Proof: Every convergent sequence is Cauchy

[zigzagdoom]

Hi,

I am trying to prove that every convergent sequence is Cauchy - just wanted to see if my reasoning is valid and that the proof is correct.

Thanks!

1. Homework Statement

Prove that every convergent sequence is Cauchy

Homework Equations

/ Theorems[/B]

Theorem 1: Every convergent set is bounded

Theorem 2: Every non-empty bounded set has a supremum (through the completeness axiom)

Theorem 3: Limit of sequence with above properties = Sup S (proved elsewhere) Incorrect - not taken as true in second attempt of proof

The Attempt at a Solution

Suppose (s_n) is a convergent sequence with limit L. Let S = {s_n: n∈ℕ}

By Theorem 1, every convergent set is bounded. By Theorem 2, sup S exists, let Sup S = M.

Since (s_n) is convergent, for every ε > 0, ∃N such that ∀n > N, | s_n - L | < ε, where L = M

Now, since M = Sup S, consider some M - φ < M for some 0 < φ < ε.

But then this means ∃s_n such that M - φ < s_n < M.

i.e. s_n + α = M for some α > 0

So M - ε < M - φ < s_n < M.

This is the same as M - ε < M - φ < M - α < M

But then | (M - ε) - (M - φ) | > | (M - φ) - (M - α) |.

I.e. the sequence is Cauchy.

 

[LCKurtz]

Hi,

I am trying to prove that every convergent sequence is Cauchy - just wanted to see if my reasoning is valid and that the proof is correct.

Thanks!

1. Homework Statement

Prove that every convergent sequence is Cauchy

Homework Equations

/ Theorems[/B]

Theorem 1: Every convergent set is bounded

Theorem 2: Every non-empty bounded set has a supremum (through the completeness axiom)

Theorem 3: Limit of sequence with above properties = Sup S (proved elsewhere)

The Attempt at a Solution

Suppose (s_n) is a convergent sequence with limit L. Let S = {s_n: n∈ℕ}

By Theorem 1, every convergent set is bounded. By Theorem 2, sup S exists, let Sup S = M.

Since (s_n) is convergent, for every ε > 0, ∃N such that ∀n > N, | s_n - L | < ε, where L = M

There is no reason to suppose L = M. And your theorem 3 above, whatever it means, is false. And you have not even stated what a Cauchy sequence is, let alone proved that property.

 

[zigzagdoom]

There is no reason to suppose L = M. And your theorem 3 above, whatever it means, is false. And you have not even stated what a Cauchy sequence is, let alone proved that property.

Thanks LCKurtz,

I think i implcitly assumed that the sequence is monotonically increasing. But the Theorem 3 is not true as a a sequence denoted by the sine function would be bounded, yet not convergent I suppose.

I will give it another crack

 

[LCKurtz]

Start by writing the definition of a Cauchy sequence. Then use:

(sn) is convergent [to L if], for every ε > 0, ∃N such that ∀n > N, | sn - L | < ε, where L = M

without the incorrect part in red, to prove it. It has nothing to do with sup.

 

[zigzagdoom]

There is no reason to suppose L = M. And your theorem 3 above, whatever it means, is false. And you have not even stated what a Cauchy sequence is, let alone proved that property.

The Attempt at a Solution

Suppose (s_n) is a convergent sequence with limit L. Let S = {s_n: n∈ℕ}

By Theorem 1, every convergent set is bounded. By Theorem 2, sup S exists, let Sup S = M.

Now, since M = Sup S, consider some M - φ < M for some 0 < φ < ε.

But then this means ∃s_n such that M - φ < s_n < M.

i.e. s_n + α = M for some α > 0

So M - ε < M - φ < s_n < M.

This is the same as M - ε < M - φ < M - α < M

But then | (M - ε) - (M - φ) | > | (M - φ) - (M - α) |.

I.e. the sequence is Cauchy.

 

[LCKurtz]

No. See my post #4 which I apparently posted the same time you were posting #5.

 

[zigzagdoom]

Start by writing the definition of a Cauchy sequence. Then use:

(sn) is convergent [to L if], for every ε > 0, ∃N such that ∀n > N, | sn - L | < ε, where L = M

without the incorrect part in red, to prove it. It has nothing to do with sup.

Attempt

A sequence is called a Cauchy sequence if the terms of the sequence eventually all become arbitrarily close to one another

That is, given ε > 0 there exists N such that if m, n > N then | a_m - a_n | < ε

(s_n) is convergent to L if for every ε > 0, ∃N such that ∀n > N, | s_n - L | < ε

which implies s_n < ε + L

Now consider three consecutive numbers in the sequence s_n, s_o, s_p. It follows that

s_n < ε + L

s_o < α + L

s_p < β + L

where β ≤ α ≤ ε by notion of convergence.

it then follows that the distance between s_n and s_o; | s_n - s_o | = | ε - α | ≥ | s_o - s_p | = | α - β |.

Hence the terms get closer and closer, so the sequence is Cauchy.

 

[zigzagdoom]

No. See my post #4 which I apparently posted the same time you were posting #5.

Thanks,

Would the argument I posted before be incorrect? I essentially just used the supremum as a point of reference to gauge distance between points.

 

[LCKurtz]

None of your arguments look good, sorry to say. You have the definitions you have given earlier (I have edited them slightly for clarity):

The sequence $\{s_n\}$ is convergent to L if for every ε > 0, ∃N such that ∀n > N, $|s_n-L|<\epsilon$. That is what you are given to work with.

