(Introductory Real Analysis) I'm trying to prove the following theorem:

- Mendicant_Bias

(Introductory Real Analysis) I'm trying to prove the following theorem:

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- Mendicant_Bias

\[\text{Theorem 1.1.8) - If a nonempty set S of real numbers is bounded below, then} \\ \text{inf{S} is a real unique number such that:}\]
\[\text{a}: \ x /\geq \alpha \ \text{for all} \ x \text{ in S;}\]\[\text{b}: \ \text{if} \ \epsilon > 0, \ \text{(no matter how small), there is an} \ x_0 \text{in S such that} \ x_<\alpha+\epsilon \]

- Mendicant_Bias

The prompt suggests applying Theorem 1.1.3, which is: http://i.imgur.com/w3awxCe.png
Along with property (I) of the field axioms regarding a field being ordered/complete, which is: http://i.imgur.com/VgX6bDE.png
Finally, it gives you the following set: http://i.imgur.com/EB7qdCv.png and says to apply Thm 1.1.3 and property (I) to the set to prove 1.1.8.
I can't see how at all, though, the set it gives you is bounded from above or beneath.

- Mendicant_Bias

At least not by a real number. I can see it bounded beneath by -infty, and if I understand correctly, bounded above by (zero?), but I'm not really sure. The hint it gives is that the set provided is bounded above is S is bounded below.

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## More answers

- Mendicant_Bias

@ganeshie8

- zzr0ck3r

Do you only have properties about supremum?

- Mendicant_Bias

I'll just take a picture of the prompt itself: http://i.imgur.com/2t9ecac.png

- Mendicant_Bias

(I): http://i.imgur.com/VgX6bDE.png
Theorem 1.1.3.) http://i.imgur.com/w3awxCe.png

- zzr0ck3r

I assume you mean, \(\alpha=\inf(S)\).
Then \(-\alpha\) is an upper bound for \(T\) and as a result we know that \(T\) has a supremum.
Now we will show that \(-\alpha\) is the supremum.
Suppose that it is not, then there is some supremum, of \(T\), \(\beta<-\alpha\) because \(-\alpha\) is an u.b. Note that \(-\beta\) is a lower bound for \(S=-T\) and \(-\beta> \alpha\). Yikes, we can't have a lower bound that is greater than the infimum. So it must be that \(-\alpha\) is the supremum of \(T=-S\).
(Now this all seemed very obvious, but those hints made me think that this is the rout they wanted you to take).
Now, if \(\exists \ x\in S\) s.t. \(x\le \alpha\) then \(-\alpha \le -x\) and \(-x\in T\) and this can't happen; we can't have an element in \(T\) that is greater than \(\sup(T)\).

- Mendicant_Bias

So, I'm having trouble even understanding what that set (I know, it's simple, I'm just bad at reading math notation) looks like graphically or on a numberline. The expression \[T=\{x|-x \in S\}\]
Does this mean that the values of x that exist in the subset S are only the values of x that are negative, e.g. -infty, all the way up to but not including zero, or is this saying that S is just a negative version of the real number line where an input of x=3 gives x=-3 in the set S, and an input of x = -3 gives x=3 in the set S?

- zzr0ck3r

It is just that you have a bunch of things you have a bunch of stuff about supremum that you have learned and now they are trying to to show you that the same things (but opposite) hold for infimum
Ok here is what is going on with that set.
Consider the set \(\{1,2,3,4,5\}\), then obviously \(5\) is the supremum.
Now let us make a "\(T\)" version of this
\(\{-5,-4,-3,-2,-1\}\) now it is obvious that \(-5\) is the infimum.
So, on the real line, the supremum of X is the infimum of -X.

- Mendicant_Bias

I'm just trying to determine how to graphically imagine that set, first and foremost. It's not like another where you have a set consisting of x, a formula for x where n is an input; It's a set where x represents all of its elements, and the "formula" for the set is just the negative of the thing that represents all of its elements. How do I interpret that? Is this half of the number line, basically-the negative half? Or is it the whole number line, just with all of its values being negative of what they are originally?

- zzr0ck3r

This set is not an arbitrary set, it must be a set in the real line or some other metric space.
We are literally just taking the negative versions of all the numbers inside.

- Mendicant_Bias

Okay, thanks for clarifying that, that's what I'm first and foremost looking for, so the set S consists of the following inputs: (-2,-1,0,1,2, ...) corresponding to the following outputs: (2,1,0,-1,-2,...). Now that that's out of the way.

