2:58

so a cannot be less than b. So they're equal.

Â Okay, well, as I said, this makes sense to me, and you may well have to work

Â pretty hard to follow that. That's just the way it is, you're not

Â being stupid. You just got a human mind, and the human

Â mind just does not like this. Okay, this one's also True.

Â This one's actually less difficult, because if a is a lower bound, any lower

Â bound, of the set a, then so is any b less than a.

Â If a, if, if little a is to the left of everything in this set, then so is

Â everything to the left of a. So, providing there is a lower bound,

Â then anything to the left of that lower bound is still a lower bound.

Â We're not talking about least upper bound or greater lower bounds, we talking about

Â just plain old lower bounds. And once you've got a lower bound,

Â anything to the left of the lower bound is a lower bound.

Â Okay, this ones False, this is third one, if a set reals has a lower bound and an

Â upper bound, its finite. And I'm going to do that by giving an

Â example. And the most obvious one, I think, it to

Â take the unit interval, zero, one. Then it has a lower bound, namely zero.

Â It has an upper bound, namely one. In fact, the lower bound is a minimum

Â element. And the upper bound is, not the least

Â upper bound. At least the one we've mentioned, is one.

Â So this element, this is, set actually has a least element and a maximum

Â element. So it's saying that it has a lower bound

Â and an upper bound. Anything to the left of zero is also a

Â lower bound. Anything to the right of zero.

Â Anything to the right of one is an upper bound.

Â So this thing is bounded below and bounded above.

Â But this is an infinite set, it's got all of the real numbers between zero and one.

Â So having though in different bounds does not mean that the, the set is, is

Â infinite, is that the set is finite. Indeed, if it's an interval it could be

Â infinite. Incidentally, this is not the same, there

Â is a phrase, finite interval. that's got a different meaning.

Â That basically means that the interval does not stretch to minus infinity or

Â plus infinity. That it is an interval between two

Â points. Well, this is a special use of the word

Â finite. It doesn't mean it's a finite set.

Â It just means it's got finite lower and upper bounds.

Â It's got endpoints if you like. so, having said that, I'm going to get

Â rid of that, because I don't want to confuse the issue.

Â As a set, this is an infinite set. Okay, and finally, let's look at the last

Â one, this is also false. let's just look at what the set is, let's

Â write it out. The set is, starts down here.

Â A long way down here, actually. Then we've got minus four, minus three,

Â minus two, minus one, or negative four, negative three, negative two, and

Â negative one, if you're pedantic on those issues.

Â And then, this one actually has a maximum element.

Â 6:17

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

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

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

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

Â one. And zero is not the least upper bound.

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

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

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

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

Â this is just tricky when you first meet it.

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

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

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

Â confused. we all are when we first meet this.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â than b. Which is impossible of course, right?

Â No number can be strictly bigger than itself.

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

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

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

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

Â It doesn't say anything meaningful at all.

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

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

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

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

Â situation, it's implicit quantification over c.

Â And whenever you've got universal quantification, implicit universal

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â minimum element, right? Why did I do that?

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

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

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

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

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

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

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

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

Â Any set with a minimal element will do this.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â this is exactly the same as this, expressed differently.

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

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

Â find its way difficult when it first meets it.

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

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

Â unless you really sort of rehearsed it out.

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

Â next question. Well, question three, deals with the

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

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

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

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

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

Â sequences. one always less than another one, and

Â you've got another sequence in between them.

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

Â limit. the thing in between them gets forced to

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

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

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

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

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

Â Then the filling will get squeezed between them.

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

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

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

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

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

Â going to have to prove the Sandwich Theorem.

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

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

Â 19:46

That's all there is to it, it's logically correct.

Â Everything's there, it's clear. There's an opening, very, you know, it's

Â not much of an opening to do in a sense. You know, arguably one should have said,

Â one could have said at the beginning we're going to use a Sandwich Theorem.

