Question

Let E = {x element of R : -1/n <= x <= 1- (1/n), n element of N} Find the supremum and infimum and justify your answer.

Answer 1

I have that the inf(E) = -1 and sup(E) = 1, but I have no idea how to justify these answers.

Answer 2

its correct

Answer 3

I know it's correct, but I dont know how to justify this, that's my issue.

Answer 4

\(\large\tt \begin{align} \color{black}{\text{for infimum}\\~\\
-\dfrac{1}{n} \leq x \leq 1-\dfrac{1}{n}\\~\\
\text{substitute n=1}\\~\\
-\dfrac{1}{1} \leq x \leq 1-\dfrac{1}{1}\\~\\
-1 \leq x \leq 0\\~\\
x\in [-1,0]\\~\\

\text{for supremum}\\~\\
-\dfrac{1}{n} \leq x \leq 1-\dfrac{1}{n}\\~\\
\text{substitute n = }\infty\\~\\
-\dfrac{1}{\infty} \leq x \leq 1-\dfrac{1}{\infty}\\~\\
0 \leq x \leq 1\\~\\
x\in [0,1]\\~\\
} \end{align}\)

Answer 5

i dont know how to justify in words

may be u can show series for n=1,2,3,4,...

Answer 6

Yeah, it would require some sort of written justification. Like, going by one of the definitions is a good place to start, just didnt know how to continue properly. So we need this:
"\(\forall \) \(\epsilon> 0\), \(\exists \) x \(\in E\) such that \(\alpha - \epsilon < x \le \alpha \iff\)sup(\(E\)) = \(\alpha\)
And then it follows for infimum, just \(\alpha \le x < \alpha + \epsilon\iff\)inf(\(E\)) = \(\alpha\)

Answer 7

@freckles Know anything about this stuff? :P

Answer 8

Would you like to try showing the inf(E)=-1 or the sup(E)=1 first?

Answer 9

Not sure one or the other is easier. Let's start with inf(E) = -1.

Answer 10

Alright. So we claim that the inf(E)=-1. That means we need to show two things.

1) That for all \(x\in E\), \(-1\le x\), and
2) If \(l\) is a real number such that \(l\le x\) for all \(x\in E\), then \(l\le -1\).

The first part shows that \(-1\) is a lower bound of the set E, while the second part shows that -1 is the greatest lowerbound, i.e the inf(E).

Answer 11

Does this make sense? or is this not quite the definition you were given?

Answer 12

It makes sense. I've seen that before, just stated a slight bit differently, so its not unfamiliar. But what I posted above was always the statement made when doing these proofs in our notes, that was the starting point I guess.

Answer 13

When we into the details of showing 1 and 2, it will probably match more with what you posted above.

Answer 14

Alright, so lets start with 1. We need to show that: $$x\in E\Longrightarrow -1\le x.$$

Answer 15

If \(x\in E\), then: $$-\frac{1}{n}\le x\le 1-\frac{1}{n}$$for all \(n\in \mathbb N\), correct?

Answer 16

RIght, as assumed by the question :P

Answer 17

Right, so in particular, since \(1\in \mathbb N\), letting \(n=1\) tells us: $$-1\le x \le 0. $$So we already see that \(-1\le x\) like we wanted.

Answer 18

So thats the proof of part 1, any questions about that?

Answer 19

Yeah, we dont need to show anything about an infimum yet, just that -1 is a lower bound.

Answer 20

Right right. Generally, the first part is really easy to show. Its the second that takes a little work. Epsilon/delta arguments always require finesse lol.

Answer 21

So I've noticed, lol.

Answer 22

So lets start with part 2. The statement we want to prove is: $$l\le x \quad\forall x\in E\Longrightarrow l\le -1.$$

Answer 23

We are going to proceed by contradiction. In other words, we are going to assume \(l\le x\) for all \(x\in E\) and \(-1<l\).

Answer 24

Hopefully we can find a contradiction by assuming those two things.

Answer 25

Okay, so like you mentioned before, if we want to do a contradiction, we specifically want to negate the conclusion.

Answer 26

Right. Since the conclusion was "\(l\le -1\)", negating that yields, "\(-1<l\)".

Answer 27

Okay, so if l <= x, then l has to be a lower bound. From there I think id have to go to my notes, haha. We don't have to do cases, do we?

Answer 28

So it turns out this part isn't too bad (which probably means the sup will get revenge haha).

\(l\le x\) for all \(x\in E\) and \(-1<l\). Can you see a particular element of \(E\) that gives us a contradiction? Try plugging in some small values of \(n\) into \(-\frac{1}{n}\le x\le 1-\frac{1}{n}\).

Answer 29

Also, as another hint, the type of contradictions we are usually looking for with inequalities is something like, "This statement tells me \(l\le c\), but this other statement says that \(l>c\), which doesn't make any sense."

Answer 30

Okay, so l <=x, but -1 < l.....So if you just say n = 1, you get -1 <=x <=0. The supposed set (just for my own visual sake) for x is [-1,0] and the set of l <= x is (-infinity, -1]. But l > -1 is not in the possible values of l if it were to be <=x, so contradiction.

I know it doesnt need that much spelling out, but putting intervals let's me see a little.

Answer 31

Yep, thats right :) So that proves that the inf(E)=-1.

Answer 32

Okay, cool. Using the other definition made that one easier. This was on an exam I had today and I didnt know how to quite justify it. I was trying to mess with epsilons like in our notes, but I didnt know how to tweek them in the right way x_x. Okay. so other way around for sup(E).

Umm....*tries not to look*....x<= 1 for all x in E.

Answer 33

Basically, its because \(-1\in E\). So \(l\) is supposed to be a number such that \(l\le -1\), but our other assumption was that \(-1<l\), which is the contradiction you brought up.

Answer 34

Yes, we need to first show that: $$x\in E\Longrightarrow x\le 1.$$

Answer 35

Are we allowed to make a limit argument for that one? lim n-> inf 1- (1/n) = 1?

Answer 36

Using a limit argument would really be useful, but a little more needs to be said.

Answer 37

Actually, now that I think about it, a limit argument would be really really good for the 2nd part.

Answer 38

1-(1/n) is increasing for all n, so taking it's limit would give a maximum value. Don't know how in-depth I would have to go on the increasing argument to say that, though.

Answer 39

That should be good enough. Its not just that \(\lim_{n\rightarrow \infty} 1-\frac{1}{n}=1\), but that is it strictly increasing to 1.

Answer 40

So basically what you are saying is: $$1-\frac{1}{n}\le 1$$for all \(n\in \mathbb N\).

Answer 41

In the end, yep, lol. And since x <= 1-(1/n), x <=1 for all n in N. So that seems to take care of that. So....we want to show that there is no value lower than 1 that is an upper bound. Was trying to think of how it would work the other way now, but I have to take a look at what we stated before and think xD

Answer 42

Oh, okay, any lower bound had to be less than -1, so any upper bound has to be greater than 1.

Answer 43

If 1 is sup(E) that is.

Answer 44

Right. The exact statement we want to prove is: $$x\le u\quad \forall x\in E\Longrightarrow 1\le u.$$

Answer 45

So another contradiction?

Answer 46

Yep. Can you say what the two things we are going to assume are?

Answer 47

So

Answer 48

Oops, no idea how that sent, lol

Answer 49

lol

Answer 50

So we'll say that x <= u implies u <=1. Looking at the direction of the inequalities didnt help me there, I just need to understand it better. I mean I get it, I just need to get it well enough where it can be more intuitive. So....if 1 is sup(E), any other upper bound must be greater than 1, so the contradiction is that theres an upper bound less than one. SO translating that to how we were writing it, x <= u implies u<=1. Okay, I think thats right then.

Answer 51

You have the right idea, but aren't saying it correctly. We want to assume that \(x\le u\) and that \(u<1\). Don't forget that negating \(1\le u\) means you need to take off the "equals to" part of the inequality.

Answer 52

You are definitely on the right path though :) Now we need to find a contradiction using these two facts.

Answer 53

Oh, you did remove the equal to part above xD Gotcha. So all I need is a single contradicition? So saying n = 1 means x <=1, but since u>=x, it cannot be <1 and that's a contradiction.

Answer 54

I think trying to comprehend what kind of argument im making and need to make is making me forget stupid things like needing to drop the equal to part.

Answer 55

Hmm... i think there is something off about what you are saying. If we let \(n=1\), then we get the interval [-1,0]. I don't see how a contradiction follows from that.

Answer 56

Oh, youre right. Im getting thing mixed up in my head, actually seeing [-1,0] makes me see why. So wait, do I need the exact sort of condition as last time to get the contradiction? Like, a situation where x <=1, but u>=x, so u cant be less than 1? Even though I didnt show that, is that what the eventual contradiction would be, or is there something else?

Answer 57

This contradiction is going to be a little harder to find. What made the inf(E) easy was that -1, the number we were claiming to be the inf(E), was actually in the set already. We aren't so lucky this time with 1.

Answer 58

However, you brought up that: $$\lim _{n\rightarrow \infty }1-\frac{1}{n}=1.$$This is going to be our strongest weapon.

Answer 59

Let me try and draw a horrible picture to spark some intuition.

Answer 60

Go for it, lol.

Answer 61

So lets assume the contradiction hypothesis, that \(u<1\).

Answer 62

Since: $$\lim _{n\rightarrow \infty} 1-\frac{1}{n}=1, $$ I should be able to get AS CLOSE TO 1 AS I WANT TO. If i mark little ticks for the elements in E, i should see:

Answer 63

(Ugh my ticks are horrible)

Answer 64

So we have to somehow show



Or something like that?