Question

In: Advanced Math

Suppose that a subset S of an ordered field F is not bounded above in F....

Suppose that a subset S of an ordered field F is not bounded above in F. Let T be a subset of F satisfying the property that, for each xS, there exists yT such that xy. Prove that T is not bounded above in F.

Solutions

Expert Solution

We will prove by contradiction that is not bounded above in   .

Suppose    is bounded above in   then there exist such that ,

for all   .

Let ,   be arbitrary .

By the given condition ,   as   .

Since   and   are subset of order field   and   ,  

Since    be arbitrary so   for all

Hence   is bounded above , a contradiction to   is not bounded above .

Hence   is not bounded above in   .

.

.

.

.

If you have doubt or need more clarification at any step please comment and don't forget to rate the answer . Your rating keeps us motivated .


Related Solutions

Let (F, <) be an ordered field, let S be a nonempty subset of F, let...
Let (F, <) be an ordered field, let S be a nonempty subset of F, let c ∈ F, and for purposes of this problem let cS = {cx | x ∈ S}. (Do not use this notation outside this problem without defining what you mean by the notation.) Assume that c > 0. (i) Show that an element b ∈ F is an upper bound for S if and only if cb is an upper bound for cS. (ii)...
Show that if S is bounded above and below, then there exists a number N >...
Show that if S is bounded above and below, then there exists a number N > 0 for which - N < or equal to x < or equal to N if x is in S
Which of the following statements are correct? a) If A is a bounded subset of the...
Which of the following statements are correct? a) If A is a bounded subset of the real line, every infinite subset of A has a limit point. b) If A is a bounded subset of the real line, every open cover of A has a finite subcover. c) If A is an infinite open subset of the real line, there is an infinite open cover with a finite subcover. d) If A is a closed subset of the real line,...
Let F be an ordered field. We say that F has the Cauchy Completeness Property if...
Let F be an ordered field. We say that F has the Cauchy Completeness Property if every Cauchy sequence in F converges in F. Prove that the Cauchy Completeness Property and the Archimedean Property imply the Least Upper Bound Property. Recall: Least Upper Bound Property: Let F be an ordered field. F has the Least Upper Bound Property if every nonempty subset of F that is bounded above has a least upper bound.​
Let F be an ordered field. We say that F has the Cauchy Completeness Property if...
Let F be an ordered field. We say that F has the Cauchy Completeness Property if every Cauchy sequence in F converges in F. Prove that the Cauchy Completeness Property and the Archimedean Property imply the Least Upper Bound Property. Recall: Least Upper Bound Property: Let F be an ordered field. F has the Least Upper Bound Property if every nonempty subset of F that is bounded above has a least upper bound.​​
Prove that if f is a bounded function on a bounded interval [a,b] and f is...
Prove that if f is a bounded function on a bounded interval [a,b] and f is continuous except at finitely many points in [a,b], then f is integrable on [a,b]. Hint: Use interval additivity, and an induction argument on the number of discontinuities.
If a set K that is a subset of the real numbers is closed and bounded,...
If a set K that is a subset of the real numbers is closed and bounded, then it is compact.
a) Suppose f:R → R is differentiable on R. Prove that if f ' is bounded...
a) Suppose f:R → R is differentiable on R. Prove that if f ' is bounded on R then f is uniformly continuous on R. b) Show that g(x) = (sin(x4))/(1 + x2) is uniformly continuous on R. c) Show that the derivative g'(x) is not bounded on R.
Suppose f : X → S and F ⊆ P(S). Show, f −1 (∪A∈F A) =...
Suppose f : X → S and F ⊆ P(S). Show, f −1 (∪A∈F A) = ∪A∈F f −1 (A) f −1 (∩A∈F A) = ∩A∈F f −1 (A) Show, if A, B ⊆ X, then f(A ∩ B) ⊆ f(A) ∩ f(B). Give an example, if possible, where strict inclusion holds. Show, if C ⊆ X, then f −1 (f(C)) ⊇ C. Give an example, if possible, where strict inclusion holds.
Suppose f maps a closed and bounded set D to the reals is continuous. Then the...
Suppose f maps a closed and bounded set D to the reals is continuous. Then the Uniform Continuity Theorem says f is uniformly continuous on D. To prove this, we will suppose it is not true, and arrive at a contradiction. So, suppose f is not uniformly continuous on D. If f is not uniformly continuous on D, then there exists epsilon greater than zero and sequences (a_n) and (b_n) in D for which |a_n -b_n| < 1/n and |f(a_n)...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT