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MODERN ANALYSIS I, Exercises of Mathematics

Real Analysis

Typology: Exercises

2014/2015

Uploaded on 09/25/2015

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MODERN ANALYSIS I: DAILY HOMEWORK #8
(DUE TUESDAY, JUNE 24)
SAHIL SINGHAL
Theorem 1. The only subsets of the continuum that are both open and closed are and C.
Proof. By Theorem 3.1 and Theorem 3.6, and Care both closed and open. We will prove
the hypothesis via contradiction: Assume there is some other subset of C,Athat is both
open and closed. As Ais neither nor C, there exists some xCsuch that xA, and
some yCsuch that y /A. Then yC\A. Then C\Ais nonempty as is A. (There
should be some reason that and Cdon’t break Axiom 4. I belive this should be that the
sets must be nonempty). Note also that A(C\A) = C. In addition, as Ais closed, C\Ais
open as is A. Then Cis not connected. This contradicts Axiom 4. Hence, the only subsets
of the continuum that are both open and closed are and C.
Theorem 2. For all x, y C, if x<y, then there exists zCsuch that zis between x
and y.
Proof. Let x, y Cwhere x < y. Assume there is no zCsuch that zis between x
and y.{pC|y < p}and {pC|p < x}are open by Corollary 3.9. Now consider
A={y} {pC|y < p}. We can construct an open interval for any aA,I=Int(x, b)
where a<b. Note the existence of bis guaranteed as the Chas no minimum nor maximum.
Then, IAas there exist no zbetween xand y. Thus, by Theorem Ais open. Similarly
B={x}∪{pC|p < x}is also open. Then AB=Cand AB=. So, Cis
disconnected. This is a contradiction. Thus, for all x, y C, if x < y, then there exists
zCsuch that zis between xand y.
Corollary 3. Every open interval is infinite.
Proof. Proof by contradiction: Let ICbe the arbitrary open interval Int(a, b). Assume
that Iis finite. Then we can order the nelements of Ias a1, a2, ..., anwhere a1< a2<
... < an. Now consider a1and a2. By the previous theorem, there exists a pCsuch that
a1<p<a2. Note that then a<a1< p so a<pand p<a2< b so p<b. Then, pI.
Also note p6=aifor ibetween 1 and nas it is less than or greater than all of these points.
Then Ihas more than nelements. This is a contradiction. Hence, every open interval is
infinite.
Corollary 4. Every point of C is a limit point of C.
Proof. Consider any point pC. Now consider IC\{p}for any open interval I=Int(a, b)
where a<p<b. Then there exists a point cCsuch that a<c<pby Theorem 4.2. Note
that c<p<bso c<b. Then, cIC\{p}so IC\{p} 6=. Then pis a limit point of
C. Hence, every point of Cis a limit point of C.
Corollary 5. Every point of the open interval Int(a,b) is a limit point of Int(a,b).
1
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MODERN ANALYSIS I: DAILY HOMEWORK

(DUE TUESDAY, JUNE 24)

SAHIL SINGHAL

Theorem 1. The only subsets of the continuum that are both open and closed are ∅ and C.

Proof. By Theorem 3.1 and Theorem 3.6, ∅ and C are both closed and open. We will prove the hypothesis via contradiction: Assume there is some other subset of C, A that is both open and closed. As A is neither ∅ nor C, there exists some x ∈ C such that x ∈ A, and some y ∈ C such that y /∈ A. Then y ∈ C\A. Then C\A is nonempty as is A. (There should be some reason that ∅ and C don’t break Axiom 4. I belive this should be that the sets must be nonempty). Note also that A ∪ (C\A) = C. In addition, as A is closed, C\A is open as is A. Then C is not connected. This contradicts Axiom 4. Hence, the only subsets of the continuum that are both open and closed are ∅ and C. 

Theorem 2. For all x, y ∈ C, if x < y, then there exists z ∈ C such that z is between x and y.

Proof. Let x, y ∈ C where x < y. Assume there is no z ∈ C such that z is between x and y. {p ∈ C|y < p} and {p ∈ C|p < x} are open by Corollary 3.9. Now consider A = {y} ∪ {p ∈ C|y < p}. We can construct an open interval for any a ∈ A, I = Int(x, b) where a < b. Note the existence of b is guaranteed as the C has no minimum nor maximum. Then, I ⊂ A as there exist no z between x and y. Thus, by Theorem A is open. Similarly B = {x} ∪ {p ∈ C|p < x} is also open. Then A ∪ B = C and A ∩ B = ∅. So, C is disconnected. This is a contradiction. Thus, for all x, y ∈ C, if x < y, then there exists z ∈ C such that z is between x and y. 

