Question

Let X be a set of real numbers with least upper bound a. Prove that if epsilon > 0, then there exists an x in X, such that (a - epsilon) < x <= a.

Answer 1

work directly from the definition of least upper bound
that is the point of the question
if what you had was not true (so i guess you are working by contradiction) then \(a\) would not be a least upper bound

Answer 2

in other words, if there was no such \(x\) with \(a-\epsilon<x\leq a\) then \(a\) would not be a LEAST upper bound

Answer 3

if that is not clear, let me know

Answer 4

Yeah, I guess I'm having trouble where to get started. For proof by contradiction would I assume that epsilon is less than 0? Or would I assume x >= a. I'm still kind of processing the definition of LUB.

Answer 5

*trouble understanding where

Answer 6

I don't quite understand where epsilon comes in.

Answer 7

ok i can see why this would be confusing so lets go slow

Answer 8

the definition of least upper bound, at least i am assuming this is the definition you are using is this:
\(a\) is a least upper bound of a set of real numbers \(X\) if
i) for every \(x\in X, x\le

Answer 9

oops

Answer 10

i) for every \(x\in X, x\leq a\) and
ii) if \(k\) is another upper bound for \(X\) then \(k\geq a\)
is that what you are using?

Answer 11

yes, those definitions are in my textbook

Answer 12

ok good

Answer 13

so lets work by contradiction and suppose that the statement above is not true, i.e it is not true that given any \(\epsilon>0\) there exists an \(x\in X\) such that \(a-\epsilon <x\leq a\)

Answer 14

do you know how to negate that statement? what it means for it not to be true?
"no" is a fine answer, just asking

Answer 15

yes, wouldn't it be for all x in X, ( a- epsilon < x <= a) is not true and then you'd just flip the inequalities?

Answer 16

oh no
so lets take this part and make it clear

Answer 17

oh lol, sorry. It's been awhile since my first proof class.

Answer 18

no problem

Answer 19

to negate that statement you do not flip any inequalities
suppose i said "for all houses in this neighbourhood, there is a room that is painted red"
what would we do to show that it is not true?
answer: find SOME house with NO red room

Answer 20

the negation of the statement "for every \(\epsilon>0\) there is some \(x\in X\) such that is
There is some \(\epsilon >0\) such that there is NO \(x\in X\) with \(a-\epsilon <x\leq a\)

Answer 21

let me know if that is clear nor not

Answer 22

okay, got it so far.

Answer 23

ok so lets assume by contradtction that the statement is not true, i.e. there is some \(\epsilon>0\) with no \(x\in X\) satisfying \(a-\epsilon<x\leq a\)
what does that mean about \(a-\spsilon\)?

Answer 24

oops about \(a-\epsilon\)?

Answer 25

it means
i) \(a-\epsilon \) is an upper bound for \(X\) since there are no \(x\)'s larger than it

Answer 26

and also since \(a-\epsilon <a\) that means \(a\) is NOT the least upper bound

Answer 27

i.e. it contradicts the fact that \(a\) was a least upper bound since we have another upper bound that is less than it

Answer 28

which would be a contradiction because the problem states that the LUB for X is a?

Answer 29

yes, exactly

Answer 30

we have found another upper bound, namely \(a-\epsilon\) that is clearly less than \(a\)

Answer 31

oh okay. and that's where that 2 condition of LUB comes in?

Answer 32

oh never mind. I'm referring to my textbook

Answer 33

yes, \(a\) was not only supposed to be the upper bound but also if \(k\) is some other upper bound then \(k\geq a\)

Answer 34

i hope the logic of that is more or less clear

Answer 35

Yeah, the logic is completely clear. The most difficult part is really understanding the definition of LUB and the role epsilon plays even though I know what a LUB is intuitively

Answer 36

thanks for your help :)

Answer 37

yeah don't get messed up by that greek letter, it just means some positive number

Answer 38

you're welcome