
Title: Cauchy Sequences and Completeness

Series: Real Analysis

Chapter: Sequences and Limits

YouTubeTitle: Real Analysis 7  Cauchy Sequences and Completeness

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Timestamps
00:00 Intro
00:14 Convergent sequences
00:54 Different property of a sequence
02:09 Definition Cauchy sequence
02:16 Connection to convergent sequences
03:14 Dedekind completeness
03:47 Sketch of proof
07:35 Application for monotonic sequences
08:44 Credits

Subtitle in English
1 00:00:00,429 –> 00:00:03,234 Hello and welcome back to real analysis
2 00:00:03,814 –> 00:00:08,720 and as always i want to thank all the nice people that support this channel on Steady or Paypal.
3 00:00:09,071 –> 00:00:13,526 Todays part 7 is about Cauchy sequences and completeness.
4 00:00:14,071 –> 00:00:19,972 For starting this topic lets recall what we already considered sequences with a special property.
5 00:00:20,529 –> 00:00:22,818 Namely convergent sequences.
6 00:00:23,314 –> 00:00:31,157 Which means there is a number “a” such that the sequence members here get arbitrarly close to this number “a” eventually.
7 00:00:31,671 –> 00:00:34,514 You already know the formal way to say this. Which is:
8 00:00:34,771 –> 00:00:40,538 for all epsilon there exists a capital N such that for all indices n greater than this N,
9 00:00:40,738 –> 00:00:44,911 the distance between “a” and “a_n” is less than the given epsilon.
10 00:00:45,111 –> 00:00:50,161 Now the problem with this definition is that you need to know the limit to show convergence.
11 00:00:50,543 –> 00:00:54,236 Simply because we measure the distance to this “a”.
12 00:00:54,436 –> 00:01:01,547 Hence there is a different idea or a different property a sequence can have, which does not need such a number “a” in definition.
13 00:01:01,747 –> 00:01:06,185 For this lets look at a number line again and at a sequence which should converge.
14 00:01:06,957 –> 00:01:10,475 So here we have a_1, a_2, a_3 and so on
15 00:01:10,743 –> 00:01:14,678 and the sequence members accumulate here, so there should be a limit here.
16 00:01:15,300 –> 00:01:19,169 However we don’t want to use this limit to describe what happens here.
17 00:01:19,557 –> 00:01:24,829 Indeed what happens here is that the sequence members themselves get closer and closer to each other.
18 00:01:25,886 –> 00:01:31,488 Hence what we want is that the sequence members lie arbitraily close to each other eventually.
19 00:01:31,771 –> 00:01:36,996 So everything is about the distance you can measure between 2 sequence members here.
20 00:01:37,196 –> 00:01:40,751 Which is the absolute value of “a_n  a_m”.
21 00:01:41,271 –> 00:01:44,663 and then they should be less than epsilon, we choose at the beginning.
22 00:01:45,414 –> 00:01:49,591 Therefore the formal way then reads: For all epsilon greater 0,
23 00:01:50,057 –> 00:01:56,578 we find a capital N such that for all indices called n and m afterwards,
24 00:01:57,043 –> 00:02:01,296 we have that the distance between the sequence members is less than epsilon.
25 00:02:02,143 –> 00:02:06,035 and now a sequence with this property we call a Cauchy sequence.
26 00:02:06,571 –> 00:02:09,666 Ok, so lets put that into a definition.
27 00:02:10,200 –> 00:02:15,586 This is exactly the definition you might have already seen in my “start learning reals” series.
28 00:02:16,229 –> 00:02:21,250 There i also showed you the important fact that for a sequence of real numbers
29 00:02:21,786 –> 00:02:27,289 we have that being a Cauchy sequence is equivalent to being a convergent sequence.
30 00:02:28,143 –> 00:02:31,900 The prove of this direction you can see in part 2 of the course
31 00:02:32,537 –> 00:02:35,758 and the other one is exactly the completeness axiom.
32 00:02:36,657 –> 00:02:41,995 So the completeness axiom tells us that there are no holes in our complete real number line.
33 00:02:42,986 –> 00:02:46,671 Now, what you really should remember is when we work in the real numbers,
34 00:02:46,743 –> 00:02:51,100 we don’t have to distinguish Cauchy sequences and convergent sequences.
35 00:02:52,086 –> 00:02:56,270 They have different definitions, but for the real numbers they mean the same thing.
36 00:02:56,914 –> 00:03:02,721 Therefore we are able to use the one or the other definition, depending what is usefull in our context
37 00:03:03,343 –> 00:03:08,455 and soon we will see that the definition of the Cauchy sequences make a lot of things easier.
38 00:03:08,655 –> 00:03:13,512 However before we apply this in examples lets discuss another important property.
39 00:03:14,186 –> 00:03:18,967 It’s called Dedekind completeness and the property for subsets of the real numbers.
40 00:03:19,457 –> 00:03:25,600 If M is such a set and also bounded from above, then we know the supremum exists.
41 00:03:25,900 –> 00:03:29,591 So there is at least upper bound as a number in R.
42 00:03:29,791 –> 00:03:33,632 Please recall that we defined the supremum in the last video.
43 00:03:34,129 –> 00:03:38,413 and maybe not so surprising, we have the same thing for the infimum as well.
44 00:03:38,414 –> 00:03:43,367 So if we have a set that is bounded from below, then the infimum exists as a real number.
45 00:03:44,014 –> 00:03:46,777 Ok, let me explain how we can prove this statement.
46 00:03:47,229 –> 00:03:51,653 and i thing it’s enough to give you the idea, how to do it for the supremum.
47 00:03:52,257 –> 00:03:56,218 For this lets consider the number line and the set M on it.
48 00:03:56,418 –> 00:04:02,058 Then we know we find an upper bound for the set M. So lets choose one and call it b_1.
49 00:04:02,900 –> 00:04:06,439 On the other hand lets choose any element in the set M
