Why does Cauchy's Convergence Theorem imply the completeness of real numbers? [closed]
$begingroup$
Lots of textbooks on Mathematical Analysis say that:
Cauchy's Convergence Theorem shows that, any convergent sequence
consisting of real numbers has a real number limit, which is called the completeness of real numbers.
In my opinion, the completeness of real numbers refers to that, if a real number sequence is convergent, it is necessarily convergent to a real number. By contrast, a rational number sequence is not necessarily convergent to a rational number,even though it converges. A classical example is that $left(1+dfrac{1}{n}right)^n to e(n to infty).$ In another word, the rational number set does not have the completeness property.
But how does the completeness of real numbers come out from Cauchy's Theorem? How to comprehend the point?
real-analysis real-numbers
asked Dec 3 '18 at 14:45
mengdie1982mengdie1982
4,897618
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closed as unclear what you're asking by user21820, Lord Shark the Unknown, KReiser, John B, Shailesh Dec 9 '18 at 3:20
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Lots of textbooks on Mathematical Analysis say that:
Cauchy's Convergence Theorem shows that, any convergent sequence
consisting of real numbers has a real number limit, which is called the completeness of real numbers.
In my opinion, the completeness of real numbers refers to that, if a real number sequence is convergent, it is necessarily convergent to a real number. By contrast, a rational number sequence is not necessarily convergent to a rational number,even though it converges. A classical example is that $left(1+dfrac{1}{n}right)^n to e(n to infty).$ In another word, the rational number set does not have the completeness property.
But how does the completeness of real numbers come out from Cauchy's Theorem? How to comprehend the point?
real-analysis real-numbers
asked Dec 3 '18 at 14:45
mengdie1982mengdie1982
4,897618
$endgroup$
closed as unclear what you're asking by user21820, Lord Shark the Unknown, KReiser, John B, Shailesh Dec 9 '18 at 3:20
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Can you explain how is your "opinion" differs from the theorem? As far as I understand the theorem also guarantee the existence of a real limit.
$endgroup$
– Yanko
Dec 3 '18 at 14:47
$begingroup$
What exactly is the statement of Cauchy's Convergence Theorem? Is it that every Cauchy sequence in $mathbb{R}$ is convergent?
$endgroup$
– Brahadeesh
Dec 3 '18 at 14:48
1
$begingroup$
You have to philosophically agree that a Cauchy sequence "should" be convergent before completeness really means anything to you. Once you do that, the point is simple: you had $mathbb{Q}$. Some Cauchy sequences in $mathbb{Q}$ don't converge, so you add limits to all the non-convergent Cauchy sequences according to the Cauchy completion construction. You get $mathbb{R}$. Now you have entirely new Cauchy sequences, so you have to ask: do any of them fail to converge in $mathbb{R}$? Completeness is telling you that they don't.
$endgroup$
– Ian
Dec 3 '18 at 14:57
add a comment |
$begingroup$
Lots of textbooks on Mathematical Analysis say that:
Cauchy's Convergence Theorem shows that, any convergent sequence
consisting of real numbers has a real number limit, which is called the completeness of real numbers.
In my opinion, the completeness of real numbers refers to that, if a real number sequence is convergent, it is necessarily convergent to a real number. By contrast, a rational number sequence is not necessarily convergent to a rational number,even though it converges. A classical example is that $left(1+dfrac{1}{n}right)^n to e(n to infty).$ In another word, the rational number set does not have the completeness property.
But how does the completeness of real numbers come out from Cauchy's Theorem? How to comprehend the point?
real-analysis real-numbers
asked Dec 3 '18 at 14:45
mengdie1982mengdie1982
4,897618
$endgroup$
Lots of textbooks on Mathematical Analysis say that:
Cauchy's Convergence Theorem shows that, any convergent sequence
consisting of real numbers has a real number limit, which is called the completeness of real numbers.
In my opinion, the completeness of real numbers refers to that, if a real number sequence is convergent, it is necessarily convergent to a real number. By contrast, a rational number sequence is not necessarily convergent to a rational number,even though it converges. A classical example is that $left(1+dfrac{1}{n}right)^n to e(n to infty).$ In another word, the rational number set does not have the completeness property.