Â But that was actually given to us in, in the question.

Â And so as long as we cite it when we use it that's fine.

Â Okay, so the opening in a sense was, was provided to us by the context of this

Â one. there was certainly a conclusion.

Â There's the conclusion, clearly stated. The reason, well, there, there's there's

Â a couple of reasons. As this is part of the reasoning.

Â But this is the key one that we're going to cite the Sandwich Theorem.

Â That's the main reason, and overall, absolutely correct, 24.

Â it's, it's not, I mean the, the complexity, the depth is actually in this

Â statement here, in the Sandwich Theorem. The Sandwich Theorem does all the work

Â for us. This is really just a corollary of the

Â Sandwich Theorem. but, you know, proofs don't have to be

Â long. So long as everything's there, so long as

Â it's clear, it's proof. Now, I, I, I just said that I didn't

Â prove this. I said it was, it was clear because in

Â the context of applying the Sandwich Theorem, you're entitled to say that this

Â is clear. Clearly one can give the epsilon proof of

Â this, you know, given that an epsilon is an n and.

Â You could use the definition of, of, of limits, but it's not necessary.

Â You know, you don't normally prove that seven plus five is, is 12, when you're

Â doing some arithmetic. You just assume that.

Â I mean, it could be proved, you could prove it from first principles.

Â you can do a various things if you want to.

Â But in the context of applying the Sandwich Theorem, you're entitled to

Â assume something like this. it is important to say if it's clear.

Â and if you'd said because of that, we can apply the Sandwich Theorem, eh, then it's

Â a, it's a little more mysterious. But saying something is clear is a valid

Â reason, so that's an okay method of saying that's something's a reason.

Â Say it's clear, if it's clear, simply saying it's clear is adequate to give a

Â reason. you don't need to have length and depth

Â in, in this proofs. So long as, well, you need to have the

Â depth but it doesn't need to be unnecessary long.

Â This is a good one, 24 max. Let's go on, let's go on now and and

Â prove the Sandwich Theorem itself. So, here we are, let's again, what it

Â says. It says you've got three sequences.

Â And that from some point onwards, the a sequence is less than everything in the b

Â sequence. The corresponding member of the b

Â sequence is less than the corresponding member of the c sequence.

Â This happens on a term by term basis, from some point onwards.

Â the nth term of this is less than the nth term of that is less than the nth term of

Â that. So from some point onwards, you've got a

Â sandwich. Where the b is the, is the filling in

Â between the two slices of bread, okay? And then, the the, the outer sequences,

Â the lower sequence and the upper sequence tend to the same limits.

Â They converge in terms of the same limits.

Â Then, the conclusion is that the sequence that's in between is convergent, and

Â moreover, converges to that self same limit.

Â 26:18

This inequality is also going to come from there, because bn is less than or

Â equal to cn. And so I could subtract the l from both

Â sides of there, and then finally this one is going to give me c n minus l is less

Â than epsilon. So what I'm doing, is for the first

Â inequality here. I'm taking this, and that feeds into that

Â inequality. Then, this here is going to feed into the

Â second inequality. This here is going to feed into the third

Â inequality, and this here, is going to feed into the last inequality.

Â So, that inequality here, gives me the first one.

Â This gives me the second, that gives me the third, and that one gives me the

Â fourth. So it's all concatenated and sort of

Â pressed in, but it's, but the reasons are all given.

Â This is there, that's there, and that's there, those are the three reasons.

Â They all combine to give me that. Then I can say that minus epsilon is less

Â than bn minus L is in the middle and plus epsilon is on the outside.

Â And I now I can put absolute values back in and say that means bn minus a, that

Â absolute value is s in epsilon. So by the definition of limits, this

Â proves that this thing is convergent and moreover it's convergent too well.

Â Well it does that providing this argument is carried out for an average epsilon.

Â So now let's give some, some, some numerical grades, logical correctness.

