Math 241

Let F be a collection of sets in Rn, and let S = UA ∈ F A and T = n A ∈ F A. For each of the following statements, either give a proof or exhibit a counterexample. (a) If x is an accumulation point of T, then x is an accumulation point of each set A in F. (b) If x is an accumulation point of S, then x is an accumulation point of at least one set A in F

(a): This statement is true. Let x be an accumulation point of T A∈F A then for each open ball B(x, r), we have B(x, r) ∩ ( T A∈F A) 6= ∅, thus by the definition of intersection, B(x, r) ∩ A 6= ∅ for each A C F. (b): This statement is false. Let F = {An} where An = (0, n−1 n ). Then S∞ n=1 An = (0, 1). Clearly 1 is an accumulation point of (0, 1), but 1 is not an accumulation point for any An; observe B(1; 1/n) ∩ An = ∅

pic b),e),g)

(b): Let x ∈ A ∩ (B ∪ C), thus x ∈ A and x ∈ B ∪ C. Without loss of generality (wlog) x ∈ B, so x ∈ A ∩ B and x ∈ (A ∩ B) ∪ (A ∩ C). Conversely, let x ∈ (A ∩ B) ∪ (A ∩ C), then wlog x ∈ A ∩ B, hence x ∈ A ∩ (B ∪ C). (e): Let x ∈ A ∩ (B \ C), then x ∈ A, x ∈ B, and x /∈ C, thus x ∈ A ∩ B and x /∈ A ∩ C, thus x ∈ (A ∩ B) \ (A ∩ C). Conversely, if x ∈ (A ∩ B) \ (A ∩ C) , then x ∈ A, x ∈ B, and x /∈ A ∩ C. We see x /∈ C, otherwise x ∈ A ∩ C, therefore x ∈ A ∩ (B \ C). (g): Suppose that (A \ B) ∪ B = A, then for every element x ∈ B we have that x ∈ A, thus B ⊆ A. Conversely, if B ⊆ A, then (A \ B) ∪ B ⊆ A. Moreover, if x ∈ A, then x ∈ B or x ∈ A \ B, thus (A \ B) ∪ B ⊇ A which implies (A \ B) ∪ B = A. Can also use venn diagrams to prove

Prove that every nonempty open set S in R' contains both rational and irrational numbers

By the representation theorem for open subsets of R 1 , we know that all such sets are disjoint unions of open intervals. Thus for solving the problem, it will suffice to restrict our attention to open intervals (a, b) with a, b ∈ R. Let n ∈ N such that 1/n < b−a. We claim that there exists some integer m ∈ Z such that m/n ∈ (a, b). Suppose not, and let m be the largest integer such that m/n < a. By assumption b < (m + 1)/n, thus b − a < (m + 1)/n − m/n = 1/n, contradicting our choice of n. The case of irrational numbers is very similar. Let n be as above, then m/( √ 2n) is irrational for each m ∈ Z\{0}. For some choice of m ∈ Z, we have m/( √ 2n) ∈ (a, b). To ensure m 6= 0, we may translate (a, b) to (a + k, b + k) with k ∈ Z so that 0 ∈/ (a + k, b + k). Then m/( √ 2n) ∈ (a + k, b + k) if and only if m/( √ 2n) − k ∈ (a, b)

Let A and B be two sets of positive numbers bounded above, and let a=supA, b=supB. Let C be the set of all products of the form xy, where xeA and yeB. Prove that ab=supC

First we prove that C has an upper bound. Let z ∈ C, then there exists x ∈ A and y ∈ B such that z = xy. By assumption, x ≤ a and y ≤ b, thus z = xy ≤ ab as x, y, a, b > 0. Hence ab is an upper bound for C. By the completeness axiom, C has a least upper bound. Let c = sup C. We know that c ≤ ab, thus for showing that c = ab, it suffices to prove that ab ≤ c. Let γ > 1 be a real number. Then by the approximation property, there exists x ∈ A and y ∈ B such that a/γ < x and b/γ < y. Therefore, as x, y, a, b, γ > 0, we know that ab < xy · γ 2 < cγ2 = c + (γ 2 − 1)c. We can now invoke Theorem 1.1 to conclude that ab ≤ c. We remark on a potential issue with this solution: in order to apply this earlier theorem as stated, we need to check that for every > 0, there exists a choice of γ such that = (γ 2 − 1)c. Indeed, γ = (/c + 1)1/2 works and this γ is a real number. Although we have not proven it yet, for each positive real number x, there exits a unique real number y such that y 2 = x. An alternate rigorous approach is to strengthen Theorem 1.1 and show that in fact one does not need the statement to hold for all > 0, only a sequence of > 0 which converges to zero

