sni=1Ai.
Proof of (b): Since B is closed and B ⊃ B ⊃ Ai, B ⊃ Ai for all i.
Hence B ⊃ SAi.
Note: My example is Ai = (1/i, ∞) for all i. Thus, Ai = [1/i, ∞), and B = (0, ∞), B = [0, ∞). Note that 0 is not in Ai for all i. Thus this inclusion can be proper.
8. Is every point of every open set E ⊂ R2 a limit point of E? Answer the same question for closed sets in R2.
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 Nr(p) of p such that Nr(p) is a subset of E. Then for every real r0, we can choose a point q such that d(p, q) = 1/2 min(r, r0). Note that q 6= p, q ∈ Nr0(p), and q ∈ Nr(p). Hence, every neighborhood Nr0(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)}. A0 = φ and thus (0, 0) is not a limit point of E.
9. Let Eo denote the set of all interior points of a set E.
(a) Prove that Eo is always open.
(b) Prove that E is open if and only if Eo = E.
(c) If G is contained in E and G is open, prove that G is contained in Eo.
(d) Prove that the complement of Eo is the closure of the complement of E.
(e) Do E and E always have the same interiors?
(f) Do E and Eo always have the same closures?
Proof of (a): If E is non-empty, take p ∈ Eo. We need to show that p ∈ (Eo)o. Since p ∈ Eo, there is a neighborhood Nr of p such that Nr is contained in E. For each q ∈ Nr, note that Ns(q) is contained in Nr(p), where s = min{d(p, q), r − d(p, q)}. Hence q is also an interior point of E, that is, Nr is contained in Eo. Hence Eo is always open.
Proof of (b): (⇒) It is clear that Eo is contained in E. Since E is open, every point of E is an interior point of E, that is, E is contained in Eo. Therefore Eo = E.
(⇐) Since every point of E is an interior point of E (Eo(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 ∈ Eo. Therefore G is contained in Eo. (Thus Eo is the biggest open set contained in E. Similarly, E is the smallest closed set containing E.)
Proof of (d): Suppose p ∈ X − Eo. 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 − Eo is contained in X − E.
Next, suppose p ∈ X − E. If p ∈ X − E, then p ∈ X − Eo 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 − Eo. Therefore X − Eo = X − E.
Solution of (e): No. Take X = R1 and E = Q. Thus Eo = φ and Eo = (R1)o = R1 6= φ.
Solution of (f ): No. Take X = R1 and E = Q. Thus E = R1, and Eo = 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 N1/2(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 = {p1, ..., pn} be finite.
Consider an open cover {Gα} of S. Since S is covered by Gα, pi is covered by Gαi, thus {Gα1, ..., Gαn} is finite subcover of S. Hence S is compact. Next, suppose S is infinite. Consider an open cover {Gp} of S, where
Gp = N1
2(p)
for every p ∈ S. Note that q is not in Gp if q 6= p. If S is compact, then S can be covered by finite subcover, say
Gp1, ..., Gpn.
Then there exists q such that q 6= pi for all i since S is infinite, a con- tradiction. Hence only every finite subset of X is compact.
11. For x ∈ R1 and y ∈ R1, define
d1(x, y) = (x, y)2, d2(x, y) = q|x − y|, d3(x, y) = |x2− y2|, 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) d1(x, y) is not a metric. Since d1(0, 2) = 4, d1(0, 1) = 1, and d1(1, 2) = 1, d1(0, 2) > d1(0, 1) + d1(1, 2). Thus d1(x, y) is not a metric.
(2) d2(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) d3(x, y) is not a metric since d3(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
d0(x, y) = d(x, y) 1 + d(x, y) is also a metric.
Proof of Claim: (a) d0(p, q) > 0 if p 6= q; d(p, p) = 0. (b) d0(p, q) = d0(q, p). (c) Let x = d(p, q), y = d(p, r), and z = d(r, q). Then x ≤ y+z.
d0(p, q) ≤ d0(p, r) + d0(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, d0 is also a metric.
12. Let K ⊂ R1 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 {Oα} is an arbitrary open covering of K. Let E ∈ {Oα} consists 0. Since E is open and 0 ∈ E, 0 is an interior point of E. Thus there is a neighborhood N = Nr(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, E1, ..., 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 Nr(p) where r = min{n1 − p, p − n+11 } 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 ∈ R1. I take
Sn = (1 − 1
2n) + K 2n+1,
S =
∞
[
n=1
Sn[{1}.
Claim: S is compact and the set of all limit points of S is KS{1}.
Clearly, S lies in [0, 1], that is, S is bounded in R1. Note that Sn ⊂ [1 − 21n, 1 − 2n+11 ]. By Exercise 12 and its note, I have that all limit points of ST[0, 1) is
0,1 2, ..., 1
2n, ...
