0:00 Introduction

0:20 Sequentially compact set

1:07 Empty set is compact

1:17 {5} is compact

1:50 R is not compact

2:10 Proof: [a,b] is compact

3:15 Heine-Borel theorem statement

3:58 Proof of Heine-Borel theorem

1 00:00:00,471 –> 00:00:03,437 Hello and welcome back to real analysis

2 00:00:03,971 –> 00:00:09,277 and as always many, many to all the nice people that support this channel on Steady or Paypal.

3 00:00:09,900 –> 00:00:14,056 In today’s part 14 we will talk about the Heine-Borel theorem.

4 00:00:14,256 –> 00:00:18,876 In order to understand this, we first have to recall what a compact set is.

5 00:00:19,429 –> 00:00:25,838 Here I can tell you, the definition we use for a compact set other people call sequentially compact.

6 00:00:26,414 –> 00:00:29,834 It just means that we use sequences to describe compactness.

7 00:00:30,586 –> 00:00:35,986 Now please recall that any subset A of the real numbers is called a compact set,

8 00:00:35,987 –> 00:00:38,786 if it forces any sequence to cluster.

9 00:00:39,171 –> 00:00:43,429 So a sequence with members in A has at least one accumulation value.

10 00:00:44,257 –> 00:00:50,359 However, the important thing is that this accumulation value also has to lie inside the set A.

11 00:00:50,700 –> 00:00:52,143 Also please note here,

12 00:00:52,186 –> 00:00:59,214 this is a short formulation for saying that we have a sequence of real numbers, where each member is an element of the set A.

13 00:00:59,700 –> 00:01:04,157 Ok, so we know the definition, therefore we now can look at examples

14 00:01:04,929 –> 00:01:07,228 and I would say we start with the simplest one.

15 00:01:07,957 –> 00:01:13,368 The empty set does not have any elements. Therefore this claim here is trivially true.

16 00:01:13,700 –> 00:01:16,047 Hence the set is compact.

17 00:01:16,247 –> 00:01:21,106 Similarly we can take another small set, so a set with just one element.

18 00:01:21,306 –> 00:01:25,743 This one is also compact, because there is only one possible sequence.

19 00:01:26,543 –> 00:01:31,015 In this case we have the constant sequence 5 with accumulation value 5.

20 00:01:31,215 –> 00:01:33,267 Ok, so this was not so hard.

21 00:01:33,329 –> 00:01:37,606 Therefore now my question would be: “What is about the largest subset we can choose?”

22 00:01:38,157 –> 00:01:41,545 there we have to look at all sequences of real numbers.

23 00:01:41,745 –> 00:01:45,863 For example I can give you the sequence that increases with n.

24 00:01:46,063 –> 00:01:52,734 So this sequence is divergent to infinity. Therefore there is no accumulation value as an element of R.

25 00:01:52,934 –> 00:01:56,197 Hence we immediately have a set that is not compact.

26 00:01:56,914 –> 00:01:59,236 Next let’s look at a closed interval.

27 00:02:00,114 –> 00:02:04,865 Such an interval with real numbers c and d is indeed a compact set.

28 00:02:05,065 –> 00:02:07,517 Ok, so let’s try to proof this.

29 00:02:08,114 –> 00:02:12,536 For this let’s assume we have a sequence with members inside this interval.

30 00:02:12,736 –> 00:02:18,419 Hence we can immediately conclude that this sequence is bounded from above and bounded from below.

31 00:02:18,957 –> 00:02:21,657 In short we say this sequence is bounded.

32 00:02:22,014 –> 00:02:26,271 However now you know, we can apply a very famous theorem.

33 00:02:26,471 –> 00:02:30,419 Namely the Bolzano-Weierstrass theorem for sequences.

34 00:02:30,871 –> 00:02:35,403 It tells us that a bounded sequence has at least one accumulation value

35 00:02:36,071 –> 00:02:39,787 and this one we can call “a” and we know, it’s a real number.

36 00:02:40,143 –> 00:02:45,808 Now the only information that is still missing is that “a” is actually an element of our set.

37 00:02:46,629 –> 00:02:52,481 However this is guaranteed, because we already know that such an interval is a closed set.

38 00:02:52,957 –> 00:02:58,892 Here please recall, closed means that we can’t leave the set by using sequences from the inside

39 00:02:59,092 –> 00:03:03,036 and with this we have proven that the set is indeed compact.

40 00:03:03,543 –> 00:03:08,401 Hence all intervals of this form we know can just call compact intervals.

41 00:03:08,971 –> 00:03:12,105 Ok, now we are ready for the Heine-Borel theorem.

42 00:03:13,071 –> 00:03:19,263 It tells us that the two properties we used in this proof here, are actually exactly what we need for a compact set.

43 00:03:19,600 –> 00:03:23,100 Hence it’s a theorem that is not hard to remember at all.

44 00:03:23,829 –> 00:03:32,893 So for any subset A of the real numbers, we have that it is compact if and only if A is bounded and closed.

45 00:03:33,457 –> 00:03:36,562 It’s essential that we have both things here together.

46 00:03:36,871 –> 00:03:43,990 This theorem is a very good result for us, because it tells us that we can substitute the complicated notion of compactness

47 00:03:44,190 –> 00:03:47,141 with the simpler notions of boundedness and closeness.

48 00:03:47,843 –> 00:03:53,911 Ok, so this is the Heine-Borel theorem, which holds for the real numbers and you really should remember it.

49 00:03:54,111 –> 00:03:57,930 For the rest of the video, I would say let’s to the proof of it.

50 00:03:58,357 –> 00:04:03,283 Indeed the direction from the right-hand side to the left-hand side, we have already proven.

51 00:04:03,343 –> 00:04:11,785 Of course you remember the only things we needed for the argument above is that A is a bounded set, such that we can use Bolzano-Weierstrass

52 00:04:11,985 –> 00:04:14,143 ans the other thing was that the set is closed.

53 00:04:14,886 –> 00:04:19,107 Which guarantees us that the accumulation value is actually in A.

54 00:04:19,500 –> 00:04:23,352 Therefore only the direction from left to right remains.

55 00:04:23,552 –> 00:04:26,835 Hence assume that our set A is compact.

56 00:04:27,214 –> 00:04:32,539 Now for showing the closeness we take any convergent sequence in the set A

57 00:04:32,739 –> 00:04:35,627 and let’s call the limit for the moment “a” tilde.

58 00:04:36,429 –> 00:04:39,777 First we only know that the limit is a real number.

59 00:04:40,214 –> 00:04:45,688 If we can show that it is also an element in A, we have shown that a set is closed.

60 00:04:45,888 –> 00:04:49,303 Ok, so at this point we can use that A is compact,

61 00:04:49,657 –> 00:04:53,763 because it tells us that the sequence a_n has an accumulation value

62 00:04:54,186 –> 00:04:57,934 and as before let’s call this accumulation value just “a”

63 00:04:58,300 –> 00:05:03,319 and now we already know, a convergent sequence can only have one accumulation value.

64 00:05:03,843 –> 00:05:07,786 Hence our limit “a” tilde is the same as “a”

65 00:05:07,986 –> 00:05:11,385 and therefore also an element in our set A.

66 00:05:11,871 –> 00:05:14,033 So the set A is closed.

67 00:05:14,643 –> 00:05:18,456 Ok, so it remains to show that A is bounded as well

68 00:05:18,986 –> 00:05:22,296 and in order to show this we use a proof by contra position.

69 00:05:22,496 –> 00:05:29,024 More concretely this means that we assume that A is not bounded and want to show that A is not compact.

70 00:05:29,514 –> 00:05:34,103 Now, whenever we have unbounded sets, we can construct the sequence in the set,

71 00:05:34,303 –> 00:05:40,080 such that all sequence member fulfill that the absolute value of a_n is greater than n

72 00:05:40,443 –> 00:05:43,361 and this should work for all natural numbers n

73 00:05:43,886 –> 00:05:49,345 and because of this property such a sequence can’t have any accumulation value.

74 00:05:49,545 –> 00:05:52,698 This is not hard to show, but maybe it’s a good exercise for you.

75 00:05:53,657 –> 00:06:00,707 Of course the important thing is the result. We have a sequence that has no accumulation value, therefore the set can’t be compact.

76 00:06:01,586 –> 00:06:04,876 and that is the proof of the Heine-Borel theorem.

77 00:06:05,157 –> 00:06:10,471 So you see, it’s not so complicated, but still the important part is that you remember the theorem.

78 00:06:11,529 –> 00:06:16,473 Ok, then I hope I see you in the next video, when we continue with another topic in real analysis.

79 00:06:17,157 –> 00:06:18,957 Have a nice day! Bye!

Q1: Which of the following sets is not compact?

A1: $[1, 2]$

A2: $\emptyset$

A3: $\mathbb{R}$

A4: ${8}$

Q2: What is the correct formulation of the Heine-Borel theorem?

A1: $\mathbb{R}$ is compact if and only if $\mathbb{R}$ is closed.

A2: $A \subseteq \mathbb{R}$ is compact if and only if $A$ is bounded and closed.

A3: $A \subseteq \mathbb{R}$ is closed if and only if $A$ is bounded and compact.

A4: $A \subseteq \mathbb{R}$ is bounded if and only if $A$ is closed and compact.

A5: $A \subseteq \mathbb{R}$ is closed if and only if $A$ is bounded.

Q3: Is the Bolzano-Weierstrass theorem used in the proof of the Heine-Borel theorem?

A1: Yes, for showing that bounded and closed sets are also compact.

A2: Yes, for showing that compact sets are also bounded and closed.

A3: No, it is not used.