Prove the comparison property for suprema

First we prove that S has an upper bound. Suppose not, then for every real number x there exists some s ∈ S such that x < s. We obtain a contradiction by letting x = t ∈ T as the condition of the theorem requires that s ≤ t. By the completeness axiom, S has a supremum. To prove that sup S ≤ sup T, suppose for contradiction that sup T < sup S. By the approximation property, there exists s ∈ S such that sup T < s ≤ sup S. Let t ∈ T, then t ≤ sup T and thus t < s, a contradiction.

Prove that every closed set in R' is the intersection of a countable collection of open sets.

For each n ∈ N, let an, bn ∈ Q such that an < p, q < bn, and both p − an < 1/n and bn − q < 1/n. First observe that \∞ n=1 (an, bn) = [p, q]. To see this, note that for any such a, b, if x ∈ [p, q] then x ∈ (a, b). Conversely, suppose that x /∈ [p, q] and without loss of generality, suppose that x < p. We may take n large enough so that an ∈ (x, p). We find that x /∈ (an, bn) and thus x is not in the intersection either. We can similarly express an unbounded closed interval as the intersection of countably many open sets. Finally, if S is a closed subset of R 1 then its complement is an open subset of R 1 and thus by the representation theorem for open sets in R 1 must be a countable collection of disjoint open intervals. Thus S must be itself the disjoint union of countably many closed intervals. Let us enumerate these intervals {Ik}. As above, we can find countably man open intervals {Ik,n} such that T∞ n=1(ak,n, bk,n) = Ik. It is not hard to check that S = T∞ n=1( S∞ k=1(ak,n, bk,n)). Note it is important here that for any x ∈ R such that x /∈ S k∈N Ik, there exists N ∈ N such that if n ≥ N then for all k ∈ N, we have x /∈ (ak,n, bk,n).

Recall the definition of an equivalence relation: https://en. wikipedia.org/wiki/Equivalence_relation . Prove that the relation A ∼ B on sets given by the exitstense of a bijection f : A → B defines an equivalence relation.

Let A, B, and C be sets. Reflexivity: To prove A ∼ A for all sets A, note that the identity function f(x) = x for all x ∈ A is a bijection from the set A to itself. Symmetry: Let A ∼ B so that there exists some bijection f : A → B. Because f is a bijection, we know that f −1 is a function. Moreover, because (f −1 ) −1 = f is a function, we know that f −1 : B → A is a bijection. Therefore B ∼ A. Transitivity: Let A ∼ B and B ∼ C so that there exists bijections f : A → B and g : B → C. We claim that g ◦ f : A → C is a bijection which implies that A ∼ C. We know from function composition that g ◦ f is a function. To check bijectivity, we will show that the composition of injective and surjective functions are injective and surjective, respectively. For surjectivity, let c ∈ C. We wish to show that c is in the image of g ◦ f. By surjectivity of g, there exists some b ∈ B such that g(b) = c. By surjectivity of f, there exits some a ∈ A such that f(a) = b. Therefore (g◦f)(a) = c. For injectivity, let a1, a2 ∈ A be distinct. For establishing injectivity of g ◦f, we must show that (g ◦f)(a1) 6= (g ◦f)(a2). By injectivity of f, f(a1) 6= f(a2), and by injectivity of g, g(f(a1)) 6= g(f(a2)), therefore g ◦ f is injective

Prove that every infinite set S contains a proper subset similar to S.

