Hello, everyone. Welcome to Analysis one, video 26.
At the end of the last video, I asked, is it the case that
every sequence has a convergent subsequence? And the answer was sort of immediately no, because
the sequence one, two, three, four, five, six, and so on doesn't have a convergent subsequence.
The next question I asked was, does every bounded sequence have a convergent subsequence?
And that's what we're going to try to resolve now. And we're going to do that by starting with
something called the Scenic Viewpoints Theorem. It's a maybe unusually poetic name. But I will
draw you picture in a moment that will hopefully help to explain it.
What it says is that if the sequence a_n is a real sequence, then it has a monotone subsequence.
And remember, monotone means increasing or decreasing.
And you might already be thinking ahead to well, what would that tell us about bounded
sequences? But we won't rush ahead just now, we'll prove the scenic viewpoints theorem first.
Just a practical note for anybody who's interested. The reason I have
gained this sort of strange glow is because I'm playing with the lighting
in my kitchen to see whether I can illuminate my handwriting a bit more helpfully for you.
So this has involved complicated things involving dismantling a bedside lamp and
propping it on a kitchen table in a strange way, but hopefully it might help with the handwriting.
So what's the idea of the scenic viewpoints theorem?
Well, let me try to draw a picture. Maybe I'll just draw that here.
So I've got my usual axes for drawing a sequence and I'll draw on some points of my sequence,
which might be positive or negative. They're going to kind of hop around
a bit. Who knows what it is they do. Here we go. Here's some points of a sequence.
And a scenic viewpoint, or a peak of the sequence, is going to be a point like this one up here,
because this point is kind of high up, and if we look forwards from this point,
so look at later terms in the sequence, nothing gets in the way of the view.
That is, all of the terms of the sequence later than this one are smaller.
So this is a peak, or a scenic viewpoint. Let me try to find another one.
So the next point in the sequence isn't because there's a point after it, that blocks the view.
But then this point here is another peak. It's not as high as the previous peak.
If you think about it for a moment, that's inevitable because the previous one was a peak.
But it is still a peak because if we look later on,
we don't have an impeded view. And then I guess, as I've drawn it here,
it looks like we've got another peak here and another peak here.
I mean, there are still another infinitely many terms of the sequence to come.
Who knows what might happen. But I hope the picture gives you
the kind of idea. And what we're going to do to prove the scenic viewpoints theorem, to
show that our sequence has a monotone subsequence, is to think about two cases.
So one case is where there are infinitely many peaks, and we will argue that they
must form a decreasing sequence. So you might be able to see from the
picture I've drawn already why they would form a decreasing sequence.
The other case is where there are only finitely many of these peaks. And that
means that after some point, no point is a peak. And so there's always a later term of the sequence
that's larger. And that's going to allow us to argue that there's an increasing subsequence.
So that's the kind of executive summary of the proof. I will write that out carefully now.
So we've got a real sequence (a_n). We want to show
it's got a monotone subsequence. And I'm going to let V be the set of all k in the natural numbers
such that if m is bigger than k, then a_m is less than a_k.
So this is the set of peaks, or scenic viewpoints. So I'll just kind of put a little note in
purple somehow. These are the kind of peaks. This is not a technical term, but hopefully
this helps you to kind of see how it relates to the picture that I showed you just now.
So there are going to be two cases. Case one is where there are infinitely many of these.
So what happens then? Well, let's say the elements of V
are k_1 less than k_2 less than k_3 and so on forever.
So I'm just listing the elements of this set in order because I want to pick out the subsequence.
So a_{k_r} as r varies is a subsequence of a_n. And it must be decreasing, and is decreasing,
because if r is less than s then k_r is less than k_s.
That's how we've constructed our ks. And so a_{k_r} is greater than a_{k_s},
because we know that k_r was a peak. So if a_{k_s} comes later then it must be smaller.
So in that case we get a decreasing subsequence.
Let me just create myself a little bit more space on my page here. So case two
is going to be the case that V is finite. And that means there's a largest element.
So then there exists an N such that if k is greater than N, then k is not in V.
And what I'm going to do is now kind of just build a sequence, because if we let
m_1 be like N plus one say, then m_1 isn't in V.
So that means that there's some point later than it, some term later than a_{m_1} in the sequence
that's bigger. So there exists an m_2 greater than well, greater than m_1, sorry.
With a_{m_2} greater than or equal to a_{m_1}. Otherwise m_1 would have been in V. And then
m_2 is not in V, so there exists an m_3 greater than m_2 with a_{m_3} greater than
or equal to a_{m_2}, and so on. So we can continue inductively,
to construct m_1 less than m_2 less than m_3 and so on, with a_{m_1} less than or
equal to a_{m_2} less than or equal to a_{m_3}, and so on. And then a_{m_r} is an increasing
subsequence of a_n.
Now that we've proved the scenic viewpoints theorem, we're in good shape to have a look
at the Bolzano-Weierstrass Theorem, which is one of the high points of this course,
it's a really important result. And it's named after Bolzano and
Weierstrass, who were two of the key figures in putting analysis on rigorous foundations,
I guess. The Bolzano-Weierstrass theorem says let the sequence a_n be a bounded real sequence.