Clearly, 1 is also a limit point of S. Therefore, the set of all limit points of S is KS{1}. Note that KS{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 {On} = {(1/n, 1)} for n = 1, 2, 3, .... The following is my proof. For every x ∈ (0, 1),
x ∈ ( 1
[1/x] + 1, 0) ∈ {On}
Hence {On} is an open covering of (0, 1). Suppose there exists a finite subcovering
( 1
n1, 1), ..., ( 1 nk, 1)
where n1 < n2 < ... < nk, respectively. Clearly 2n1
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 R1, for ex- ample) if the word ”compact” is replaced by ”closed” or by ”bounded.”
Theorem 2.36: If {Kα} is a collection of compact subsets of a metric space X such that the intersection of every finite subcollection of {Kα} is nonempty, then TKα is nonempty.
Corollary: If {Kn} is a sequence of nonempty compact sets such that Kn contains Kn+1 (n = 1, 2, 3, ...), then TKn 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 < p2 < 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:
{Gα} = {Nr(p) : p ∈ E and (p − r, p + r) ⊂ S}
Surely, {Gα} is a open covering of E. If E is compact, then there are finitely many indices α1, ..., αn such that
E ⊂ Gα1[...[Gαn.
For every Gαi = Nri(pi), take p = max1≤i≤npi. Thus, p is the nearest point to √
3. But Nr(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
an
10n : an = 4 or an = 7o
Claim: E is uncountable.
Proof of Claim: If not, we list E as follows:
x1 = 0.a11a12...a1n...
x2 = 0.a21a22...a2n...
... ...
xk = 0.ak1ak2...akn...
... ...
(Prevent ending with all digits 9) Let x = 0.x1x2...xn... where
xn=
4 if ann = 7 7 if ann = 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 ET(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.p1p2...pn...
Since p /∈ E, there exists the smallest k such that pk 6= 4 and pk 6= 7.
When pk = 0, 1, 2, 3, select the smallest l such that pl = 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.p1...pl−14pl+1...pk−17 < p < 0.p1...pk−14.
Thus there is a neighborhood of p such that contains no points of E, a contradiction.
When pk = 5, 6,
0.p1...pk−147 < p < 0.p1...pk−174.
Thus there is a neighborhood of p such that contains no points of E, a contradiction.
When pk = 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.p1p2...pn... Let
xk = 0.y1y2...yn...
where
yn =
pk if k 6= n 4 if pn = 7 7 if pn = 4
Thus, |xk− p| → 0 as k → ∞. Also, xk 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 R1 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 {In} such that
S ⊂[In and X|In| < 1.
Consider E = R1 −SIn. 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 TS = 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 ATB and ATB are empty. Since A and B are closed sets, A = A and B = B. Hence ATB = ATB = ATB = φ. Hence A and B are separated.
Proof of (b): Suppose ATB is not empty. Thus there exists p such that p ∈ A and p ∈ B. For p ∈ A, there exists a neighborhood Nr(p) of p contained in A since A is open. For p ∈ B = BSB0, if p ∈ B, then p ∈ ATB. Note that A and B are disjoint, and it’s a contradiction.
If p ∈ B0, 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 Nr(p)of p containing a point q 6= p such that q ∈ B. Note that Nr(p) ⊂ A, thus q ∈ A. With A and B are disjoint, we get a contradiction. Hence AT(B) is empty.
Similarly, ATB is also empty. Thus A and B are separated.
Proof of (c): Suppose ATB is not empty. Thus there exists x such
that x ∈ A and x ∈ B. Since x ∈ A, d(p, x) < δ. x ∈ B = BSB0, 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 Nr(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 ATB is empty. Similarly, ATB 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 = δt = 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 = ASB, then X is not connected, a contradiction. Thus there exists yt∈ X such that y /∈ ASB. Let
E = Et= {x ∈ X : d(x, p) = δt} 3 yt.
For any real t ∈ (0, 1), Et is non-empty. Next, Et and Es are disjoint if t 6= s (since a metric is well-defined). Thus X contains a uncount-
able set {yt : 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 ATB = φ, and BTA = φ. In order to do these, it is sufficient to show ATB = φ. Let x ∈ ATB = φ, then we have:
(1) x ∈ A ⇒ d(x, p) ≤ r(2) x ∈ B ⇒ d(x, p) > r It is impossible. So, ATB = 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, xi) : xi ∈ C}, and thus let r ∈ R − S and inf S < r < sup S.
And construct A and B as in (a), we have C = ASB, 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 R2.)
Solution: Closures of connected sets is always connected, but interiors of those is not. The counterexample is
S = N1(2, 0)[N1(−2, 0)[{x − axis} ⊂ R2.