But how does the completeness of real numbers come out from Cauchy's Theorem? How to comprehend the point?
real-analysis real-numbers
real-analysis real-numbers
asked Dec 3 '18 at 14:45
mengdie1982mengdie1982
4,897618
asked Dec 3 '18 at 14:45
mengdie1982mengdie1982
4,897618
asked Dec 3 '18 at 14:45
mengdie1982mengdie1982
4,897618
asked Dec 3 '18 at 14:45
mengdie1982mengdie1982
4,897618
asked Dec 3 '18 at 14:45
mengdie1982mengdie1982
4,897618
4,897618
closed as unclear what you're asking by user21820, Lord Shark the Unknown, KReiser, John B, Shailesh Dec 9 '18 at 3:20
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by user21820, Lord Shark the Unknown, KReiser, John B, Shailesh Dec 9 '18 at 3:20
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Can you explain how is your "opinion" differs from the theorem? As far as I understand the theorem also guarantee the existence of a real limit.
$endgroup$
– Yanko
Dec 3 '18 at 14:47
$begingroup$
What exactly is the statement of Cauchy's Convergence Theorem? Is it that every Cauchy sequence in $mathbb{R}$ is convergent?
$endgroup$
– Brahadeesh
Dec 3 '18 at 14:48
1
$begingroup$
You have to philosophically agree that a Cauchy sequence "should" be convergent before completeness really means anything to you. Once you do that, the point is simple: you had $mathbb{Q}$. Some Cauchy sequences in $mathbb{Q}$ don't converge, so you add limits to all the non-convergent Cauchy sequences according to the Cauchy completion construction. You get $mathbb{R}$. Now you have entirely new Cauchy sequences, so you have to ask: do any of them fail to converge in $mathbb{R}$? Completeness is telling you that they don't.
$endgroup$
– Ian
Dec 3 '18 at 14:57
add a comment |
$begingroup$
Can you explain how is your "opinion" differs from the theorem? As far as I understand the theorem also guarantee the existence of a real limit.
$endgroup$
– Yanko
Dec 3 '18 at 14:47
$begingroup$
What exactly is the statement of Cauchy's Convergence Theorem? Is it that every Cauchy sequence in $mathbb{R}$ is convergent?
$endgroup$
– Brahadeesh
Dec 3 '18 at 14:48
1
$begingroup$
You have to philosophically agree that a Cauchy sequence "should" be convergent before completeness really means anything to you. Once you do that, the point is simple: you had $mathbb{Q}$. Some Cauchy sequences in $mathbb{Q}$ don't converge, so you add limits to all the non-convergent Cauchy sequences according to the Cauchy completion construction. You get $mathbb{R}$. Now you have entirely new Cauchy sequences, so you have to ask: do any of them fail to converge in $mathbb{R}$? Completeness is telling you that they don't.
$endgroup$
– Ian
Dec 3 '18 at 14:57
$begingroup$
Can you explain how is your "opinion" differs from the theorem? As far as I understand the theorem also guarantee the existence of a real limit.
$endgroup$
– Yanko
Dec 3 '18 at 14:47
$begingroup$
Can you explain how is your "opinion" differs from the theorem? As far as I understand the theorem also guarantee the existence of a real limit.
$endgroup$
– Yanko
Dec 3 '18 at 14:47
$begingroup$
What exactly is the statement of Cauchy's Convergence Theorem? Is it that every Cauchy sequence in $mathbb{R}$ is convergent?
$endgroup$
– Brahadeesh
Dec 3 '18 at 14:48
$begingroup$
What exactly is the statement of Cauchy's Convergence Theorem? Is it that every Cauchy sequence in $mathbb{R}$ is convergent?
$endgroup$
– Brahadeesh
Dec 3 '18 at 14:48
1
1
$begingroup$
You have to philosophically agree that a Cauchy sequence "should" be convergent before completeness really means anything to you. Once you do that, the point is simple: you had $mathbb{Q}$. Some Cauchy sequences in $mathbb{Q}$ don't converge, so you add limits to all the non-convergent Cauchy sequences according to the Cauchy completion construction. You get $mathbb{R}$. Now you have entirely new Cauchy sequences, so you have to ask: do any of them fail to converge in $mathbb{R}$? Completeness is telling you that they don't.
