Regarding Real numbers as limits of Cauchy sequences

In summary: The proposition is correct. You just need to correct your proof. I think that you are trying to be too clever with the construction of a sequence ##(a'_n)##. Just use the fact that ##x_n## is a Cauchy sequence. You should be able to use the definition of a Cauchy sequence to show that there exists an ##N## such that ##|x_n - x_m| < \epsilon## for all ##n,m \ge N##. That will give you the bound away from zero.You should also note that the definition of a Cauchy sequence does not require the elements of the sequence to be rational. They can be reals.In summary, the proposition states
  • #1
Terrell
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Homework Statement


Let ##x\in\Bbb{R}## such that ##x\neq 0##. Then ##x=LIM_{n\rightarrow\infty}a_n## for some Cauchy sequence ##(a_n)_{n=1}^{\infty}## which is bounded away from zero.

2. Relevant definitions and propositions:
stackx tao analysis.png

epsilon-closeness.png


3. The attempt at a proof:
Proof:(by construction)
Let ##x\in\Bbb{R}## such that ##x\neq 0##. Then by definition 5.3.1., ##x=LIM_{n\rightarrow\infty}a_n## such that ##(a_n)_{n=1}^{\infty}## is a Cauchy sequence; That is \begin{align}\forall\epsilon\in\Bbb{Q^+}\exists N\in \Bbb{Z^+}\forall j,k\geq N(\vert a_j-a_k\vert\leq\epsilon)\end{align}
and ##(a_n)_{n=1}^{\infty}## is also bounded. Suppose ##x=LIM_{n\rightarrow\infty}a_n## is not bounded away from zero. Thus, ##\exists j\in\Bbb{N}## such that ##a_j=0##.

Consider ##x'=LIM_{n\rightarrow\infty}a'_n## where ##\forall n\in\Bbb{N}(n\neq j\Leftrightarrow a_n=a'_n)##. Consequently due to definition 5.3.4. and Lemma 5.3.6.
\begin{align}x-x'=LIM_{n\rightarrow \infty}a_n - LIM_{n\rightarrow \infty}a'_n=LIM_{n\rightarrow \infty}(a_n-a'_n)\end{align}
Thus, ##\forall n\in\Bbb{N}(n\neq j\Leftrightarrow a_n-a'_n=0)##. Keep in mind that ##x,x'\in\Bbb{R}## and ##\forall\in\Bbb{N},a_n,a'_n\in\Bbb{Q^+}##. Since ##a'_j\neq a_j=0##, then by trichotomy of rationals ##a'_j>0## or ##a'_j<0##. Without loss of generality, suppose ##a'_j>0## then ##\vert a_j-a'_j\vert =\vert 0-a'_j\vert =\vert -a'_j\vert=-(-a'_j)=a'_j##. Hence, ##LIM_{n\rightarrow\infty}(a_n-a'_n)## where ##(a_n-a'_n)_{n=1}^{\infty}## represents the sequence \begin{align}a_1-a'_1=0,...,a_{j-1}-a'_{j-1}=0,a_j-a'_j=a'_j,a_{j+1}-a'_{j+1}=0,0,...\end{align}
Clearly, for all ##\epsilon'\in\Bbb{Q^+}## there exists ##\overline{n}\in\Bbb{Z^+}## such that ##\forall k\geq\overline{n},\vert a_k-a'_k\vert\leq\epsilon'##. Since ##x## is equivalent to ##x'## and the sequence ##(a'_n)_{n=1}^{\infty}## representing ##x'## is bounded away from zero, then ##(a'_n)_{n=1}^{\infty}## is a cauchy sequence representing ##x## that is bounded away from zero because by definition 5.3.1. equivalent cauchy sequences represent the same real number.
 