Since S is path-connected, S is connect. But So = N1(2)SN1(−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 ATB = ATB = φ. Note that
S = S − T
= A[B − T
= (A[B)\Tc
= (A\Tc)[(B\Tc)
where T = S − S. Thus
(A\Tc)\B\Tc ⊂ (A\Tc)\B\Tc
⊂ A\B
= φ.
Hence (ATTc)TBTTc = φ. Similarly, ATTcT(BTTc) = φ.
Now we claim that both ATTc and BTTc are nonempty. Suppose that BTTc = φ. Thus
A\Tc = S ⇔ A\(S − S)c = S
⇔ A\(A[B − S)c = S
⇔ A\((A[B)\Sc)c = S
⇔ A\((Ac\Bc)[S) = S
⇔ (A\S)[(A\Ac\Bc) = S
⇔ A\S = S.
Thus B is empty, a contradiction. Thus BTTc is nonempty. Similarly, ATTc 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 Rk, suppose a ∈ A, b ∈ B, and define
p(t) = (1 − t)a + tb
for t ∈ R1. Put A0 = p−1(A), B0 = p−1(B). [Thus t ∈ A0 if and only if p(t) ∈ A.]
(a) Prove that A0 and B0 are separated subsets of R1.
(b) Prove that there exists t0 ∈ (0, 1) such that p(t0) /∈ ASB.
(c) Prove that every convex subset of Rk is connected.
Proof of (a): I claim that A0TB0 is empty. (B0TA0 is similar). If not, take x ∈ A0TB0. x ∈ A0 and x ∈ B0. x ∈ B0 or x is a limit point of B0. x ∈ B0 will make x ∈ A0TB0, that is, p(x) ∈ ATB, a contradiction since A and B are separated.
Claim: x is a limit point of B0 ⇒ p(x) is a limit point of B. Take any neighborhood Nr 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 B0, 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 ∈ B0, that is, p(y) ∈ B. Also, p(y) ∈ Nr. Therefore, p(x) is a limit point of B. Hence p(x) ∈ ATB, a contradiction since A and B are separated.
Hence A0TB0 is empty, that is, A0 and B0 are separated subsets of R1.
Proof of (b): Suppose not. For every t0 ∈ (0, 1), neither p(t0) ∈ A nor p(t0) ∈ B (since A and B are separated). Also, p(t0) ∈ ASB for all t0 ∈ (0, 1). Hence (0, 1) = A0SB0, a contradiction since (0, 1) is connected. I completed the proof.
Proof of (c): Let S be a convex subset of Rk. If S is not connected, then S is a union of two nonempty separated sets A and B. By (b), there exists t0 ∈ (0, 1) such that p(t0) /∈ ASB. But S is convex, p(t0) must lie in ASB, a contradiction. Hence S is connected.
22. A metric space is called separable if it contains a countable dense sub- set. Show that Rk 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 = (x1, x2, ..., xk) ∈ Rk, we can find a rational sequence {rij} → xj for j = 1, ..., k since Q is dense in R1. Thus,
ri = (ri1, ri2, ..., rik) → x
and ri ∈ S for all i. Hence S is dense in Rk. Also, S is countable, that is, S is a countable dense subset in Rk, Rk is separable.
23. A collection {Vα} 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 ∈ Vα ⊂ G for some α. In other words, every open set in X is the union of a subcollection of {Vα}.
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 {Vα} = { all neighborhoods with rational radius and center in S }. We claim that {Vα} is a base for X.
For every x ∈ X and every open set G ⊂ X such that x ∈ G, there exists a neighborhood Nr(p) of p such that Nr(p) ⊂ G since x is an interior point of G. Since S is dense in X, there exists {sn} → x. Take a rational number rn such that rn < r2, and {Vα} 3 Nrn(sn) ⊂ Nr(p) for enough large n. Hence we have x ∈ Vα ⊂ G for some α. Hence {Vα} 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 x1 ∈ X. Having chosen x1, ..., xj ∈ X, choose xj+1, if possible, so that d(xi, xj+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 x1 ∈ X. Having chosen x1, ..., xj ∈ X, choose xj+1, if possible, so that d(xi, xj+1) ≥ δ for i = 1, ..., j. If this process cannot stop, then consider the set A = {x1, x2, ..., xk}. If p is a limit point of A, then a neighborhood Nδ/3(p) of p contains a point q 6= p such that q ∈ A. q = xk for only one k ∈ N . If not, d(xi, xj) ≤ d(xi, p) + d(xj, p) ≤ δ/3 + δ/3 < δ, and it contradicts the fact that d(xi, xj) ≥ 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 Nδ(x1), Nδ(x2), ..., Nδ(xk), 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 Nr(p). Note that Nr/2(p) can be covered by finite many neighborhoods Ns(x1), ..., Ns(xk) of radius s = 1/n where n = [2/r] + 1 and xi ∈ A for i = 1, ..., k. Hence, d(x1, p) ≤ d(x1, q) + d(q, p) ≤ r/2 + s < r where q ∈ Nr/2(p)TNs(x1). Therefore, x1 ∈ Nr(p) and x1 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 Nr(x) such that Nr(x) ⊂ G since x is an interior point of G. Hence x ∈ Nm(p) ∈ {Vα} for some p where m = [2/r] + 1. For every y ∈ Nm(p), we have
d(y, x) ≤ d(y, p) + d(p, x) < m + m = 2m < r.