- zzr0ck3r

|dw:1441571930772:dw|

- Mendicant_Bias

Jesus. That picture is confusing in comparison to what you just said, given half the number line is missing. So yeah, all of the values switch, but you still have the entirety of the number line, right? The whole thing. Not just half. Not just the side with two and five on it originally, like it appears in the picture. Just, final word, and then moving on without introducing anymore confusion.

- zzr0ck3r

nah, I am only talking about the numbers between [ ]
I am just saying that [2,5] becomes [-5,-2]

- zzr0ck3r

-[a,b]=[-b,-a]

- Mendicant_Bias

Okay, so then let's talk about all the numbers between (-infty,infty). That entire....yeah, everything in between switches, and EVERYTHING is in between, right? Cool.

- zzr0ck3r

ok but you get the point, make all the number their opposite, and things that were at the top are now at the bottom.

- zzr0ck3r

yes, but \(-\mathbb{R}=\mathbb{R}\)

- zzr0ck3r

\[-(-\infty, \infty)=(-\infty, \infty)\]

- zzr0ck3r

If you take the negative versions of all the real numbers, you just get back all of the real numbers.

- zzr0ck3r

There is no point in doing this unless you are bounded above or below or both..

- Mendicant_Bias

Alright. And since the set doesn't have a "different" lettered input, like say we had \[S=\{x|x=-n, n \in R\}\], there isn't any meaning to it.
I guess the nail in the coffin to my confusion on this is whether having a clearly different input, in which n=1 maps to x = -1, is significant.
Because in our case, the numbers flipping is arbitrary; the input is the same as the output, it's not putting it *through* an operation or anything, it's just already negative or positive.
So in our case, x=-3 corresponds to -3. While in the other case, n=-3 corresponds to 3, is that right?
There's a difference in the way the two sets are counted.

- zzr0ck3r

correct, you can think of it as a mapping, but you are just flipping it

- Mendicant_Bias

I'm confused by the variable constituting your input and output being the same. For the set x|-x in S or something like that, I'd just have to imagine it as the line y=x if graphed.
So if I graphed the first set, the problem prompt, would I get the line y=-x, or y=x?

- zzr0ck3r

I would write the set you suggested as \[\{-x\mid x\in \mathbb{R}\}\]

- zzr0ck3r

you dont graph sets.

- Mendicant_Bias

Why not?

- zzr0ck3r

um, you graph relations

- zzr0ck3r

do you mean fill in the real line?

- Mendicant_Bias

Sometimes your set is defined by a relation, where you have an input and an output and it makes sense to graph it. Right?

- zzr0ck3r

I think you are making this more complicated. Stop for one sec and consider this question
Say I take a cup and fill it with water and put a lid on it. All of the water in the cup is below that lid (the lid is the supremum)
Now if we flip the cup upside down, all of the water is above the lid (the lid is now the infimum).

- zzr0ck3r

if X is the cup right side up, then -X is the cup upside down.

- zzr0ck3r

You are not wrong, we can think of this as a bijective map, but it makes it harder. and we can think of it as a mapping from R to R, and that would look like the line y=-x, but I dont think this is a good idea, nor have i ever seen it talked about in this way.

- Mendicant_Bias

I understand the supremum and infimum, but I also want to understand what I'm talking about conceptually so I'm able to answer questions like this and not just ignore them. If I am having trouble understanding more fundamental things regarding how sets work, I have to answer those questions before I'm able to consistently apply that to solving other problems.
It looks to you (reasonably) like I'm failing to write well-defined questions probably, but they're still there for me. Nonetheless, it's good to be talking about this.

- zzr0ck3r

we are simply saying that if the set has an order, and we flip the set, then the order gets reversed

- zzr0ck3r

agreed

- Mendicant_Bias

Alright. I can go with that. So, now the other questions.

- zzr0ck3r

I think if we get through the other stuff, this earlier stuff will be more clear.

- zzr0ck3r

And I know this is confusing :) it is the most failed class where I teach.

- Mendicant_Bias

I'm confused by the notion that the set of real numbers is a "complete" ordered set, or that any set extending to infinity or negative infinity is complete, because to my understanding, the Supremum has to be a real number, and infty/-infty are not. I might just misunderstand the definition of a supremum.

- Mendicant_Bias

(And since I'm confused about the set of real numbers being a "complete" ordered set, I'm consequently confused about this one.)

- zzr0ck3r

what do you mean by complete?

- zzr0ck3r

The real numbers are not bounded above or below, there is no supremum.
you may define a supremum, but that comes later.