Then the sequence a_n has a convergent subsequence. So any bounded real sequence
has a convergent subsequence, which is pretty cool when you think about it.
So how are we going to prove this? Well, I'm going do this one on a slide,
because actually it's really a case of gluing together pieces we've got already.
There aren't really any new ideas, apart from how do we assemble everything we've done so far.
So we've got a bounded real sequence. By the Scenic Viewpoints Theorem, that has a monotone
subsequence, because we've just proved that every real sequence has a monotone subsequence.
So let's take a monotone subsequence, and then it must be bounded,
because the whole sequence is bounded and consequently any subsequence is bounded.
So now we have a monotone and bounded subsequence, that must converge by
the monotone sequences theorem. That was the one that says that a
bounded monotone sequence converges. Oh. And there is our convergent subsequence.
It was there almost before we knew it. So the key thing was really that we'd proved
the scenic viewpoints theorem, and we'd proved the monotone sequences theorem,
and gluing those together meant that we could prove Bolzano-Weierstrass very quickly.
But don't be misled. It's a deep result building on work that we've done earlier.
I have two favourite proofs of the Bolzano-Weierstrass theorem. And
that's one of them. With the other one,
I thought what I do is make it into a kind of proofsorter quiz activity thing on Canvas.
So I have a look at that if you've got a few minutes and you'd like to give that go.
I hope that will be quite a fun activity as well as quite a useful one for you.
Another remark is that the monotone sequences theorem and the scenic viewpoints theorem
make sense for real sequences, but not for complex sequences in general.
Just think about why that is.
But the Bolzano-Weierstrass theorem, that a bounded sequence has a convergent subsequence
there's nothing intrinsic about the statement of that result that means it has to be about reals.
There are no inequalities between terms of the sequences there. So maybe the Bolzano-Weierstrass
theorem could hold for complex sequences. Dot dot dot.
I've called this corollary 43. That might be a bit controversial because it is a theorem,
but I sort of want to highlight that it's a consequence of the theorem that we've
already proved rather than a whole new thing. So this is Bolzano-Weierstrass for complex
sequences, and it says let the sequence z_n be a bounded, complex sequence.
Then the sequence z_n has a convergent subsequence.
So basically, Bolzano-Weierstrass holds for complex sequences too.
And the reason I've called it a corollary is that I want to highlight that we're going to
deduce it from the result for real sequences rather than going back to the drawing board.
So let's take the sequence z_n a bounded complex sequence, and let's write
z_n is x_n plus i y_n, where x_n and y_n are reals.
Because what I want to do is think about the real sequences x_n and y_n, and argue with them.
So we know that the sequence z_n is bounded. If we say the modulus of z_n is less than equal to
M for all n greater or equal to one. So we're saying that M is a bound for our
sequence. Well then certainly the sequence x_n and the sequence y_n are bounded by M as well.
That actually goes back to something that we did, it feels like a long time ago
now, in the complex numbers course where we commented that when we
looked at the definition of the modulus, we could see if the modulus of a complex
number is at most M then the modulus of the real part and the modulus of the imaginary
part are each at most M. OK, so x_n and y_n these
sequences, they're bounded by M,. and they're real, so by Bolzano-Weierstrass.
Let's just focus on one at a time here. Because the plan
is to kind of reduce repeatedly to subsequences.
So the sequence x_n has a convergent subsequence,
which we might call, say, x_{n_r}.
So that's a convergent subsequence in the reals. And what I want to do now is focus on the y_n,
but in fact, only focus on the y_{n_r},
so only pick out the terms corresponding to this subsequence.
So then y_{n_r} is a bounded real sequence,
because it's a subsequence of a bounded real sequence,
so by Bolzano-Weierstrass, which I will abbreviate to B-W the second time, it has
a convergent subsequence say y_{n_{r_s}}
kind of stacking up these subscripts. This is not super-convenient notation, but it'll do the job.
So the real part, we took a convergent subsequence.
Then we looked at the relevant subsequence, the imaginary part,
to refine again to a convergent subsequence.
If we look back at the real part of that refined subsequence, so x_{n_{r_s}}
s is a subsequence of the convergent sequence
x_{n_r}. I'm trying to make my r and n look different for you.
So it converges. We proved previously that a subsequence of a convergent sequence converges.
So the real and imaginary parts, if you like, in these subsequences converge.
And so z_{n_{r_s}} converges.
I think that might have been theorem 33, the one that says that a complex sequence converges
if and only if the real and imaginary parts converge. And we've shown that in this case,
the real and imaginary parts do converge,
and so the whole sequence converges.
I'm not going to show you any applications right
away of Bolzano-Weierstrass. They will come in due course.
I'm going to wrap up this video in a moment.
Just before I go, something for you to think about.
If I have a sequence a_n, and it converges, does that mean that a_n minus a_{n-1} tends to zero?
Question.
Second question. If I've got a sequence a_n, and I tell you that a_n minus
a_{n-1} goes to zero as n goes to infinity,
does that mean that the sequence converges? So there's sort of two directions there.
So that's something for you to have a think about. And we
will start having a look at that in the next video. See you soon.