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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}.
