Write down the following, if they exist (a) all minimal elements (b) all maximal elements (c) all upper bounds of {d, f}
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- Describe the kernel of epimorphism in Exercise 20. Consider the mapping :Z[ x ]Zk[ x ] defined by (a0+a1x++anxn)=[ a0 ]+[ a1 ]x++[ an ]xn, where [ ai ] denotes the congruence class of Zk that contains ai. Prove that is an epimorphism from Z[ x ] to Zk[ x ].Let A = {2, 3, 4, . . .} be ordered by “x divides y”.(i.) Determine the minimal elements of A.(ii.) Determine the maximal elements of ALet A ⊆ R be nonempty and bounded above, and let S be the union of all A ∈ A. (a) First, prove that S ∈ R by showing that it is a cut. (b) Now, show that S is the least upper bound for A. This finishes the proof that R is complete. Notice that we could have proved that least upper bounds exist immediately after defining the ordering on R, but saving it for last gives it the privileged place in the argument it deserves. There is, however, still one loose end to sew up. The statement of Theorem 8.6.1 mentions that our complete ordered field contains Q as a subfield. This is a slight abuse of language. What it should say is that R contains a subfield that looks and acts exactly like Q.
- The topic is boundedness and fields in calculus.1. Suppose that A is not an empty set, and is a set of real numbers bounded below. Let -A be the set of all numbers -x with x ∈ A. Prove that infA = -sup(-A).PS: inf means least upper bound and sup means the greatest lower bound.2. Suppose that 0 < s ∈ R and fix s. Put a,b,c,d ∈ Q where b,d > 0 and f = a/b = c/d such that sr = (sa )(1/b). Now if x ∈ R, define S(x) = {St : t ∈ Q and t less than or equal x} Prove that St = supS(f)PS: sup is the same as the greatest lower bound3. Prove that if the multiplicative axiom of a field implies that if x≠0 and xy = x, then y =1.a)Let A = {x ∈ R + : x 2 < 2} . Explain why A i bounded above and below and find sup(A) and inf(A). b) Prove and give an example that if A and B are two bounded non-empty subset of R, then A ∪ B is also bounded3. a) We know that the set S = {1/n : n ∈ N} is not compact because 0 is a limit point of S that is not in S. To see the non-compactness of S in another way, find an open cover of S that does not have a finite subcover! b) Is the Cantor set closed? Is it compact? Explain!
- Let σpo := (<) be the usual signature for strict partial orders. A linear order (A, <) is called dense if for any a, b ∈ A with a < b, there is c ∈ A such that a < c < b. a. Write down a σpo -theory DLO axiomatizing the class of all dense linear order without endpoints (i.e. without a least / largest elements). b. Verify that (ℚ, <) is a countable (if it is finite or admits a bijection to ℕ) model of DLO. You may use that ℚ is countable without proof. Note: It is a theorem that DLO has only one countable model up to isomorphism, i.e. all countable models are isomorphic to each other.Say that X ⊆ R is closed under multiplication if for all x,y ∈ X, we have that xy ∈ X. Give three examples of finite subsets of R which are closed under multiplication. Give an example of an infinite proper subset of R which is closed under multiplication. Prove your answer. Give an example of an infinite proper subset of R which is not closed under multiplication. Prove your answer.This is under Topology class:Answer thisa.) Show that if A is closed in Y and Y is closed in X, then A is closed in X.
- Give an example for an open set A ⊂ R with the following property: The boundary of the closure of A has positive Lebesgue measureShow that R^3 is a banach space with explanation.a)To prove this directly, you'd suppose that A is a subset of B. Then you'd need to show that sup(A)≤sup(B)sup(A)≤sup(B). Since A is bounded above, sup(A) is the least upper bound for A which, by definition of l.u.b., is less than or equal to all upper bounds for A. b)Note that, max{sup(A), sup(B)} is either sup(A) or sup(B). So you have a couple of natural cases to break things down into. c) I am not sure of