Question

Proofs question

Answer 1

If sup A < sup B, show that there exists an element b ∈ B that is an upper bound for A.

Used contradiction:

Answer 2

Suppose that is no such element b ∈ B that is an upper bound for A.
This means that for any b ∈ B, ∃ an element a ∈ A s.t. a > b.
In other words, for every element b ∈ B, we can find an element a ∈ A that is greater than b.

By definition of Supremum, sup A is an upper bound for A, and is the least such upper bound.
However, since for every b∈ B we can find an element a ∈ A s.t. a > b, this means that sup A cannot be less than or equal to any element B.

But we are given that sup A < sup B, which means that sup A is strictly less than sup B.
Contradicting that sup A is an upper bound for B.
Thus, our initial assumption that there is no element b∈ B that is an upper bound for A must be incorrect.
Implying that there exists at least one element b∈ B that is an upper bound for A.

Answer 3

@jhonyy9 @surjithayer

Answer 4

The intuition behind the proof is this. The strict inequality first of all is very important. The result need not be true if the condition is supA≤supB
. If supA<supB
what you need to see is there are real numbers (uncountably many in fact) between supA
and supB
. Why because r=(supB−supA)>0
and forms an interval on the line. Think what happens if none of these real numbers are in B
. Then none of the elements that are greater than supA
are in B
. That is there are no elements in B
which are greater than supA
. Then supA
is an upper bound for B
. Can this be possible. NO!! Why? Since supB
is the least upper bound of B
and supA
is a value less than it. This is the story in layman's terms.

I should add what the above contradiction proves. It says there are numbers in B
which are greater than supA
and hence are an upper bound for A
.

Here is a Lemma that I have always found useful in understanding the intuition behind suprema. u=supA⟺x∈A⟹x≤u
and b<u⟹∃ x′∈A
such that b<x′
. This particular Lemma makes a lot of proofs simple.

A direct application for your question states that ∃b∈B
such that b>supA
.I HOPE THIS HELP YOU

Answer 5

See what i said to chic and trendy

Answer 6

imagine taking analysis kek