00:00 Intro

00:20 Bolzano-Weierstrass theorem

01:13 Proof

05:42 Credits

1 00:00:00,414 –> 00:00:03,104 Hello and welcome back to real analysis

2 00:00:04,014 –> 00:00:09,337 and as always many, many thanks to all the nice people that support this channel on Steady or Paypal.

3 00:00:09,537 –> 00:00:14,716 In today’s part 10 we will talk about the so called Bolzano-Weierstrass theorem.

4 00:00:15,014 –> 00:00:19,010 It’s a fundamental result you should learn in any analysis course.

5 00:00:19,614 –> 00:00:23,857 The statement itself is really short, such that i can immediately tell you it.

6 00:00:24,057 –> 00:00:27,997 For this we consider a sequence a_n, which is bounded

7 00:00:28,197 –> 00:00:33,392 and then we can conclude that this sequence has at least one accumulation value

8 00:00:33,700 –> 00:00:38,734 or to put it in other words, a bounded sequence of real numbers has to cluster somewhere.

9 00:00:39,300 –> 00:00:46,542 For this please recall, having an accumulation value or cluster point just means the sequence has a convergent subsequence.

10 00:00:47,043 –> 00:00:52,890 One possible visualization would be on the number line, where you have a lower bound and an upper bound.

11 00:00:53,514 –> 00:00:57,447 Therefore all the sequence member lives between does values.

12 00:00:57,647 –> 00:01:02,203 Of course they may jump around, but we still have infinitely many of them.

13 00:01:03,014 –> 00:01:07,559 Therefore you find at least 1 point, where infinitely many sequence members cluster.

14 00:01:08,471 –> 00:01:12,100 Indeed that’s the whole statement of the Bolzano-Weierstrass theorem.

15 00:01:13,043 –> 00:01:16,527 Ok, then I would say let’s start proving it.

16 00:01:17,100 –> 00:01:20,211 In order to do this let’s take the picture from above again.

17 00:01:20,886 –> 00:01:25,280 Let’s call the lower bound c_0 and the upper bound d_0.

18 00:01:26,214 –> 00:01:31,429 Now, the overall idea for the whole proof is that we just bisect the interval here.

19 00:01:32,314 –> 00:01:37,857 Then we have 2 intervals, 2 parts and please remember we want to find a subsequence here.

20 00:01:38,700 –> 00:01:43,757 Therefore we will continue with the interval that has infinitely many sequence members in it

21 00:01:43,986 –> 00:01:45,403 and ignore the other one.

22 00:01:45,800 –> 00:01:52,745 Therefore the procedure is that we first look at the left interval and observe, if we have infinitely many sequence members in it.

23 00:01:53,300 –> 00:01:59,552 Of course it could happen that both intervals have infinitely many members in it, but then we would choose the left-hand side.

24 00:02:00,186 –> 00:02:04,802 Ok, so what we get is a new interval that has half the length of the one before.

25 00:02:05,457 –> 00:02:09,012 So either the upper bound or the lower bound changed.

26 00:02:09,212 –> 00:02:13,534 Therefore we choose new names, namely c_1 and d_1.

27 00:02:13,734 –> 00:02:17,685 Then maybe not so surprising, we bisect the new interval again

28 00:02:17,885 –> 00:02:22,013 and then we repeat the selecting and the bisecting over and over again.

29 00:02:22,729 –> 00:02:26,067 Hence what we get in the end are nested intervals.

30 00:02:26,814 –> 00:02:35,421 We have our original interval c_0 to d_0 and the new interval c_1 to d_1 is a proper subset of this one

31 00:02:35,957 –> 00:02:41,515 and then after repeating the whole process here, we get the next interval c_2 to d_2.

32 00:02:41,715 –> 00:02:45,936 Which is once again a proper subset of the former interval here.

33 00:02:46,136 –> 00:02:49,490 Hence we have this inclusion as often as we want.

34 00:02:50,014 –> 00:02:55,952 Now the point of the statement is that the length of the interval gets smaller and smaller in each step.

35 00:02:56,857 –> 00:03:05,671 In particular if we calculate (d_1 - c_1), we get 1/2 of the original distance, because we bisected the interval.

36 00:03:06,171 –> 00:03:11,828 In the same way, for (d_2 - c_2) we get 1/2 of (d_1 - c_1)

37 00:03:12,657 –> 00:03:16,059 or in other words 1/4 of the original distance.