Let S be an infinite set. By the previous homework problem 2.12, there exists a countable set T such that T ⊆ S. Let T = {ti}i∈N, and let f : T → T \ {t1} be the function such that for each i ∈ N, f(ti) = ti+1. Clearly f is a bijection. Let F : S → S\{t1} be the function such that F(s) = s if s ∈ S \ T f(s) if s ∈ T . It is evident that F is a bijection and this completes the proof

Show sup and inf are unique sets

Let S ⊂ R and suppose that s1 and s2 are two different suprema for S. Without loss of generality s1 < s2, but this contradicts condition (b) from Definition 1.13 for s2; observe that s1 is a smaller upper bound for S than s2

Let sn = a0.a1. . . an be a finite decimal representation. Suppose that we have an infinite set of finite decimal representations S = {sn} which are compatible in the sense that for each n, sn − sn−1 = an · 10−n . We can define s = sup S. Let us define the object a0.a1a2 . . . associated to S to be an infinite decimal representation of s. Notice I am not assuming that a0.a1a2 . . . is obtained from the algorithm described in the book in Section 1.16, so this definition is more general than the one given. Suppose that a0.a1 . . . and b0.b1 . . . are two infinite decimal representations which determine real numbers x and y, respectively. Prove that x = y if and only if there exits some n ∈ Z≥0 such that ak = bk for k < n, an = bn + 1, and am = 0 and bm = 9 for n < m.

Let sn be as in the statement, and let rn = b0.b1 . . . bn. For one direction, suppose that there exits some n ∈ Z≥0 such that ak = bk for k < n, an = bn + 1, and am = 0 and bm = 9 for n < m. We wish to show that x = y. Obviously, y ≤ x, so we just need to show that x ≤ y. Suppose for contradiction that y < x. Then similar to the solution given to part b) above, it suffices that to prove that we can find some m such that x − rm < y − rm. We know that x − rm = 10−m when m > n and the statement follows. For the converse, suppose that x = y, then we wish to prove there exits some n ∈ Z≥0 such that ak = bk for k < n, an = bn + 1, and am = 0 and bm = 9 for n < m (or the same statement with the roles of a and b reversed, or ak = bk for all i). Suppose not and let n be the first integer for which an 6= bn. If bn − 2 ≤ an, then let z = a0.a1 . . .(an + 1), and observe x < z < y, a contradiction. The case where an − 2 ≤ bn is identical. Without loss of generality, assume now that an = bn + 1. If there exists some m > n for which an 6= 0, then x ≤ a0.a1 . . . am−10 . . . < y, a contradiction. Finally, suppose that there exists some m > n for which bm 6= 9, then x < b0.b1 . . . bm−19 ≤ y, a contradiction

If S and T are subsets of R", prove that (intS)n(intT)=int(SnT), and (intS)v(intT) subset int(SUT).

Let x ∈ (int S) ∩ (int T), then there exists radii r1, r2 > 0 such that B(x; r1) ⊆ S and B(x, r2) ⊆ T. Let r = min{r1, r2}, then B(x; r) ⊆ S ∩ T, hence x ∈ int (S ∩ T). Conversely, let x ∈ int (S ∩ T) then there exists some radius r > 0 such that B(x; r) ⊆ S ∩T, hence B(x; r) ⊆ S and B(x; r) ⊆ T, thus x ∈ (int S)∩(int T), which implies that int (S ∩ T) ⊆ (int S) ∩ (int T). This completes the proof. For the second containment from the question, without loss of generality let x ∈ (int S), then there exists some radius r > 0 such that B(x; r) ⊆ S. Because B(x; r) ⊆ S ∪ T as well, it follows that (int S) ⊆ int (S ∪ T). By symmetry of S and T we conclude that (int S) ∪ (int T) ⊆ int (S ∪ T)

Prove that the set of open disks in the xy - plane with center (x, x) and radius x > 0, x rational, is a countable covering of the set {(x, y) : x > 0, y > 0}.