50 00:04:06,771 –> 00:04:09,351 and this one we call just a_1.
51 00:04:09,771 –> 00:04:15,125 Now the idea is that we then go to the point that is exactly in the middle of these 2 points.
52 00:04:15,829 –> 00:04:19,130 and this one of course i now call c_1.
53 00:04:19,500 –> 00:04:22,980 So we just calculate the sum and then we divide by 2.
54 00:04:23,343 –> 00:04:26,888 So in this case we find this mittle point somewhere here.
55 00:04:27,557 –> 00:04:34,143 Of course the idea is that by doing this calculation we get closer to this point. Which is the supremum of M in the picture.
56 00:04:34,629 –> 00:04:40,571 Therefore now i explain how we get the whole sequence. Which gives us an approximation of the supremum.
57 00:04:41,014 –> 00:04:44,743 Ok, you might have already seen, we can have 2 cases for c_1.
58 00:04:44,971 –> 00:04:48,400 Either it’s still an upper bound like b_1 or not.
59 00:04:48,930 –> 00:04:55,071 In the first case we just have a smaller upper bound. So a better one. So we can just substitute b_1.
60 00:04:55,500 –> 00:05:00,793 Or to put it to other words, we introduce b2 to be the new upper bound given by c_1.
61 00:05:01,214 –> 00:05:05,707 On the other hand the left hand side we don’t have to change. We just stay at a_1
62 00:05:06,443 –> 00:05:12,710 and then afterwards we do the same thing again now with a_2 and b_2, define the middel point as c_2.
63 00:05:13,443 –> 00:05:18,668 In this case you then would to see that the middle point lies left to some points of M.
64 00:05:18,868 –> 00:05:22,041 Therefore then we have to think what we do in the second case.
65 00:05:22,986 –> 00:05:29,100 Now, as i said here it’s possible to find a point x in M, which is larger than c_1.
66 00:05:29,943 –> 00:05:33,865 Then of course this new larger point should be our new a_2.
67 00:05:34,314 –> 00:05:37,607 So with this we shifted the left point to the right
68 00:05:37,943 –> 00:05:41,386 and that we don’t need to change the right hand side with b_1.
69 00:05:42,600 –> 00:05:47,055 Now having this we have the whole procedure how we can form the sequences.
70 00:05:47,786 –> 00:05:55,803 So in general when we have a_n, b_n, we define cn and then as before we just define the 2 next numbers here.
71 00:05:56,386 –> 00:06:02,399 So this is a recursive definition, which gives us 2 sequences, an and bn.
72 00:06:02,671 –> 00:06:08,731 and the sequence members bn are always upper bounds, which approximate the supremum.
73 00:06:09,286 –> 00:06:15,229 Therefore the only thing that remains to show is that the sequence bn is indeed a Cauchy sequence.
74 00:06:16,114 –> 00:06:20,282 However this is not hard to see, because we have the following estimate.
75 00:06:20,914 –> 00:06:25,310 Namely for 2 indices m and n, where m is greater than n
76 00:06:25,510 –> 00:06:29,324 we can calculate the distance b_n to b_m.
77 00:06:29,524 –> 00:06:36,124 I used the absolute value here, but honestly we don’t need it, because b_n is greater or equal than b_m
78 00:06:36,771 –> 00:06:41,900 and now the distance gets larger when we substitute any “a” for this b_m
79 00:06:42,319 –> 00:06:45,922 and the best “a” we can choose would be “a” with index n.
80 00:06:46,122 –> 00:06:51,833 Now we know by our construction with the middle point, that we always cut the distance in half.
81 00:06:52,157 –> 00:07:00,191 Therefore we know that this is 1/2 to the power “n  1” times the starting distance b_1  a_1
82 00:07:00,529 –> 00:07:04,302 and now this whole number here we can get arbitrarily small.
83 00:07:04,986 –> 00:07:11,214 This is what you can show and then you can formalize the conclusion that b_n is indeed a Cauchy sequence
84 00:07:11,578 –> 00:07:17,998 and then the last step is just using the completeness axiom to show that b_n is a convergent sequence.
85 00:07:18,814 –> 00:07:23,300 Also the only possibility for the limit then is the supremum of M.
86 00:07:23,620 –> 00:07:26,545 Ok, so that’s the overall idea of the whole prove.
87 00:07:26,745 –> 00:07:29,016 The missing details you can easily fit in.
88 00:07:29,216 –> 00:07:35,396 However now knowing that the supremum and infimum always exists has an important application.
89 00:07:36,114 –> 00:07:40,359 We immediately get a nice criterion to show convergence of a sequence.
90 00:07:40,800 –> 00:07:44,208 If the sequence a_n is monotonically decreasing
91 00:07:44,671 –> 00:07:49,135 which simply means that each sequence member is less or equal than the predecessor
92 00:07:50,057 –> 00:07:53,001 and if the sequence is also bounded from below
93 00:07:53,457 –> 00:07:57,745 which simply means that the set, given by the squence member has a lower bound.
94 00:07:58,486 –> 00:08:03,088 Then we can finally conclude that the sequence is indeed a convergent sequence.
95 00:08:03,629 –> 00:08:08,946 There you see, this is a very useful criterion, because you only have to check 2 properties.
96 00:08:09,657 –> 00:08:12,914 Which could be easier to check then the definition of convergence.
97 00:08:13,614 –> 00:08:18,865 In this formulation you might already see we used the existence of the supremum of this set.
98 00:08:19,543 –> 00:08:23,883 However if we use the existence of the infimum we get another criterion.
99 00:08:24,543 –> 00:08:30,899 Of course it’s very similar. There we just need a monotonically increasing sequence, which is bounded from above.
100 00:08:31,586 –> 00:08:34,386 and i then we can also conclude that we have a convergent sequence.
101 00:08:35,043 –> 00:08:39,526 Ok and then in the next video i show you some examples for this application.
102 00:08:40,386 –> 00:08:43,686 Therefore i hope i see you there and have a nice day. Bye!