Hence Nm(p) ⊂ G, that is, Vα ⊂ G for some α, and therefore {Vα} is a countable base of K. Next, collect all of the center of Vα, say D, and we claim D is dense in K (D is countable since Vα is countable). For all p ∈ K and any > 0 we can find Nn(xn) ∈ {Vα} where n = [1/] + 1.
Note that xn ∈ D for all n and d(p, xn) → 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 {Gn}, n = 1, 2, 3, .... If no finite subcollection of {Gn} covers X, then the complement Fn of G1S...SGn is nonempty for each n, but TFn is empty. If E is a set contains a point from each Fn, 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 {Gn}, n = 1, 2, 3, ....
If no finite subcollection of {Gn} covers X, then the complement Fn of G1S
...SGn is nonempty for each n, but TFn is empty. If E is a set contains a point from each Fn, consider a limit point of E.
Note that Fk ⊃ Fk+1 ⊃ ... and Fn is closed for all n, thus p lies in Fk for all k. Hence p lies in TFn, but TFn 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 ⊂ Rk, 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 PcTE is at most countable. Hint: Let {Vn} be a countable base of Rk, let W be the union of those Vn for which ETVn is at most countable, and show that P = Wc.
Proof: Let {Vn} be a countable base of Rk, let W be the union of those Vn for which ETVn is at most countable, and we will show that P = Wc. Suppose x ∈ P . (x is a condensation point of E). If x ∈ Vn for some n, then ETVn is uncountable since Vn is open. Thus x ∈ Wc. (If x ∈ W , then there exists Vn such that x ∈ Vn and ETVn
is uncountable, a contradiction). Therefore P ⊂ Wc.
Conversely, suppose x ∈ Wc. x /∈ Vn for any n such that ETVn is countable. Take any neighborhood N (x) of x. Take x ∈ Vn ⊂ N (x), and ETVn is uncountable. Thus ETN (x) is also uncountable, x is a condensation point of E. Thus Wc ⊂ P . Therefore P = Wc. Note that W is countable, and thus W ⊂ W TE = PcTE 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 NTE = 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 Rkhas 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 PcTE is at most countable by Exercise 27. Since E is closed, P ⊂ E. Also, PcTE = 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 R1 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 Ix = (a, b) is an open interval containing x.
Now Ix ⊂ O, for if w ∈ Ix, 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 {Ix}, x ∈ O. Since each x ∈ O is contained in Ix, and each Ix⊂ O, we have O = SIx.
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 {Ix} 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 Rk =S∞1 Fn, where each Fn is a closed subset of Rk, then at least one Fn has a nonempty interior.
Equivalent statement: If Gn is a dense open subset of Rk, for
n = 1, 2, 3, ..., then T∞1 Gn is not empty (in fact, it is dense in Rk).
(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 Gn be a dense open subset of Rk for n = 1, 2, 3, .... I need to prove that T∞1 Gn intersects any nonempty open subset of Rk is not empty.
Let G0 is a nonempty open subset of Rk. Since G1 is dense and G0 is nonempty, G0TG1 6= φ. Suppose x1 ∈ G0TG1. Since G0 and G1 are open, G0TG1 is also open, that is, there exist a neighborhood V1 such that V1 ⊂ G0TG1. Next, since G2 is a dense open set and V1 is a nonempty open set, V1TG2 6= φ. Thus, I can find a nonempty open set V2 such that V2 ⊂ V1TG2. Suppose I have get n nonempty open sets V1, V2, ..., Vn such that V1 ⊂ G0TG1 and Vi+1 ⊂ ViTGn+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 Vn+1 such that Vn+1⊂ VnTGn+1. By induction, I can form a sequence of open sets {Vn: n ∈ Z+} such that V1 ⊂ G0TG1 and Vi+1 ⊂ ViTGi+1 for all n ∈ Z+. Since V1 is bounded and V1 ⊃ V2 ⊃ ... ⊃ Vn ⊃ ..., by Theorem 2.39 I know that
∞
\
n=1
Vn6= φ.
Since V1 ⊂ G0TG1 and Vn+1 ⊂ Gn+1, G0T(T∞n=1Gn) 6= φ. Proved.
Note: By Baire’s theorem, I’ve proved the equivalent statement. Next, Fn has a empty interior if and only if Gn = Rk− Fn is dense in Rk. Hence we completed all proof.
