Question

Help finish the stuff, please.
I don't know how to get the final answer.

Answer 1

I need conclusion of \(s \geq U_A-I_B\) to finish the proof

Answer 2

https://www.math.ucdavis.edu/~hunter/m125b/ch2.pdf
what you did is similar to what these guys did

Answer 3

does this help
\[U_A-I_B-\epsilon <x-y \\U_A-I_B<x-y+\epsilon \\ \]
If k is upper bound for A-B, then x-y<=k
So we have
\[U_A-I_B<k+\epsilon \\ U_A-I_B \le k\]

Answer 4

\[\text{ Suppose } U_A \text{ is } \sup(A) \\ ,I_A \text{ is } \inf(A) \\,U_B \text{ is } \sup(B) \\ \text{ , and } I_B \text{ is } \inf(B) .\\ \text{ Therefore we have the following things: } \\ U_A \ge x \text{ for all } x \in A \\I_A \le \text{ for all } x \in A \\U_B \ge y \text{ for all } y \in B \\ I_B \le y \text{ for all } y \in B \\ \text{ We only need to use the parts with } U_A \text{ and } I_B . \\ I_B \le y \text{ \implies } -I_B \ge -y \\ U_A+(-I_B) \ge x+(-y)=x-y \]

Answer 5

All of this sounds really good to me.

Answer 6

The second part is what we are having issues with.

Answer 7

yes,

Answer 8

\[\text{ Let } S=\sup(A-B) \\ \text{ But we also have or just found that } U_A-I_B \text{ is an upper found for the set } A-B \\ \text{ since we showed } U_A-I_B \ge x-y \text{ for all } x \in A , y \in B \\ \text{ Since S is a least upper bound , this means one of the two things , that } \\ U_A-I_B=S \text{ or } U_A-I_B > S \\ \text{ or we can just say } U_A-I_B \ge S \]

Answer 9

How do you feel about this part?

Answer 10

My second line got cut off A-B is suppose to be there

Answer 11

does it repeat part a?

Answer 12

I have might have said some things more than once.
I'm just trying to draw a conclusion from the part a to continue into our part b

Answer 13

We want to show now that \[U_A-I_B \le S \\ \text{ In doing this we will be able to conclude } U_A-I_B=S\]

\[\text{ So let } \epsilon>0 \text{ then there exists } x \in A \text{ such } \\ U_A-\frac{\epsilon}{2}<x \]
I believe this part is true, there should be a number x such you can take away a certain positive number away from the upper bound such that is true.

Answer 14

Agree

Answer 15

\[\text{ there exists } y \in B \text{ such that } I_B-\frac{\epsilon}{2} >y \]same thing for this statement

Answer 16

\[\text{ which implies } \frac{\epsilon}{2}-I_B<-y \]

Answer 17

no, for infimum, we must + epsi, not - eps

Answer 18

yeah that was typed wrong

Answer 19

\[\text{ there exists } y \in B \text{ such that } y \le I_B+\frac{\epsilon}{2} \text { which implies } -y \ge -I_B-\frac{\epsilon}{2}\]

Answer 20

that should be better

Answer 21

Agree

Answer 22

\[x-y=x+(-y) \ge U_A-\frac{\epsilon}{2}-I_B-\frac{\epsilon}{2} \\x-y \ge U_A-I_B-\epsilon \]

Answer 23

I just added those two inequalities together

Answer 24

yup

Answer 25

\[x-y+\epsilon \ge U_A-I_B \]
added epsilon both sides
x-y is an element of A-B
S=sup(A-B)
if we let there exist k such that k is also an upper bound for A-B
\[\text{ so we have } k \ge x-y \text { for all } x \in A \text{ and for all } y \in B \]\[k \ge x-y \\ k+\epsilon \ge x-y +\epsilon \\ k+\epsilon \ge U_A-I_B \\\]

Answer 26

Now I think you were saying something like
suppose a,b,c are positive.
If we have a+b>=c where a>=b, then a>=c.
And the only reason I knew in the a>=b part is because k should bigger than epsilon since is the upper bound for {(x-y)}

Is tht the part we are stuck at?

Answer 27

I wonder if we can prove that statement.

Answer 28

I am waiting. :)

Answer 29

5+3>=7
but 5>=7 not true
so let me think maybe i'm missing something

Answer 30

I forgot why k was greater than or equal to U_A-I_B lol

Answer 31

hahaha.... U are funny!!

Answer 32

but i know it should be

Answer 33

somehow

Answer 34

too bad you can't just put that in your proof
like well somehow we know this is true :p

Answer 35

like i was thinking epsilon would have to be a positive positive number
so small it doesn't if matter if we had the epsilon there

Answer 36

My prof is picky, so that I would like to have a flawless proof to not get his rejection. hihihi

Answer 37

\[x>U_A-\frac{\epsilon }{2} \text{ and } -y > -I_B-\frac{\epsilon}{2} \\ x-y=x+(-y) > U_A-I_B-\epsilon \\ x-y+\epsilon > U_A-I_B \\ k \ge x-y \\ k+\epsilon > U_A-I_B \\ k \ge U_A-I_B \]
I wonder if it makes more sense if we have chosen to use > instead of ge to begin with

Answer 38

anyways I think I have wasted plenty of your time

Answer 39

backward!! I do appreciate what you did for me. I waste your time. I owe you. :)

Answer 40

omg no whatever i wasted your time

Answer 41

i thought i knew what i was talking about at the beginning
then i don't

Answer 42

is k>=x-y+epsilon for all x and y if k is an upperbound for {(x-y)}

Answer 43

where epsilon >=0

Answer 44

I think that is true
if so we are done

Answer 45

\[k \ge x-y+ \epsilon \ge U_A-I_B \]

Answer 46

x-y is an arbitrary element of (A-B) , let define P = A -B
so that if x -y is sup P , then x -y+ep > sup P

Answer 47

and we are stuck at that point.
x -y+ ep > sup P
x -y+ep > sup A -Inf (B)
no way to compare sup P and sup A- Inf (B)

Answer 48

i give up

Answer 49

me too. hihihi.. anyway, thanks a ton.