Quiz Content
Q1: Let $(a_n){n \in \mathbb{N}}$ be a sequence of real numbers. What is the correct definition of ‘The sequence $(a_n){n \in \mathbb{N}}$ is a Cauchy sequence’.
A1: $\forall \varepsilon > 0 ~~ \exists N \in \mathbb{N} ~~ \forall n,m \leq N ~:~ a_n  a_m < \varepsilon$.
A2: $\forall \varepsilon > 0 ~~ \exists N \in \mathbb{N} ~~ \forall n,m \geq N ~:~ a_n  a_m < \varepsilon$.
A3: $\exists \varepsilon > 0 ~~ \exists N \in \mathbb{N} ~~ \forall n,m \leq N ~:~ a_n  a_m < \varepsilon$.
A4: $\exists \varepsilon > 0 ~~ \forall N \in \mathbb{N} ~~ \forall n,m \leq N ~:~ a_n  a_m < \varepsilon$.
Q2: Which of the following sequences is not a Cauchy sequence in $\mathbb{R}$?
A1: $ \left( \frac{1}{n} \right)_{n \in \mathbb{N}} $
A2: $ \left( 2^{n} \right)_{n \in \mathbb{N}} $
A3: $ \left( \frac{1}{3} n \right)_{n \in \mathbb{N}} $
A4: $ \left( \frac{1}{3n} \right)_{n \in \mathbb{N}} $
Q3: Which implication is guaranteed by the completeness axiom (C)?
A1: $(a_n){n \in \mathbb{N}}$ Cauchy sequence $~\Rightarrow~$ $(a_n){n \in \mathbb{N}}$ convergent sequence.
A2: $(a_n){n \in \mathbb{N}}$ convergent sequence $~\Rightarrow~$ $(a_n){n \in \mathbb{N}}$ Cauchy sequence.
A3: $(a_n){n \in \mathbb{N}}$ not a Cauchy sequence $~\Rightarrow~$ $(a_n){n \in \mathbb{N}}$ not a convergent sequence.
Q4: The sequence $ \left( (1 + \frac{1}{n})^n \right)_{n \in \mathbb{N}} $ is monotonically increasing and bounded from above by $3$. What is the correct conclusion?
A1: The sequence is divergent.
A2: The sequence is convergent with limit $3$.
A3: The sequence is convergent with limit greater than $3$.
A4: The sequence is convergent with limit less or equal than $3$.
A5: The sequence is not bounded from below.