$endgroup$
– Ian
Dec 3 '18 at 14:57
$begingroup$
You have to philosophically agree that a Cauchy sequence "should" be convergent before completeness really means anything to you. Once you do that, the point is simple: you had $mathbb{Q}$. Some Cauchy sequences in $mathbb{Q}$ don't converge, so you add limits to all the non-convergent Cauchy sequences according to the Cauchy completion construction. You get $mathbb{R}$. Now you have entirely new Cauchy sequences, so you have to ask: do any of them fail to converge in $mathbb{R}$? Completeness is telling you that they don't.
$endgroup$
– Ian
Dec 3 '18 at 14:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There is a subtlety here regarding convergence that perhaps you have missed. When you say that
In my opinion, the completeness of real numbers refers to that, if a real number sequence is convergent, it is necessarily convergent to a real number.
you are making a slight (but crucial) error.
Firstly, it is improper to suggest that a sequence in a space $X$ converges, but its limit lies outside $X$. Because, if a sequence is convergent then by definition its limit must lie in the same space.
However, there are sequences that appear to converge: their tails lie in intervals of smaller and smaller length, tending to zero. Such sequences are called Cauchy sequences.
Now, a Cauchy sequence in a space $X$ may or may not have a limit in the space $X$. For instance, there are Cauchy sequences in $mathbb{Q}$, such as the one you have chosen,
$$
a_n = left( 1 + frac1n right)^{n},
$$
which do not converge in $mathbb{Q}$.
The important theorem here is that every Cauchy sequence in $mathbb{R}$ converges. This is perhaps what you meant by the Cauchy Convergence Theorem? Note the difference between the above statement and the one you have quoted in your post.
So, if you want to know why the Cauchy Convergence Theorem implies that $mathbb{R}$ is complete, it is because completeness is defined that way: a space is said to be complete if every Cauchy sequence in $X$ is convergent. And the Cauchy Convergence Theorem says that $mathbb{R}$ is complete.
This might be unsatisfactory, because it might appear that all we have done is make a clever definition. But I'll be happy to address any concerns if you have in the comments.
answered Dec 3 '18 at 14:58
BrahadeeshBrahadeesh
6,46942363
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add a comment |
$begingroup$
well, if you mean by cauchy convergence theorem that any cauchy sequence converges, then yes, because this is precisely the definition of being complete and $mathbb{R}$ often gets defined as the completion of $mathbb{Q}$. However, your second statement is wrong, since convergence demands a limit point! hence the sequence you wrote down does not converge in $mathbb{Q}$, but it defines a cauchy sequence there! The existence of the limit is essential! so please revise your definition of convergence.
answered Dec 3 '18 at 14:51
EnkiduEnkidu
1,36719
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is a subtlety here regarding convergence that perhaps you have missed. When you say that
In my opinion, the completeness of real numbers refers to that, if a real number sequence is convergent, it is necessarily convergent to a real number.
you are making a slight (but crucial) error.
Firstly, it is improper to suggest that a sequence in a space $X$ converges, but its limit lies outside $X$. Because, if a sequence is convergent then by definition its limit must lie in the same space.
However, there are sequences that appear to converge: their tails lie in intervals of smaller and smaller length, tending to zero. Such sequences are called Cauchy sequences.
Now, a Cauchy sequence in a space $X$ may or may not have a limit in the space $X$. For instance, there are Cauchy sequences in $mathbb{Q}$, such as the one you have chosen,
$$
a_n = left( 1 + frac1n right)^{n},
$$
which do not converge in $mathbb{Q}$.
The important theorem here is that every Cauchy sequence in $mathbb{R}$ converges. This is perhaps what you meant by the Cauchy Convergence Theorem? Note the difference between the above statement and the one you have quoted in your post.
So, if you want to know why the Cauchy Convergence Theorem implies that $mathbb{R}$ is complete, it is because completeness is defined that way: a space is said to be complete if every Cauchy sequence in $X$ is convergent. And the Cauchy Convergence Theorem says that $mathbb{R}$ is complete.
