Regarding Real numbers as limits of Cauchy sequences

[Terrell]

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:

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.

 

[PeroK]

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:

$a_n = 1/n$

 

[Orodruin]

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.

 

[Terrell]

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

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."

 

[Terrell]

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!

 

[Orodruin]

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.

 

[Terrell]

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.

 

[PeroK]

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 $x \ne 0$.

 

[Orodruin]

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.

 

[PeroK]

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.

 

[Orodruin]

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 ...

 

[Terrell]

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?

 

[Terrell]

I was trying to avoid spelling it out so bluntly ...

sorry I misunderstood because you wrote "series" instead of "sequence".

 

[Orodruin]

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.

 

[PeroK]

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?

 

[Terrell]

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 :(

 

[Terrell]

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.

 

[PeroK]

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.

 

[PeroK]

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!

 

[Terrell]

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 $x$.

 

[Terrell]

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!

 

[PeroK]

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 $x$.

 

[Terrell]

There might be no zeroes in the sequence for $x$.

Can I use cases with induction? Would that work?

 

[PeroK]

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.

 

[Terrell]

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.

 

[PeroK]

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.

 

[Terrell]

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?

 

[PeroK]

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!

 

[Terrell]

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!