That's also the least upper bound. Okay, if there is a maximum element, it

has to be the least upper bound. So, there is a least upper bound, it's

the maximum element, and it's negative one, it's not zero.

Zero is well to the right of this thing. So, the least upper bound is negative

one. And zero is not the least upper bound.

Okay, well how did you do on that one? you know, if it's, if I'm reading it soon

now, you throw your for, you throw your, your fist to your forehead.

And say oh, golly, how could I be so stupid.

Well, you weren't being stupid, you were being human.

this is just tricky when you first meet it.

you know, and if it, if it looks simple now, then good.

And if it still looks complicated, then you're going to have to work for a bit

more, but you're not you're not at all unique, if having seen this you're still

confused. we all are when we first meet this.

Okay, well let's go to number two now. Well as with the previous question the

approach is to move slowly and painstakingly and be careful about what

things say. Just do some positive parts, so that we

know exactly what it says. So let's do that one step at a time.

B less than or equal to a for all a in a, that says b is a lower bound of the set

a. If c is, so that says, c is a lower bound

of the set a. So b is given as a lower bound of the set

a, and if c is another lower bound of the set a, then b is greater than or equal to

c. In other words, b is the biggest one, b

is bigger than any other one, or at least it's greater than or equal to any other

one. So b is a lower bound and moreover, b is

at least as big as any other lower bound. So he answer is yes, that one does say

it's the greatest lower bound. What about the second one?

That says b less than, so this says again it's the same thing, b is a lower bound.

And this is, if c is any other lower bound, then b is bigger than c.

there's a problem here, because b itself, satisfies that, that would mean, b is

bigger than, b. This statement, would imply, b, is bigger

than b. Which is impossible of course, right?

No number can be strictly bigger than itself.

So this thing, it's just, it's not only is this, does it not say that.

So this is, in fact, a False statement. It doesn't say anything sensible.

What it implies is that b is bigger than itself as some b.

So certainly, this does not say as the greatest lower bound.

It doesn't say anything meaningful at all.

Let me say a little bit more. What's going on in here and here?

It means we've got implicit quantification over c, implicit

quantification over c. This is really saying for all c, if c

less than or equal to with... This is a quantification, in the, in this

situation, it's implicit quantification over c.

And whenever you've got universal quantification, implicit universal

quantification, whenever you've got universal quantification, you can hit

every possible target. So, whenever there's something like this

going on with universal quantifier, among the possible c's, is the b itself.

I mean the b isn't really been quantified within this.

B is just some number that we're trying to classify.

So the b is just a number, it's a fixed number if you like.

The a's are being quantified over, for all a's, for all a's, and the c here is

being quantified over. Now, we haven't used the traditional

language of for alls here. But it is a universal quantifier, and

there, it, because it, it just says any c less than or equal to a, and so the c can

be anything. And in particular, the c can be the

original b. In the first case, it was no problem.

Because b is greater than or equal to b. This case its a big problem it makes it

false in fact, because the number can't be bigger then itself.

Okay, lets look at the next one, well lets see lets why don't we take A to be

the closed unit interval zero, one, okay? The greatest lower bound of A is 0, the

minimum element, right? Why did I do that?

Well, because zero is not less, strictly less than everything in A.

In the case of this set, you would have to have zero strictly less than zero,

which is impossible, okay? So, it cannot be the case, I mean, this

can't [LAUGH], it just can't be, I mean, it's just not true.

This, this thing is not true because to be a, to be a greatest lower bound you

would have to, this tells you you'd have to be strictly less than everything in

the set. Then in the case of a closed set, it

doesn't have to be the unit interval zero, one.

Any set with a minimal element will do this.

All we need is a minimium element. And if the set is a minimum element.

Then the greatest lower bound cannot be strictly less, than everything in, in the

set, because if there's a minimum element, it is the greatest lower bound.

So that rules out this one. Okay, what about this one?

well, essentially the same reasoning. You can't classify your greatest lower

bound, bas, by including the requirement that is stricter that's in everything in

the set. If there's a minimum element, it is the

greatest lower bound, and it's not strict or less than this.

So, now it doesn't capture it. Okay, let's look at this one.

This says, b is a lower bound of the set A.

And what this one says, by the way, there's a reason why it's expressed in

terms of these epsilons. A lot of the theory of the real numbers

that was developed in the late 19th or the early 20th century.

it turned out that doing this stuff with, with epsilons was very powerful.

But if you look at what it says, it's going to turn out to actually say the

same as the one right at the top. That says, if you combine that with that.

This says, forget that bit for a minute. This says, any number slightly bigger

than b, bit bigger than a, any number bigger than b.

Good grief, okay. You're going to add an epsilon to b, so

you gotta go slightly to the right of b. And what it says is, if you go slightly

to the right of b, you can find an a to the left of that.

Let me just draw a little picture, okay? We've got this set, a, here, so okay.

So a is somewhere in here, okay? And I've got a b, just to the left of

everything in A, or at least not to the right of everything in A when it, it's

less than or equal to everything in A. And then what I'm saying is, if I go

slightly to the right, to a number b plus epsilon, which we'll put in here.

Then already I've gone to the right of some element A, of the, the, the, of A of

the set, here. So, b is certainly to the left of

everything in A, but if I go slightly to the right by adding an epsilon onto b,

then I will be able to find an element in the set to the left of it.

So, this can't be a lower bound. So b's are low bound, but if I go, if I

pick anything to the right of b, and this is saying, I really shouldn't of wanted

that, maybe. This is saying everything to the right,

anything to the right of b, anything to the right of b, already dominates

something in the set A. So anything in the right of b will fail

to be a A lower bound. So b is the greatest lower bound.

Okay, so you just have to parse apart that combined with that.

For any epsilon greater than zero, b plus epsilon has a property.

So these two together is basically saying, for any number bigger than b.

That's really what I was saying, for any number bigger than b, there is an a such

that. In other words, it's saying these thing,

this is exactly the same as this, expressed differently.

Okay, well, you probably have to think a little bit about this.

As, as I've, mentioned a couple of times already now, the human mind just seems to

find its way difficult when it first meets it.

An as I, demonstrated by a couple of misspoke, misspoken remarks I've made.

Even when you're familiar with material it's tricky to say these things precisely

unless you really sort of rehearsed it out.

And it's just the way the mind works. Okay, let's move ahead to the, to the

next question. Well, question three, deals with the

delightfully named Sandwich Theorem. incidentally, this is not to be confused

with the Ham Sandwich Theorem, which is a very different theorem altogether.

I'm not kidding you, there is a thing called the Ham Sandwich Theorem.

You can look it up on the web and find out what it is.

This something quite different. and what it says is that if you have two

sequences. one always less than another one, and

you've got another sequence in between them.

Then when one sequence and another sequence when they're sent to the same

limit. the thing in between them gets forced to

the same limit. it's called the Sandwich Theorem, because

the a sequence and the c sequence are slices of bread, and the b sequence is in

between the slices of bread. And what this says is, if you make a, a

sandwich with two slices of bread and some meat in between them, or some, some

filling, and you squeeze the two slices of bread together.

Then the filling will get squeezed between them.

actually, of course, what happens is is that mayonnaise runs out all over your

lap, or ketchup and things, so it's kind of messy in real life.

But in principle, what we're saying is if we squeeze the two slices of bread

together, then the then the fillings get squeezed in between them to the same

thing. Okay, now in, in, in question four, we're

going to have to prove the Sandwich Theorem.

Here we're just going to apply it. And it's a very useful theorem.

It's a simple theorem. It's, it's intuitively obvious.