This might be unsatisfactory, because it might appear that all we have done is make a clever definition. But I'll be happy to address any concerns if you have in the comments.
answered Dec 3 '18 at 14:58
BrahadeeshBrahadeesh
6,46942363
$endgroup$
add a comment |
$begingroup$
There is a subtlety here regarding convergence that perhaps you have missed. When you say that
In my opinion, the completeness of real numbers refers to that, if a real number sequence is convergent, it is necessarily convergent to a real number.
you are making a slight (but crucial) error.
Firstly, it is improper to suggest that a sequence in a space $X$ converges, but its limit lies outside $X$. Because, if a sequence is convergent then by definition its limit must lie in the same space.
However, there are sequences that appear to converge: their tails lie in intervals of smaller and smaller length, tending to zero. Such sequences are called Cauchy sequences.
Now, a Cauchy sequence in a space $X$ may or may not have a limit in the space $X$. For instance, there are Cauchy sequences in $mathbb{Q}$, such as the one you have chosen,
$$
a_n = left( 1 + frac1n right)^{n},
$$
which do not converge in $mathbb{Q}$.
The important theorem here is that every Cauchy sequence in $mathbb{R}$ converges. This is perhaps what you meant by the Cauchy Convergence Theorem? Note the difference between the above statement and the one you have quoted in your post.
So, if you want to know why the Cauchy Convergence Theorem implies that $mathbb{R}$ is complete, it is because completeness is defined that way: a space is said to be complete if every Cauchy sequence in $X$ is convergent. And the Cauchy Convergence Theorem says that $mathbb{R}$ is complete.
This might be unsatisfactory, because it might appear that all we have done is make a clever definition. But I'll be happy to address any concerns if you have in the comments.
answered Dec 3 '18 at 14:58
BrahadeeshBrahadeesh
6,46942363
$endgroup$
add a comment |
$begingroup$
There is a subtlety here regarding convergence that perhaps you have missed. When you say that
In my opinion, the completeness of real numbers refers to that, if a real number sequence is convergent, it is necessarily convergent to a real number.
you are making a slight (but crucial) error.
Firstly, it is improper to suggest that a sequence in a space $X$ converges, but its limit lies outside $X$. Because, if a sequence is convergent then by definition its limit must lie in the same space.
However, there are sequences that appear to converge: their tails lie in intervals of smaller and smaller length, tending to zero. Such sequences are called Cauchy sequences.
Now, a Cauchy sequence in a space $X$ may or may not have a limit in the space $X$. For instance, there are Cauchy sequences in $mathbb{Q}$, such as the one you have chosen,
$$
a_n = left( 1 + frac1n right)^{n},
$$
which do not converge in $mathbb{Q}$.
The important theorem here is that every Cauchy sequence in $mathbb{R}$ converges. This is perhaps what you meant by the Cauchy Convergence Theorem? Note the difference between the above statement and the one you have quoted in your post.
So, if you want to know why the Cauchy Convergence Theorem implies that $mathbb{R}$ is complete, it is because completeness is defined that way: a space is said to be complete if every Cauchy sequence in $X$ is convergent. And the Cauchy Convergence Theorem says that $mathbb{R}$ is complete.
This might be unsatisfactory, because it might appear that all we have done is make a clever definition. But I'll be happy to address any concerns if you have in the comments.
answered Dec 3 '18 at 14:58
BrahadeeshBrahadeesh
6,46942363
$endgroup$
There is a subtlety here regarding convergence that perhaps you have missed. When you say that
In my opinion, the completeness of real numbers refers to that, if a real number sequence is convergent, it is necessarily convergent to a real number.
you are making a slight (but crucial) error.
Firstly, it is improper to suggest that a sequence in a space $X$ converges, but its limit lies outside $X$. Because, if a sequence is convergent then by definition its limit must lie in the same space.
However, there are sequences that appear to converge: their tails lie in intervals of smaller and smaller length, tending to zero. Such sequences are called Cauchy sequences.
Now, a Cauchy sequence in a space $X$ may or may not have a limit in the space $X$. For instance, there are Cauchy sequences in $mathbb{Q}$, such as the one you have chosen,
$$
a_n = left( 1 + frac1n right)^{n},
$$
which do not converge in $mathbb{Q}$.
The important theorem here is that every Cauchy sequence in $mathbb{R}$ converges. This is perhaps what you meant by the Cauchy Convergence Theorem? Note the difference between the above statement and the one you have quoted in your post.