Note that confusion can sometimes be caused by the fact that the notation for open intervals in R and coordinates for points in R 2 is identical, so be careful. Let (x, y) ∈ R 2 such that x, y > 0. It suffices to prove that the point (x, y) belongs to some open ball B(q, q) with q ∈ Q such that q > 0. Without loss of generality, suppose y ≤ x, then ||(x, y) − (y, y)|| = y − x < y, so (x, y) ∈ B(y; y). By previous homework, we know that every open interval contains a rational point. Let q be a rational number in the interval (y − x, y). We claim that the point (x, y) ∈ B((q, q); q). By the triangle inequality, ||(x, y) − (q, q)|| ≤ ||(x, y) − (q, y)|| + ||(q, y) − (q, q)|| = (q − x) + (y − q) = y − x < q. This proves the claim.

Let x be a positive real number and let a0.a1 . . . be an infinite decimal representation for x. Prove that x is rational if and only if there exists m, t ∈ N such that for all n ≥ m, we have an+t = an. That is, the decimal representation of x is eventually periodic.

Suppose there exists m, t ∈ N such that for all n ≥ m, we have an+t = an. Then, let y = 10n−1 · (x − a0.a1 . . . an−1). Observe that if y is rational then x is rational. Let z = b1 . . . bt where bi = 9 for all i, and note that z is a positive integer. Let w = c0.c1 . . . be an infinite decimal representation where ckt = 1 for all k ∈ N, and 0 otherwise. Then z · w = 0.999 . . . = 1 by problem 2. Hence w = 1 z . Moreover, y = an . . . an+t−1 · w = an . . . an+t−1 · 1 z , hence y is rational. Conversely, let a, b ∈ N. Suppose we calculate a decimal expansion for a b using the algorithm from the book. At the nth step, we have constucted a number rn = a0.a1 . . . an so that a b −rn < 10−n . If we calculate this difference a b −rn as a fraction (without canceling any common factors in the numerator and the denominator), the denominator becomes b · 10n . Because a b − rn < 10−n , the numberator must be at most b. Therefore there are only b possible numerators at each step of the algorithm. If the algorithm does not cease, then there must exist some point at which you obtain a numerator you have already seen and this will cause the algorithm to enter a loop, which will result in a decimal expansion which is periodic onwards. On the other hand, if the algorithm does cease, then the infinite decimal representation is eventually all 0's and this satisfies the desired conditions.

Let S be the collection of all sequences whose terms are the integers 0 and 1.Show that S is uncountable.

This problem is essentially asking you to work through Cantor's diagonalization argument. Let T be the set of all sequences with entries 0 or 1 and let S ⊂ T be the set of all such sequence which are infinite. By Theorem 2.16, it suffices to prove that S is uncountable, so suppose for contradiction that S = {Si}i∈N is an enumeration of these sequences. Futhermore let us denote Si = si,1si,2si,3 . . . be the entries of Si . Let us define a sequence x = x1x2x3 . . . with entries 0 or 1 as follows: xi = 1 si,i = 0 0 si,i = 1 . Suppose that there exists some j ∈ N such that Sj = x. Then sj,j = xj , but this contradicts the definition of x. Thus x is an infinite sequence with entries 0 or 1 which does not appear in our list contradicting our assumption that we had a complete list of all infinite sequences with entries 0 or 1. Hence S is uncountable and so is T.

Prove that the set S of rational numbers in the inerval (0, 1) cannot be expressed as the intersection of a countable collection of open sets. Hint: Write S = {x1, x2,... }, assume that S =n k=1 k=infinity Sk, where each Sk is open, and construct a sequence {Qn} of closed intervals such that Qn+1 ⊆Qn⊆Sn and such that xn /∈ Qn. Then use the Cantor intersection theorem to obtain a contradiction

We follow the hint and recursively construct the Qn. Let R1 = R and for n ∈ N, let Rn+1 = Qn∩Sn+1. First we observe that Rn is non empty because Qn is a closed interval which contains some rational point from (0, 1) which by assumption must also be in Sn+1. Because Qn is a closed interval and Sn is a disjoint union of open intervals and they have a common point, there is an open interval contained in Rn+1. For n ∈ Z≥0, we define 1

if S is an infinite set, prove that S contains a countably infinite subset

We inductively construct an infinite sequence of distinct elements belonging to S. Suppose that for k ∈ N we have defined Sk = S \ Ak, where Ak = {a1, . . . , ak} is a collection of k distinct elements from S, and suppose that Sk is non empty. There exists some element ak+1 ∈ Sk, hence can define Ak+1 = Ak ∪ {ak+1}, and Sk+1 = S \ Ak+1. Because S is infinite, we know that S 6= Ak+1 and thus Sk+1 is non empty. This shows that we may construct an infinite sequence of elements {ak} contained in S.

