Question

Can anyone pleaseeee helppp

Let A,B⊆R be nonempty. Define A−B={a−b|a∈A,b∈B}. Prove that inf(A−B)=inf(A)−sup(B

Answer 1

Have to prove inf(A-B)=inf(A)+inf(-B)

Answer 2

Then have to show inf(-B)=-sup(B)

Answer 3

Just not sure how

Answer 4

@wio

Answer 5

\(\color{blue}{\text{Originally Posted by}}\) @cbrezovs
Then have to show inf(-B)=-sup(B)
\(\color{blue}{\text{End of Quote}}\)
If this is true, you'd have to go back to definitions to show it.

Answer 6

Do u know how to do the problem a diff way?

Answer 7

I don't remember all the properties of inf and sup.

Answer 8

inf(A-B)=inf(A)+inf(-B)

Do you know this is true, or is it intuitive?

Answer 9

I believe its true @Zarkon was helping me, he said he knew how to prove it and he told me that was true

Answer 10

Let's consider the definition.

Let \(X\) be defined as \[
\{x| \forall b\in B\quad x<b\}
\]That is, all elements below \(B\).\[
x=\inf (B)\iff x\in X\wedge \forall y \in X\quad x\geq y
\]

Answer 11

Here is my attempt at an algebraic definition. It is something we can work with.

Answer 12

Ok cool

Answer 13

So we can say \(x_A =\inf (A)\) and \(x_{B^-}=\inf(-B)\)

Answer 14

Using my definition:\[
\{x| \forall a\in A\quad x<a\}
\]\[
x_A=\inf (A)\iff x_A\in X\wedge \forall x \in X\quad x_A\geq x
\]And \[
\{x| \forall b\in B\quad x<-b\}
\]\[
x_{B^-}=\inf (-B)\iff x_{B^-}\in X\wedge \forall x \in X\quad x_{B^-}\geq x
\]

And we need to use these facts to show that:
\[
\{x| \forall c\in A-B\quad x<c\}
\]\[
x_A+x_{B^-}=\inf (A-B)\iff x_A+x_{B^-}\in X\wedge \forall x \in X\quad x_A+x_{B^-}\geq x
\]

Answer 15

We could start by saying that \(c=a-b\)

Answer 16

Hmmm, I should distinguish those \(X\) sets as well.

Answer 17

This is a bit messy though, let's start off with something simpler.

Answer 18

This definition is less messy. We just need to proof stuff about max and lower bounds.\[
X=\{x|\forall a\in A\implies x<a\}\quad x_A=\inf(A)\iff x_A=\max (X)
\]

Answer 19

I think we can show:\[
\max(A-B) =\max(A)+\max(-B)
\]By starting with:\[
m_a=\max(A)\iff \forall a\in A\quad m_a\geq a\\
m_b=\max(-B)\iff \forall b'\in -B\quad m_b\geq b'
\]First I think we just let \(b'=-b\) and get:\[
\forall b\in B\quad m_b\geq -b
\]Then we add these inequalities:\[

\forall a\in A,\;\forall b\in B\quad m_a+m_b\geq a-b
\]Whis simplifies to: \[
\forall c\in A-B\quad m_a+m_b\geq c\iff m_a+m_b=\max(A-B)
\]

Answer 20

Is that an m? I'm not sure what that notation is in the last line

Answer 21

Yes, I use \(m_x\) to be like the max of \(X\) sort of thing.

Answer 22

I was just trying to show that if you add the max together, you have the max of the sum.

Answer 23

Let's just get this out of the way \(a\in A,b\in B,a-b=c\in A-B\)
I don't like having to type those out all the time.

Answer 24

Let \(
X_A, X_B, X_C
\) denote the lower bound of \(A, B, A-B\) respectively.

Answer 25

You have to show \(x_A=\inf(A),x_B=\inf(-B), x_A+x_B\in X_C\)

Answer 26

The concept is not hard, but it's just such a tedious proof, unless you are allowed to assume certain properties.

Answer 27

This is a hint I was g iven use the definition of convergence for each of the sequences, and for convergence of (zn) consider the two cases for n even and odd

Answer 28

I have to head to bed right now. I appreciate all your help so far.. you've been awesome.So I'm gunna just give u the medal now..maybe we can work on it more tomorrow