So, if you want to know why the Cauchy Convergence Theorem implies that $mathbb{R}$ is complete, it is because completeness is defined that way: a space is said to be complete if every Cauchy sequence in $X$ is convergent. And the Cauchy Convergence Theorem says that $mathbb{R}$ is complete.
This might be unsatisfactory, because it might appear that all we have done is make a clever definition. But I'll be happy to address any concerns if you have in the comments.
answered Dec 3 '18 at 14:58
BrahadeeshBrahadeesh
6,46942363
answered Dec 3 '18 at 14:58
BrahadeeshBrahadeesh
6,46942363
answered Dec 3 '18 at 14:58
BrahadeeshBrahadeesh
6,46942363
answered Dec 3 '18 at 14:58
BrahadeeshBrahadeesh
6,46942363
6,46942363
add a comment |
add a comment |
$begingroup$
well, if you mean by cauchy convergence theorem that any cauchy sequence converges, then yes, because this is precisely the definition of being complete and $mathbb{R}$ often gets defined as the completion of $mathbb{Q}$. However, your second statement is wrong, since convergence demands a limit point! hence the sequence you wrote down does not converge in $mathbb{Q}$, but it defines a cauchy sequence there! The existence of the limit is essential! so please revise your definition of convergence.
answered Dec 3 '18 at 14:51
EnkiduEnkidu
1,36719
$endgroup$
add a comment |
$begingroup$
well, if you mean by cauchy convergence theorem that any cauchy sequence converges, then yes, because this is precisely the definition of being complete and $mathbb{R}$ often gets defined as the completion of $mathbb{Q}$. However, your second statement is wrong, since convergence demands a limit point! hence the sequence you wrote down does not converge in $mathbb{Q}$, but it defines a cauchy sequence there! The existence of the limit is essential! so please revise your definition of convergence.
answered Dec 3 '18 at 14:51
EnkiduEnkidu
1,36719
$endgroup$
add a comment |
$begingroup$
well, if you mean by cauchy convergence theorem that any cauchy sequence converges, then yes, because this is precisely the definition of being complete and $mathbb{R}$ often gets defined as the completion of $mathbb{Q}$. However, your second statement is wrong, since convergence demands a limit point! hence the sequence you wrote down does not converge in $mathbb{Q}$, but it defines a cauchy sequence there! The existence of the limit is essential! so please revise your definition of convergence.
answered Dec 3 '18 at 14:51
EnkiduEnkidu
1,36719
$endgroup$
well, if you mean by cauchy convergence theorem that any cauchy sequence converges, then yes, because this is precisely the definition of being complete and $mathbb{R}$ often gets defined as the completion of $mathbb{Q}$. However, your second statement is wrong, since convergence demands a limit point! hence the sequence you wrote down does not converge in $mathbb{Q}$, but it defines a cauchy sequence there! The existence of the limit is essential! so please revise your definition of convergence.
answered Dec 3 '18 at 14:51
EnkiduEnkidu
1,36719
answered Dec 3 '18 at 14:51
EnkiduEnkidu
1,36719
answered Dec 3 '18 at 14:51
EnkiduEnkidu
1,36719
answered Dec 3 '18 at 14:51
EnkiduEnkidu
1,36719
1,36719
add a comment |
add a comment |
$begingroup$
Can you explain how is your "opinion" differs from the theorem? As far as I understand the theorem also guarantee the existence of a real limit.
$endgroup$
– Yanko
Dec 3 '18 at 14:47
$begingroup$
What exactly is the statement of Cauchy's Convergence Theorem? Is it that every Cauchy sequence in $mathbb{R}$ is convergent?
$endgroup$
– Brahadeesh
Dec 3 '18 at 14:48
1
$begingroup$
You have to philosophically agree that a Cauchy sequence "should" be convergent before completeness really means anything to you. Once you do that, the point is simple: you had $mathbb{Q}$. Some Cauchy sequences in $mathbb{Q}$ don't converge, so you add limits to all the non-convergent Cauchy sequences according to the Cauchy completion construction. You get $mathbb{R}$. Now you have entirely new Cauchy sequences, so you have to ask: do any of them fail to converge in $mathbb{R}$? Completeness is telling you that they don't.
$endgroup$
– Ian
Dec 3 '18 at 14:57