Â I'm going to say four for that, the, the logic was absolutely brilliantly laid

Â down. I mean, this, this is very succinct,

Â nicely done, very impressive. clarity, absolutely clear, absolutely

Â clear. Opening, there wasn't an opening.

Â The opening has to be that epsilon greater than zero be given.

Â That's how the story starts, this is the equivalent of, once upon a time, the,

Â the, in a fairy story. Yeah, I mean, you have to start somehow,

Â somehow. You have to open the story.

Â This is the opening for this kind of argument.

Â It's equivalent to saying, we've gone to use induction for an induction proof.

Â and that plum proof has to being by stating that the epsilon's epsilon, that

Â the epsilon is an arbitrary positive number.

Â This is crucial, first of all, it, it tells you it's positive.

Â You've got to make, make that, that explicit.

Â And you've gotta say that it's arbitrary. So there's no opening here.

Â state the conclusion. Oh yes, the conclusion is beautifully

Â stated. Four marks for the conclusion, what about

Â reasons? I'm going to give four marks for reasons

Â in particular. I love this one here, this is really nice

Â that they, that this has been written here.

Â Let's put a little smiley stick here, that this has been done here.

Â That's nice to put that in here and rather than expect people to sort of pull

Â it back out of the hypothesis. yeah, in fact, without that, I, I think I

Â would have deducted my actual reason. this is the the, that's there and that's

Â there. You know, arguably, one could have put a

Â bit more detail in, the way I explained it when I went through it.

Â But in terms of expressing what needs to expressed, this does it very well.

Â Overall evaluation, in all conscience as a mathematician, I can't go any higher

Â than zero for overall. Without that, the whole thing falls

Â apart. This is not a triviality.

Â This is what, this is what makes the theorem work.

Â It hinges on the fact that it works where epsilon greater than zero be given, okay,

Â and you can't simply say, well, you know, we all know that.

Â Well, we don't know that. This is a proof, and then, this is part

Â of a proof. It's not just part of a proof.

Â It's an absolutely critical part of a proof.

Â So, can't go beyond beyond that. Not only did it have a critical , not

Â only is it an incredibly important part of the proof, but this approach, this way

Â of dealing with things. Came after the result of, of hundreds of

Â years research in mathematics. It was only late in the 19th century,

Â when people figured out how to do this kind of thing and make sense of limits.

Â So, this not just important within this proof, this was a, a major advance in

Â mathematics in the late 19th century. incredibly powerful.

Â So I've got four lots of four, I've got 16 for this one.

Â and I don't think that's unduly harsh. even though I feel sorry for the person

Â because this is very clever. You know, manipulating inequalities and

Â getting them right and that, that's tricky, and people have difficulty with

Â that. but I've given credit for that.

Â Okay, that was in the correctness and the clarity and given the reasons.

Â 31:32

Okay, so we're proving that this particular sequence, n plus one over two

Â n plus one, tends to the limit one half. Okay, I, I will tell you that this is

Â actually my proof. And the reason I'm doing that rather than

Â taking the proof that I have from students or composite from students.

Â Is that these proofs tend to be sort of longish because you're manipulating

Â inequalities and fractions and things and they wouldn't fit neatly onto a single

Â slide. So I know that they fit onto a slide, I

Â actually give my own proof, and so this is a very succinct proof, okay?

Â and I'll tell you up front, that I'm going to give myself four marks, which

Â I'm likely to do, aren't I? But, in fact, this is a, is a correct

Â proof. So let's take away all of the all of the

Â suspense, are saying I'm going to give myself four marks on this one.

Â instructors can do that, right? but in fact it, it, this is fine, but I,

Â this is not the kind of proof I would expect to see from a student.

Â and certainly not with a beginning student.

Â I'd expect to see quite a bit more. But let's just, see what's going on.

Â Okay, so, we begin by saying that epsilon is greater than zero be given.

