Question

Suppose S ⊂ R is nonempty and M is an upper bound for S. Show M
= sup S if and

only if for every Ɛ > 0, there exists x ∈ S so that x > M −
Ɛ.

Answer #1

Suppose S and T are nonempty sets of real numbers such that for
each x ∈ s and y ∈ T we have x<y.
a) Prove that sup S and int T exist
b) Let M = sup S and N= inf T. Prove that M<=N

Real Analysis I
Prove the following exercises (show all your work)-
Exercise 1.1.1: Prove part (iii) of Proposition
1.1.8. That is, let F be an ordered field and x, y,z ∈ F. Prove If
x < 0 and y < z, then xy > xz.
Let F be an ordered field and x, y,z,w ∈ F. Then:
If x < 0 and y < z, then xy > xz.
Exercise 1.1.5: Let S be an ordered set. Let A
⊂...

Suppose A ⊆ R is nonempty and bounded above and β ∈ R. Let A + β
= {a + β : a ∈ A}
Prove that A + β has a supremum and sup(A + β) = sup(A) + β.

Show that every nonempty subset of the real numbers with a lower
bound has a greatest lower bound.

Suppose A is a subset of R (real numbers) sucks that both infA
and supA exists. Define -A={-a: a in A}.
Prive that:
A. inf(-A) and sup(-A) exist
B. inf(-A)= -supA and sup(-A)= -infA
NOTE:
supA=u defined by: (u is least upper bound of A) for all x in A,
x <= u, AND if u' is an upper bound of A, then u <= u'
infA=v defined by: (v is greatest lower bound of A) for all y in...

Suppose ? ⊂ R^? , ? ⊂ R^? are nonempty and open and ? : ? → R^?
and ? : ? → R^? . Let ℎ : ? × ? → R ?+? be defined by ℎ(u, v) =
(?(u), ?(v)). If ? is continuous at x ∈ ? and ? is continuous at y
∈ ? , then show that ℎ is continuous at (x, y) ∈ ? × ? .
Hint: Note that for any vectors z...

Let S and T be nonempty subsets of R with the following
property: s ≤ t for all s ∈ S and t ∈ T.
(a) Show that S is bounded above and T is bounded below.
(b) Prove supS ≤ inf T .
(c) Given an example of such sets S and T where S ∩ T is
nonempty.
(d) Give an example of sets S and T where supS = infT and S ∩T
is the empty set....

Using the completeness axiom, show that every nonempty set E of
real numbers that is bounded below has a greatest lower bound
(i.e., inf E exists and is a real number).

Let A ={1-1/n | n is a natural number}
Prove that 0 is a lower bound and 1 is an upper
bound: Start by taking x in A. Then x = 1-1/n
for some natural number n. Starting from the fact that 0 <
1/n < 1 do some algebra and arithmetic to get to 0 < 1-1/n
<1.
Prove that lub(A) = 1: Suppose that
r is another upper bound. Then wts that r<= 1.
Suppose not. Then r<1. So 1-r>0....

Let S be a nonempty set in Rn, and its support
function be
σS = sup{ <x,z> : z ∈ S}.
let conv(S) denote the convex hull of S. Show that σS
(x)= σconv(S) (x), for all x ∈ Rn

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