Let f: S-+T be a function. Prove that the following statements are equivalent. a)f is one-to-one on S. b)f(AnB)=f(A) nf(B) for all subsets A, BofS. c)f-'[f(A)]=A for every subset A of S. d)For al disjoint subsets A and B of S, the images f(A)and f'(B) are disjoint. e) For all subsets A and B of S with B-A, we have f(A-B)=f(A)-f(B).

We prove (a) ⇒ (b) ⇒ (c) ⇒ (d) ⇒ (e) ⇒ (a). Each individual implication is proven by the contrapositive. (a) ⇒ (b): First observe that f(A ∩ B) ⊆ f(A) ∩ f(B) for all functions. Suppose f(A ∩ B) 6= f(A) ∩ f(B), then there exists y ∈ f(A) ∩ f(B) such that y /∈ f(A ∩ B). 1 2 Therefore y ∈ f(A \ B) and y ∈ f(B \ A), thus there exits x1 ∈ A \ B and x2 ∈ B \ A, such that f(x1) = f(x2) = y. Therefore f is not injective. (b) ⇒ (c): For an arbitrary function f, we have A ⊆ f −1 (f(A)). Therefore if A 6= f −1 (f(A)) then f −1 (f(A)) \A 6= ∅. Let B = f −1 (f(A)) \A, and observe A∩B = ∅. Hence f(A ∩ B) = f(∅) = ∅. On the other hand, f(A) ∩ f(B) = f(B) 6= ∅ as f(B) ⊆ f(A). Thus condition (b) fails to hold. (c) ⇒ (d): Suppose that there exists disjoint set A, B such that f(A) and f(B) are not disjoint, then there exits some x1 ∈ A and x2 ∈ B such that f(x1) = f(x2). Then x2 ∈ f −1 (f({x1}), hence f −1 (f({x1}) 6= {x1}. (d) ⇒ (e): We first note that for all sets B ⊆ A, we have f(A) \ f(B) ⊆ f(A \ B). Suppose that f(A) \ f(B) 6= f(A \ B), then there exists some y ∈ f(A \ B) such that y /∈ f(A) \ f(B). Hence y ∈ f(A) and y ∈ f(B). Therefore y ∈ f(A \ B) ∩ f(B). As A \ B and B are disjoint, we see that condition (d) fails to hold. (e) ⇒ (a): Suppose there are x1, x2 ∈ S such that x1 6= x2 and f(x1) = f(x2). Let A = {x1, x2} and B = {x2}, then f(A \ B) = f(x1), but f(A) \ f(B) = ∅, so condition (a) fails to hold.

pic

a) Given the recursive definition of the ai we need to prove that for each n ∈ N, we have 0 ≤ an ≤ k − 1. Let rn−1 be as in part (b). If an = 0, then rn = rn−1 ≤ x. Thus because we choose an maximal with rn ≤ x, we may conclude that 0 ≤ an. To see that an < k, suppose for contradiction that an ≥ k. Then rn−1 + 1 kn−1 = rn−1 + k kn ≤ rn ≤ x, but this contradicts our assumption that we chose an−1 maximal. b) Clearly x is an upper bound for {rn}. To see that it is the least upper bound, suppose for contradiction that there exists some y ∈ R such that y = sup {rn}, and y < x. We claim that we can find some n such that y < rn ≤ x. If suffices to prove that there exists some n such that 0 ≤ x − rn < x − y; in this case we may subtract x from each term to obtain −x ≤ −rn < −y and then multiply through by −1 to obtain the desired inequalities. By construction, we know that 0 ≤ x − rn. Furthermore, x − rn < k−n otherwise we could have chosen an larger. Therefore, simply take n large enough so that k −n < x − y and this completes the proof