38 00:03:16,600 –> 00:03:22,102 Ok, by knowing this we can use induction to show the general statement for (d_n - c_n).

39 00:03:22,629 –> 00:03:27,686 We simply get 1/2 to the power n times the length of the original interval

40 00:03:28,200 –> 00:03:32,783 and there you immediately see, this is a nice sequence that converges to 0.

41 00:03:32,983 –> 00:03:39,574 However we also get some information about the sequences d_n and c_n when we look at the picture again.

42 00:03:40,243 –> 00:03:45,924 First of all we know they are bounded sequences, because all the sequence members also lie in this interval

43 00:03:46,329 –> 00:03:52,852 and secondly we know the sequence c_n can only increase, where the sequence d_n can only decrease

44 00:03:53,457 –> 00:03:57,707 and now these things imply that the sequences are also convergent.

45 00:03:58,186 –> 00:04:03,746 Please recall, this is exactly the monotonicity criterion we discussed in former videos.

46 00:04:03,946 –> 00:04:07,400 It’s important because we can use it very often, as you can see.

47 00:04:08,100 –> 00:04:11,980 What we also can use very often are the limit theorems.

48 00:04:12,500 –> 00:04:16,785 From above we can use the fact that we already know this limit, which is 0

49 00:04:16,985 –> 00:04:21,004 and now we have learned that we can push the limit inside the difference.

50 00:04:21,386 –> 00:04:24,619 Simply because both limits here also exist.

51 00:04:25,286 –> 00:04:31,074 So please note our conclusion here is that the limits of d_n and c_n are the same.

52 00:04:31,529 –> 00:04:37,181 Therefore finally we are able to define a subsequence for our original sequence a_n.

53 00:04:37,381 –> 00:04:44,010 We simply do that by choosing for a_n_k one of the infinitely many elements inside one of the intervals.

54 00:04:44,514 –> 00:04:49,986 With this we then know that a_n_k lies always between c_k and d_k.

55 00:04:50,686 –> 00:04:55,835 Please recall here again, if we increase k the interval here gets smaller and smaller

56 00:04:56,257 –> 00:05:00,154 and we already know in the limit the boundaries are the same

57 00:05:00,686 –> 00:05:04,935 or to put it in other words here we can now use the sandwich theorem.

58 00:05:05,471 –> 00:05:10,827 It simply tells us that the sequence in the middle is also convergent with the same limit

59 00:05:11,027 –> 00:05:14,827 and exactly this limit is our wanted accumulation value.

60 00:05:15,457 –> 00:05:18,187 Now this means that our proof is finished here.

61 00:05:18,614 –> 00:05:21,768 Ok and that’s the Bolzano-Weierstrass theorem.

62 00:05:22,514 –> 00:05:26,479 Every bounded sequence has at least one accumulation value

63 00:05:27,029 –> 00:05:33,057 and now i can also tell you this statement also holds when you consider sequences of complex numbers.

64 00:05:33,257 –> 00:05:36,290 So it’s useful in many different situations.

65 00:05:36,643 –> 00:05:38,676 Ok, I think that’s good enough for today.

66 00:05:38,876 –> 00:05:40,065 So I hope I see you next time.

67 00:05:40,265 –> 00:05:41,086 Bye!

Q1: What is the correct formulation of the Bolzano-Weierstrass theorem?

A1: Every convergent sequence has at least one accumulation value.

A2: Every accumulation value for a sequence is also an upper bound.

A3: Every bounded sequence is also convergent.

A4: Every bounded sequence has at least one accumulation value.

A5: Every bounded sequence has at least one divergent subsequence.

Q2: Can you apply the Bolzano-Weierstrass theorem to the sequence $(a_n)_{n\in \mathbb{N}}$ given by $a_n = n^2$?

A1: Yes, the sequence only has one accumulation value.

A2: No, the sequence is not bounded.

Q3: Consider the sequence $(a_n)_{n\in \mathbb{N}}$ given by $$a_n = \begin{cases} 0 \text{ if } n \text{ even}\ 1 \text{ if } n \text{ odd and divisible by } 3\ 2 \text{ if } n \text{ odd and not divisible by } 3\\end{cases} $$ Is the Bolzano-Weierstrass theorem applicable?

A1: Yes!

A2: No!