Â Okay, so I start off well. Okay, I give the, the, the opener.

Â So that's the, that's the opener right here, taken care of.

Â Choose N large enough so that N greater than or equal to two epsilon.

Â Well, if epsilon then the idea, the intuition, of course, is that.

Â Is what makes the proof where. Is that the epsilon can be arbitrarily

Â small. Which means one over two epsilon will be

Â arbitrarily large. But of course I can always pick an n

Â bigger than that. that actually is a, is a mathematical

Â principle known as the Archimedean Principle.

Â But we won't go into that now. Okay, intuitively it's clear that you can

Â always pick an integer bigger than any, any real number.

Â Okay, so why am I picking that? Well, because I'm looking ahead to what I

Â want to the end, and after doing this kind of thing for years I can look ahead

Â and see what I've got. In practice what you've probably got to

Â do, is just work through with, with, without saying what the N is.

Â And then see what you have to do at the end, you know, so I could break this down

Â in the beginning because I could see ahead.

Â But typically you might have to go through the logic first before you can

Â decide on this steps. On this sentence for most people, comes

Â at the end of the manipulation. Okay, then for N great or equal to, and

Â I'll come back to that point in a minute. For N greater or equal to N we've got the

Â following, okay? we need to show that this is within

Â epsilon, okay? So were going to show that this guy is

Â less the epsilon. Well let's just work it out.

Â Write it as a single fraction, common denominator is two into two n plus one,

Â so then I've got two, two n plus one, minus two n plus one.

Â do the algebra, that just becomes one over two, two n plus one.

Â everything's positive now, so I don't need the absolute value symbol.

Â so and I can get rid of the half now, by saying it's less than that over that.

Â Just simplifying things and I can say that that's less than one over two n

Â because the denominator is bigger here so it makes it smaller.

Â 35:43

So it was doing this manipulation that gives me the, the choice of the big N.

Â As I say, I've been doing this for so long that I can see ahead to see what's

Â going on. But even if you can't, you can just do.

Â Just start with what you're trying to prove, you're trying to prove that this

Â is less than epsilon. See what it is you have to prove, and

Â make simplifications. Get rid of the two, you don't need it.

Â Because that makes it bigger. Get rid of the, the plus one, you don't

Â need it and then you get to something simple, and you can pick your N that way.

Â Okay, and then by definition of a limit. So let's see why I gave this, the logical

Â correctness, well. It has a start in there, the opener.

Â the, [INAUDIBLE], let's take them one by one.

Â The logical correctness. These are just manipulations.

Â This is just inequalities. Okay, it's pretty slick, I did this in a

Â few steps. You might make a few more steps.

Â but it's logically correct. It's absolutely clear.

Â There is an opening, there is a strong conclusion.

Â the definition of a limit, reasons. Yeah, well there's a reason there.

Â and there's sort of reasons, embedded in here, in what I'm doing.

Â I didn't write reasons down here in my explanation, because this is just

Â straightforward manipulation. You know, you could say, you could say

Â what I've just articulated. You could give explanations and reasons

Â but you don't need to. This is, this is, this is transparently

Â clear which is basic algebraic manipulation of inequalities.

Â so reasons are okay I mean, this is an important reason, I think, it's

Â definitely worth saying that. and if you go back to previous question,

Â question four, I didn't make a big deal about it.

Â But if you look at question four, it also ended by, by citing the fact that we're

Â using the definition for a limit. And then overall yeah, you, you give four

Â for this, because there's absolutely nothing missing.

Â It's, it's a clear, concise, vigorous, logical complete proof, okay?

Â They, these take a while to get used to, though.

Â Because you are working backwards. you have to sort of play with the thing

Â until you get it into a form where you know how to pick the N.

Â But it comes down to picking an N so big that everything happens that you want to

Â happen. Okay, well that's the end of problem set

Â eight which is the last problem set of this course.

Â How about that.

Â