- zzr0ck3r

That is no longer real analysis, it would be on the "extended reals"

- zzr0ck3r

I am not sure what you mean by complete, there is a mathematical term for a space to be complete, but you must first lean of cauchy sequences, and I dont think you are that far yet.

- zzr0ck3r

@Mendicant_Bias you still here?

- Mendicant_Bias

Yes, sorry, was writing an e-mail for a moment. I'll quote the passage of my book talking about it, but it has to do with the real numbers being the only complete ordered field, one sec.

- Mendicant_Bias

http://i.imgur.com/zx2C0u0.png

- zzr0ck3r

ok, yes. This is the same comlete

- zzr0ck3r

It is just saying that this is a defining relationship of the real numbers. It is not something we prove, it is something we start with. That is, that every non empty bounded set has a supemum, i.e. if there is a top element in the set, there is one and only one that is bigger than all the rest.

- zzr0ck3r

if it is bounded. This does not say that the real line has a supremum, it says that any bounded subset does.

- Mendicant_Bias

The confusion also stems from a couple of book problems, which I've solved right, but have, as a definite answer, -infty or infty defined as a supremum or infimum:
(Part a) http://i.imgur.com/VKz3QgJ.png
(Answer, focusing on part a again) http://i.imgur.com/1ckz1Kn.png
By the very definition of a supremum or infimum, requiring it to be a real number, I don't understand how -infty can be the infimum.

- zzr0ck3r

I am confused by your question. A supremum must be an element in the set, there is no element infinity.

- zzr0ck3r

do you just want to know the answer to part a?

- zzr0ck3r

when I say it must be in the set, I mean the universal set...

- Mendicant_Bias

Well I have the answer, but I don't understand how it can be Inf(S)=-infty is possible, by the same reasoning you wrote.
Also, I have yet to learn what the infinite set is, so maybe I don't have to concern myself with it yet

- Mendicant_Bias

Sorry, misspoke, the answer to (a) being Sup(S)=infty, I don't understand how that's possible/correct.

- Mendicant_Bias

*universal set, not infinite set

- zzr0ck3r

ok I know where the confusion stems from.
When you write \(\sup(R)=\infty\) this means that there is no supremum. It is just very bad notation.

- zzr0ck3r

but your answer says no next to infinity

- Mendicant_Bias

Yeah, whether it's a member of the set or not, but that's not my confusion, e.g. I understand that if I have a set consisting of all of the negative numbers up to but not including zero, Sup(S)=0, but 0 isn't a member of the set;
While having a set consisting of all of the negative numbers up to and including zero, Sup(S)=0, and 0 is a member of the set.
0 is a real number, while infty is not; that's where my confusion lies, if that makes any sense

- zzr0ck3r

it does, if someone handed me a piece of paper that said the supremum of the even numbers is infinity, I would assume that to mean that it is unbounded above.

- Mendicant_Bias

Alright, I'll ignore the book question for now, maybe that was just bad writing?

- zzr0ck3r

I am guessing if you read all of it, they will make a small note about what they mean by \(\sup(X) = \infty\).
If you would like to learn a little more about this, read the following link
https://en.wikipedia.org/wiki/Extended_real_number_line

- zzr0ck3r

It is very much like this
\[\lim_{x\rightarrow \infty}x=\infty\]
We dont actually mean infinity, but we write it...

- Mendicant_Bias

Alright. Now, keeping myself from digressing: I don't understand how the set we are given has an upper bound or lower bound since it covers the whole real line.

- zzr0ck3r

It just says it is a set or real numbers, not THE set of real numbers.

- Mendicant_Bias

Alright, that makes far more sense. So it was just me misreading the set notation.

- zzr0ck3r

So we know that \(S\subset \mathbb{R}\) such that \(\exists \ M\) s.t. for all \(x\) in \(S\) we have that \(M\le x\)
Makse sense?

- Mendicant_Bias

I read that as, "S contains some (but not all) elements of R in itself/is a subset of R, such that there exists some M (what is s.t.?) for all x in this subset S we have that M is lesser than or equal to x."

- Mendicant_Bias

Oh, such that

- zzr0ck3r

s.t. = such that
yes you are correct.
This is just saying that \(S\) is a set of real numbers, and there is some number that is amsller than all numbers in S.

- zzr0ck3r

Just to let you know, I got to go pretty soon.

- Mendicant_Bias

So now I can definitely understand it being bounded, and it having a least upper bound/greatest lower bound.
And sure thing, thank you so much for your help so far. I've learned a lot that was just difficult to learn considering my misunderstanding stemmed from not reading notation right, and neither of us really knowing that for quite a while, heh.
I'll call this off for now, take what I have, and come back to this.

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