And the sequence $\{s_n\}$ is Cauchy means that given ε > 0 there exists N such that if m, n > N then $|s_m-s_n|<\epsilon$. That is what you have to prove.

Now if you think about that, you are informally saying that $s_n$ gets close to L for large n because the sequence converges to it. So for large n and m, both $s_n$ and $s_m$ get close to L. That is what you are given. Now, if they both get close to L, wouldn't they have to get close to each other? Intuitively it makes sense. Your problem is to write down the appropriate inequalities to prove it. I hope that's enough of a hint to get you going.

 

[zigzagdoom]

None of your arguments look good, sorry to say. You have the definitions you have given earlier (I have edited them slightly for clarity):

The sequence $\{s_n\}$ is convergent to L if for every ε > 0, ∃N such that ∀n > N, $|s_n-L|<\epsilon$. That is what you are given to work with.

And the sequence $\{s_n\}$ is Cauchy means that given ε > 0 there exists N such that if m, n > N then $|s_m-s_n|<\epsilon$. That is what you have to prove.

Now if you think about that, you are informally saying that $s_n$ gets close to L for large n because the sequence converges to it. So for large n and m, both $s_n$ and $s_m$ get close to L. That is what you are given. Now, if they both get close to L, wouldn't they have to get close to each other? Intuitively it makes sense. Your problem is to write down the appropriate inequalities to prove it. I hope that's enough of a hint to get you going.

OK so from the above, I would do as follows:

Third Attempt

Since $|s_n-L| < \epsilon$

$|s_n| < |L + \epsilon|$

$|s_n - s_m| < |L + \epsilon - s_m|$

$| (s_n - L) - (s_m - L) | < |L + \epsilon - s_m|$

But $|s_n-L| < \epsilon$ and $|s_m-L| < \epsilon$ as the both $s_m$ and $s_n$ converge to L, for $∀m,n$

So $| (s_n - L) - (s_m - L) | < 2\epsilon$ (by triangle inequality)

i.e. $| s_n - s_m | < 2\epsilon$, that is $s_n$ and $s_m$ get arbitrarily close. The sequence is therefore Cauchy.

 

[Samy_A]

OK so from the above, I would do as follows:

Third Attempt

Since $|s_n-L| < \epsilon$

$|s_n| < |L + \epsilon|$

$|s_n - s_m| < |L + \epsilon - s_m|$

This last inequality isn't necessarily true.
Take $s_n=L-\frac{9}{10}\epsilon$ and $s_m=L+\frac{9}{10}\epsilon$
Then $|s_n - s_m|=\frac{18}{10}\epsilon$, and $|L + \epsilon - s_m|=\frac{1}{10}\epsilon$

But it doesn't really matter because you don't use that inequality.

The core argument is here:

But $|s_n-L| < \epsilon$ and $|s_m-L| < \epsilon$ as the both $s_m$ and $s_n$ converge to L, for $∀m,n$

So $| (s_n - L) - (s_m - L) | < 2\epsilon$ (by triangle inequality)

i.e. $| s_n - s_m | < 2\epsilon$, that is $s_n$ and $s_m$ get arbitrarily close. The sequence is therefore Cauchy.

Now you have to word this correctly (for example, writing "both $s_m$ and $s_n$ converge to L" is sloppy).

 

[zigzagdoom]

This last inequality isn't necessarily true.
Take $s_n=L-\frac{9}{10}\epsilon$ and $s_m=L+\frac{9}{10}\epsilon$
Then $|s_n - s_m|=\frac{18}{10}\epsilon$, and $|L + \epsilon - s_m|=\frac{1}{10}\epsilon$

But it doesn't really matter because you don't use that inequality.

The core argument is here:

Now you have to word this correctly (for example, writing "both $s_m$ and $s_n$ converge to L" is sloppy).

Thanks Samy,

I did not realize a sequence can converge to a Limit from both the left and the right - will keep that handy fact in mind.

So in terms of wording I would approach as follows;

$∀\epsilon > 0, ∃N ∋$ $∀n,m > N, | s_n - L | < \epsilon$ and $| s_m - L | < \epsilon$.

By the trianle inequality, $| s_n - L + L - s_m | ≤ | s_n - L | + | s_m - L | < \epsilon + \epsilon = 2\epsilon$

Since $2\epsilon$ is arbitrarily large, we can make both $s_n$ and $s_m$ as close together as we like, for $∀n,m > N$.

Therefore, the sequence is Cauchy.

Would this be a more rigid argument?

 

[Samy_A]

So in terms of wording I would approach as follows;

$∀\epsilon > 0, ∃N ∋$ $∀n,m > N, | s_n - L | < \epsilon$ and $| s_m - L | < \epsilon$.

By the trianle inequality, $| s_n - L + L - s_m | ≤ | s_n - L | + | s_m - L | < \epsilon + \epsilon = 2\epsilon$

Since $2\epsilon$ is arbitrarily large, we can make both $s_n$ and $s_m$ as close together as we like, for $∀n,m > N$.

Therefore, the sequence is Cauchy.

Would this be a more rigid argument?

There is a typo, as you write "arbitrarily large", while you clearly mean "arbitrarily small".
One can nitpick about details, but I think this proof is correct.

 

[zigzagdoom]

There is a typo, as you write "arbitrarily large", while you clearly mean "arbitrarily small".
One can nitpick about details, but I think this proof is correct.

Thank you for the help

 

[LCKurtz]

And if you want to spiff it up a little, pick N so that if n,m > N then $|s_n-L|<\frac \epsilon 2$ and $|s_m-L|<\frac \epsilon 2$ in the first place, so $|s_m-s_n|<\epsilon$.