Corollary 3. Every open interval is infinite.

Proof. Proof by contradiction: Let I ⊂ C be the arbitrary open interval Int(a, b). Assume that I is finite. Then we can order the n elements of I as a 1 , a 2 , ..., an where a 1 < a 2 < ... < an. Now consider a 1 and a 2. By the previous theorem, there exists a p ∈ C such that a 1 < p < a 2. Note that then a < a 1 < p so a < p and p < a 2 < b so p < b. Then, p ∈ I. Also note p 6 = ai for i between 1 and n as it is less than or greater than all of these points. Then I has more than n elements. This is a contradiction. Hence, every open interval is infinite. 

Corollary 4. Every point of C is a limit point of C.

Proof. Consider any point p ∈ C. Now consider I ∩C{p} for any open interval I = Int(a, b) where a < p < b. Then there exists a point c ∈ C such that a < c < p by Theorem 4.2. Note that c < p < b so c < b. Then, c ∈ I ∩ C{p} so I ∩ C{p} 6 = ∅. Then p is a limit point of C. Hence, every point of C is a limit point of C. 

Corollary 5. Every point of the open interval Int(a,b) is a limit point of Int(a,b). 1

2 SAHIL SINGHAL

Proof. Let Int(a, b) be an open interval and let p be an arbitrary point of Int(a, b). Now consider I ∩ (Int(a, b){p}) for any open interval I = Int(c, d) Then there exists a point x such that max(a, c) < x < min(b, d). Then x ∈ Int(a, b) and x ∈ Int(c, d) where x 6 = p. Therefore, x ∈ I ∩ (Int(a, b){p}) so I ∩ (Int(a, b){p}) 6 = ∅. Hence, p is a limit point of Int(a, b). 

Exercise 6. Construct an infinte collection of open sets whose intersection is not open.

Consider all the sets Int(a, b) where a < 1 < b. Their intersection is { 1 }, which is closed since C = (C{ 1 }) ∪ { 1 } so 1, a limit point of C by the previous corollary, must be a limit point of either of these sets. However, { 1 } is finite so has no limit points. Therefore, 1 must be a limit point of the C{ 1 }.. Therefore the set has a limit point of 1 but does not contain it. Therefore, it is not closed. Therefore { 1 }, its complement, is not open.

Exercise 7. If sup X exists, then it is unique, and similarly for inf X.

Proof. Let X be a subset of C and let sup X exist. Assume that a = supX and b = supX. Then b ≤ a and a ≤ b by Definition 35 since both are upper bounds. Then, a = b.

Let X be a subset of C and let inf X exist. Assume that a = inf X and b = inf X. Then b ≤ a and a ≤ b by Definition 35 since both are upper bounds. Then, a = b. 

Theorem 8. Let a < b. The least upper bound and greatest lower bound of the open interval Int(a, b) are: sup Int(a, b) = b and inf Int(a, b) = a.

Proof. By definition a < x for all x ∈ Int(a, b). Therefore a is a lower bound. Now consider l, any other lower bound of Int(a, b). If a < l, then there exists a c such that a < c < min(l, b) by Theorem 4.2. a < c so c ∈ Int(a, b). Therefore, l is not a lower bound. Thus, l ≤ a therefore inf Int(a, b) = a.

By definition x < b for all x ∈ Int(a, b). Therefore b is a upper bound. Now consider u, any other upper bound of Int(a, b). If u < a, then there exists a c such that a < c < max(a, u) by Theorem 4.2. c < b so c ∈ Int(a, b). Therefore, u is not an upper bound. Thus, l ≤ a therefore inf Int(a, b) = a. 

Theorem 9. Let X be a subset of C. Suppose that sup X exists and sup X /∈ X. Then sup X is a limit point of X. The same holds for inf X.

Proof. Let X be a subset of C and assume that supX exists such that supX /∈ X. Now consider any open inteval I = Int(a, b) where supX ∈ I. Then, a < supX < b. Note that X{supX} = X. Then I ∩ (X{supX}) = I ∩ X. We will now prove the hypothesis via contradiction: Assume there exists an I such that I ∩ X = ∅. Then, there exists no x ∈ X such that x ∈ I. Then for all x ∈ X, x < a. Then a is an upper bound for X. Then, a ≥ supX. However, a < supX. This is a contradiction. Therefore supX is a limit point of X. This proof is the same for the infinum. 

Corollary 10. Both a and b are limit points of the open interval Int(a, b)

Proof. Let I = Int(a, b) be an open interval. Then, inf I = a and supI = b by Theorem 4.8. Then, a and b are limit points by the previous theorem. 

Corollary 11. [a, b] = {x ∈ C|a ≤ x ≤ b}.