Question

Let S be defined as S = {x|x^2 < x}. I'm supposed to show that sup(S) = 1. Now I've shown that 1 is an upper bound, but I'm not sure how to show that it's the least upper bound. It makes intuitive sense, but I'm having trouble finishing it off.

Answer 1

let \(1-\epsilon\) be an upper bound of \(S\) for real \(0<\epsilon<1\). it follows then that \((1-\epsilon)^2\ge1-\epsilon\) but observe:$$(1-\epsilon)^2=(1-\epsilon)+(\epsilon^2-\epsilon)$$but we have \(\epsilon^2<\epsilon\) hence \(\epsilon^2-\epsilon<0\) and thus \((1-\epsilon)^2<(1-\epsilon)\) i.e. \((1-\epsilon)\in S\) etc. you should be able to do something like this

Answer 2

ignore the 'it follows then ...' that's not necessarily true but the idea is that you can pick always pick a smaller epsilon so none can be upper bounds

Answer 3

Ah, I see now. @satellite73 was helping me with this yesterday, but I couldn't get that middle step where you separated the (1-epsilon)^2. Thanks a lot!

Answer 4

i think the mechanics are easy, what is hard is to figure out exactly what you have to cprove

Answer 5

@jabberwock well consider \((1-\epsilon)^2=(1-\epsilon)(1-\epsilon)=(1-\epsilon)-\epsilon(1-\epsilon)=(1-\epsilon)+(\epsilon^2-\epsilon)\)

Answer 6

the definition is one thing, but you don't show it by using the definition
i think that is the point of the exercise

Answer 7

write down exactly what it is you have to show
i mean precisely

Answer 8

@oldrin.bataku Oh, I understood the step, but I was just saying that I couldn't find it on my own.

Answer 9

the definition says \( q\) is a lub for \(S\) if
1) \(q\) is an upper bound and
2) if \(r\) is also an upper bound, then \(r\geq q\)

Answer 10

@satellite73 Yeah, I took a real analysis class years, ago, but it's frustrating because I get hung up on chapter 1 stuff.

Answer 11

but in practical value, you show this by one of two equivalent definitions

Answer 12

Yeah, I didn't have any problem showing that it was AN upper bound. Just the second part was throwing me.

Answer 13

a) if \(r\in \mathbb{R}\) and \(r<q,\) then there exists \(a\) such that \(a\in S\) and \(r<a\)

Answer 14

or more likely this one
b) for all \(\epsilon >0\) there exists \(a\in S\) such that \(q-\epsilon<a\)

Answer 15

Yeah, I had tried manipulating that for a while too, but I couldn't figure out what to do with it before oldrin's explanation.

Answer 16

if you write down exactly what this means in your example, it means for all \(\epsilon>1\) there is an \(a:a^2-a<1\) and \(1-\epsilon <a\)

Answer 17

oops i mean \(a^2-a<0\) sorry

Answer 18

in fact in this example you can probably pick \(a=1-\frac{\epsilon}{2}\)

Answer 19

mechanics should be ok
if \(\epsilon>0\) you have \((1-\epsilon )(1-(1-\epsilon))=(1-\epsilon)(-\epsilon)<0\) as \(1-\epsilon >0\) and \(-\epsilon<0\)

Answer 20

now if you replace \(\epsilon\) by \(\epsilon/2\) you will also get a true inequality, meaning that since \(1-\epsilon \in S\) than also \(1-\epsilon/2\in S\)

Answer 21

*then

Answer 22

I hadn't considered the epsilon/2 idea to get a strict inequality. That's a good idea.

Answer 23

again i think the mechanics, i.e. the algebra, should be straight forward (not always, usually that is the hard part) but for a problem like this, where it is completely obvious that 1 is the lub, the hard part is writing down precisely what you have to show in general, and then translating it to the specific example you have
that is usually the hard part in any introductory problem working from a definition

Answer 24

Agreed. I kept getting some end results like this one (calling epsilon the number itself):
\[\epsilon^2>x \epsilon\]that I thought would be useful, but I just couldn't figure out how to make it work without assuming what I was trying to prove.

Answer 25

btw if you get stuck with the algebra, try it with a number
for example, why isn't \(.9\) the lub of \(x^2<x\)?
it is certainly true that \(.9^2-.9=.9(1-.9)<0\) even without computing, but here you see that \(.95(1-.95)<0\) as well
in this case my \(\epsilon\) was \(.1\) and \(\epsilon/2=.05\)

Answer 26

lol i had that backwards!

Answer 27

i meant "it is certainly true that \(.9^2-.9=.9(.9-1)<0\) etc

Answer 28

Yeah, I didn't have a problem seeing why that result I came up with should be true, and I decided to try substituting epsilon into x to get
\[\epsilon^2>\epsilon^2\]and then realizing that I was assuming that epsilon had to be a member of S, which would be true for anything smaller than 1, but again the problem came down to proving it.

Answer 29

Sorry to keep bugging you with this, but I was trying this, and I'm not sure it's alright to assume that \[1-\frac{\epsilon}{2}\in S\]or that 1-epsilon is in S either.