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  • #2
Terrell said:
Proof:(by contradiction)
Let ##x\in\Bbb{R}## such that ##x\neq 0##. Then by definition 5.3.1., ##x=LIM_{n\rightarrow\infty}a_n## such that ##(a_n)_{n=1}^{\infty}## is a Cauchy sequence; That is \begin{align}\forall\epsilon\in\Bbb{Q^+}\exists N\in \Bbb{Z^+}\forall j,k\geq N(\vert a_j-a_k\vert\leq\epsilon)\end{align}
and ##(a_n)_{n=1}^{\infty}## is also bounded. Suppose ##x=LIM_{n\rightarrow\infty}a_n## is not bounded away from zero. Thus, ##\exists j\in\Bbb{N}## such that ##a_j=0##.

You have tried to show that all sequences ##(a_n)## representing ##x## are bounded away from ##0##. This cannot be true. A simple counterexample is ##(a_n) = 0, 1, 1, 1 \dots##, which clearly is a sequence representing the number ##1##.

The question requires that you show that there exists a Cauchy sequence for ##x## bounded away from ##0##. Not that all such sequences are bounded away from ##0##.

Also, if ##(a_n)## is not bounded away from ##0## it doesn't mean that ##a_j = 0## for any ##j##. For example:

##a_n = 1/n##
 
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  • #3
To add to what @PeroK said, you can take any Cauchy sequence that represents ##x##. If it is bounded away from 0 you are done. If it is not, then you can construct a one that is based on it, which will be a proof of existence.
 
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  • #4
PeroK said:
You have tried to show that all sequences ##(a_n)## representing ##x## are bounded away from ##0##. This cannot be true. A simple counterexample is ##(a_n) = 0, 1, 1, 1 \dots##, which clearly is a sequence representing the number ##1##.

The question requires that you show that there exists a Cauchy sequence for ##x## bounded away from ##0##. Not that all such sequences are bounded away from ##0##.

Also, if ##(a_n)## is not bounded away from ##0## it doesn't mean that ##a_j = 0## for any ##j##. For example:

##a_n = 1/n##
Using these definitions
epsilon-closeness.png

What if I end my argument with "since ##x## is equivalent to ##x'## and the sequence ##(a'_n)_{n=1}^{\infty}## representing ##x'## is bounded away from zero, then ##(a'_n)_{n=1}^{\infty}## is a cauchy sequence representing ##x## that is bounded away from zero because by definition 5.3.1. equivalent cauchy sequences represent the same real number."
 

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  • #5
Orodruin said:
To add to what @PeroK said, you can take any Cauchy sequence that represents ##x##. If it is bounded away from 0 you are done. If it is not, then you can construct a one that is based on it, which will be a proof of existence.
I changed the last paragraph and added some new preliminary definitions. Thank you for the feedback!
 
  • #6
PeroK said:
Also, if ##(a_n)## is not bounded away from ##0## it doesn't mean that ##a_j = 0## for any ##j##. For example:

##a_n = 1/n##

Yes, but if the series does not converge to zero, it does. However, that there exists a ##j## for which ##a_j = 0## does not mean that there does not exist other values of ##n## for which ##a_n = 0## as well. Thus, just removing the zero at ##j## does not guarantee that the new sequence will be bounded away from zero.

To OP: I think you are seriously overthinking this problem. Ask yourself the question if infinite sub-sequences of Cauchy series are also Cauchy series.
 
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  • #7
Orodruin said:
To OP: I think you are seriously overthinking this problem. Ask yourself the question if infinite sub-sequences of Cauchy series are also Cauchy series.
I would not know since I haven't reach that part of the textbook I'm using yet. I would like to follow the text axiomatically so I am avoiding using any theorems that are ahead of the chapter. But do you think my changes to my original proof verifies the proposition to be proven? Thank you! P.S. I am currently using Terrence Tao, Analysis I.
 
  • #8
Terrell said:
I would not know since I haven't reach that part of the textbook I'm using yet. I would like to follow the text axiomatically so I am avoiding using any theorems that are ahead of the chapter. But do you think my changes to my original proof verifies the proposition to be proven? Thank you! P.S. I am currently using Terrence Tao, Analysis I.

Your proof is not repairable and you need to rethink. Post #2 explains why. Note that you will need to use the definition of equivalent Caucky sequences at some point, as that is the key to the condition ##x \ne 0##.
 
  • #9
Terrell said:
I would not know since I haven't reach that part of the textbook I'm using yet.
You must have. It is an easy thing to show from the basic definition of a Cauchy sequence.
 
  • #10
Orodruin said:
You must have. It is an easy thing to show from the basic definition of a Cauchy sequence.

And, in fact, all you need here is that if you chop off the first part of a Cauchy sequence, you still have a Cauchy sequence. That, I would say, is obvious enough that it can be stated without proof.
 
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  • #11
PeroK said:
And, in fact, all you need here is that if you chop off the first part of a Cauchy sequence, you still have a Cauchy sequence. That, I would say, is obvious enough that it can be stated without proof.
I was trying to avoid spelling it out so bluntly ... :rolleyes:
 
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  • #12
PeroK said:
And, in fact, all you need here is that if you chop off the first part of a Cauchy sequence, you still have a Cauchy sequence. That, I would say, is obvious enough that it can be stated without proof.
I have shown this in an earlier exercise. How could this be used?
 
  • #13
Orodruin said:
I was trying to avoid spelling it out so bluntly ... :rolleyes:
sorry I misunderstood because you wrote "series" instead of "sequence".
 
  • #14
Terrell said:
sorry I misunderstood because you wrote "series" instead of "sequence".
Ooops. I do that mistake from time to time. Of course, I intended to write Cauchy sequence, there are no series here.
 
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  • #15
Terrell said:
I have shown this in an earlier exercise. How could this be used?

The first thing you need is a strategy for this proof. What is your strategy here?
 
  • #16
PeroK said:
Your proof is not repairable and you need to rethink. Post #2 explains why. Note that you will need to use the definition of equivalent Caucky sequences at some point, as that is the key to the condition ##x \ne 0##.
I don't understand why "since ##x## is equivalent to ##x'## and the sequence ##(a'_n)_{n=1}^{\infty}## representing ##x'## is bounded away from zero, then ##(a'_n)_{n=1}^{\infty}## is a cauchy sequence representing ##x## that is bounded away from zero because by definition 5.3.1. equivalent cauchy sequences represent the same real number." would not repair the proof :(
 
  • #17
PeroK said:
What is your strategy here?
my strategy is to construct an equivalent sequence for ##x## which is constructed as the sequence of ##x'## as shown above. Then I will show that ##(a_n)_{n=1}^{\infty}## is equivalent to ##(a'_n)_{n=1}^{\infty}##. Then use definition 5.3.1.
 
  • #18
Terrell said:
I don't understand why "since ##x## is equivalent to ##x'## and the sequence ##(a'_n)_{n=1}^{\infty}## representing ##x'## is bounded away from zero, then ##(a'_n)_{n=1}^{\infty}## is a cauchy sequence representing ##x## that is bounded away from zero because by definition 5.3.1. equivalent cauchy sequences represent the same real number." would not repair the proof :(

Post #2 showed that simply removing any zero term from the sequence is not sufficient.
 
  • #19
Terrell said:
my strategy is to construct an equivalent sequence for ##x## which is constructed as the sequence of ##x'## as shown above. Then I will show that ##(a_n)_{n=1}^{\infty}## is equivalent to ##(a'_n)_{n=1}^{\infty}##. Then use definition 5.3.1.

Your strategy was:

Take a Cauchy sequence representing ##x##. Assume it is not bounded away from 0. This implies it has a single zero term. Remove the single zero term from the sequence ...

That is ill-conceived as a sequence that is not bounded away from 0 might have an infinite number of 0 terms; or, it might have none!
 
  • #20
PeroK said:
Post #2 showed that simply removing any zero term from the sequence is not sufficient.
But what I think I have shown is despite ##x## being represented as a sequence ##(a_n)_{n=1}^{\infty}## with a zero term, we can construct an equivalent sequence ##(a'_n)_{n=1}^{\infty}## without zeroes that represents ##x##.
 
  • #21
PeroK said:
might have an infinite number of 0 terms
what if I use the method of constructing ##x'## above then use induction to 'cover' all zeroes? If this still would not work, can I have any other ideas? Thank you!
 
  • #22
Terrell said:
what if I use the method of constructing ##x'## above then use induction to 'cover' all zeroes?

There might be no zeroes in the sequence for ##x##.
 
  • #23
PeroK said:
There might be no zeroes in the sequence for ##x##.
Can I use cases with induction? Would that work?
 
  • #24
Terrell said:
Can I use cases with induction? Would that work?

I think you need to start again. In particular you need to think about:

##x \ne 0##, therefore, ##(a_n)## does not converge to zero. And what you can say about ##(a_n)## from this.
 
  • #25
PeroK said:
I think you need to start again. In particular you need to think about:

##x \ne 0##, therefore, ##(a_n)## does not converge to zero. And what you can say about ##(a_n)##.
It converges to a nonzero number...? And there is a lower bound to all terms in the sequence which could or not be the nonzero number it converges to.
 
  • #26
Terrell said:
It converges to a nonzero number..?

You need to rethink the whole proof.

Also, this whole subject you are studying is a development of the real numbers as Cauchy sequences. Even my hint about not converging to 0 is a bit of a short-cut that doesn't really respect the formality of this mathematics. Technically, one Cauchy sequence does not converge to a number, it is that number.

The only numbers you have at this stage are the rationals. And ##(a_n)## does not necessarily converge to a non-zero rational. All you know is that it doesn't converge to 0.

You need to rethink the proof; otherwise, I'll just be taking you through what would be my proof step at a time.
 
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  • #27
PeroK said:
You need to rethink the proof; otherwise, I'll just be taking you through what would be my proof step at a time.
What general strategy do you have in mind?
 
  • #28
Terrell said:
What general strategy do you have in mind?

I would take a Cachy sequence representing ##x## and use the fact that it doesn't converge to 0 to produce an equivalent Cauchy sequence that is bounded away from zero.

You don't need a proof by contradiction.

Now, it's definitely up to you. I've given you more than enough!
 
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  • #29
PeroK said:
I would take a Cachy sequence representing ##x## and use the fact that it doesn't converge to 0 to produce an equivalent Cauchy sequence that is bounded away from zero.

You don't need a proof by contradiction.

Now, it's definitely up to you. I've given you more than enough!
This will do. Thank you again!
 

Related to Regarding Real numbers as limits of Cauchy sequences

1. What are Cauchy sequences?

A Cauchy sequence is a sequence of real numbers that converges to a limit as the number of terms in the sequence increases. In other words, as the terms in the sequence get closer and closer together, they eventually get arbitrarily close to a single number known as the limit.

2. How are Cauchy sequences related to real numbers?

Real numbers can be seen as limits of Cauchy sequences. This means that every real number can be approximated by a Cauchy sequence, and conversely, every Cauchy sequence has a limit that is a real number.

3. How do Cauchy sequences differ from other types of sequences?

Cauchy sequences are different from other types of sequences because they are defined in terms of the distance between consecutive terms, rather than the terms themselves. This allows for a more abstract and general definition, and makes Cauchy sequences useful in many areas of mathematics.

4. What is the importance of Cauchy sequences in mathematics?

Cauchy sequences are important because they provide a rigorous way to define real numbers and to make sense of their properties. They are also used in many areas of mathematics, such as analysis and topology, to prove theorems and solve problems.

5. Can all real numbers be represented as Cauchy sequences?

Yes, all real numbers can be represented as Cauchy sequences. This is known as the completeness property of the real numbers, which states that every Cauchy sequence has a limit that is a real number. This property is what makes the real numbers a complete and well-defined mathematical system.

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