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By learning these material it is very easy to understand the topic.

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

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  • October 29, 2022
  • 5
  • 2021/2022
  • Class notes
  • Devadath
  • Cauchy sequences
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Course Material 2.4 Cauchy Sequences.

We know that if a sequence{𝑥𝑛 } converges to some point say 𝑥, then |𝑥𝑛 − 𝑥 | → 0
or the distance between 𝑥𝑛 and 𝑥 becomes smaller for sufficiently large values of 𝑛. Then
what happens to the distance between the terms of the sequence say 𝑥𝑛 and 𝑥𝑚 for
sufficiently large values of 𝑚 and 𝑛. We may observe that |𝑥𝑛 − 𝑥𝑚 | will also become
smaller for sufficiently large values of 𝑚 and 𝑛. This is the concept behind a Cauchy
sequence, named after a famous (French) mathematician Augustin-Louis Cauchy, who
was a pioneer in Real Analysis and Complex Analysis.

Definition -Cauchy Sequence

A real sequence {𝑥𝑛 } is said to be a Cauchy Sequence if for each 𝜖 > 0 there exists a
positive integer 𝑁 such that for all 𝑛 ≥ 𝑁 and 𝑚 ≥ 𝑁, |𝑥𝑛 − 𝑥𝑚 | < 𝜖.

If {𝑥𝑛 } is a Cauchy Sequence, we may say that {𝑥𝑛 } is Cauchy

Theorem

Every convergent sequence is a Cauchy sequence.

Proof. Suppose {𝑥𝑛 } converges to the limit say 𝑥. Then for each 𝜖 > 0 there exists a
𝜖
positive integer 𝑁 such that for all 𝑛 ≥ 𝑁 , |𝑥𝑛 − 𝑥 | < 2.

𝜖
Then for all 𝑚 ≥ 𝑁 also we have |𝑥𝑚 − 𝑥 | < 2.

Then |𝑥𝑛 − 𝑥𝑚 | = |𝑥𝑛 − 𝑥 + 𝑥 − 𝑥𝑚 | ≤, |𝑥𝑛 − 𝑥 | + |𝑥𝑚 − 𝑥 |

𝜖 𝜖
So that for all 𝑛 ≥ 𝑁 and 𝑚 ≥ 𝑁 we have |𝑥𝑛 − 𝑥𝑚 | < 2 + 2 = 𝜖

Thus proving that {𝑥𝑛 } is a Cauchy Sequence.

Remark:

It is an important question whether the converse of this result is true. That is, is it true
that every Cauchy sequence is convergent? In the space ℝ of real numbers it is true. This
may not be true in other metric spaces. If a metric space 𝑋 satisfies this property, we say
that 𝑋 is complete. Thus the real line ℝ is a complete metric space.

, Theorem

A real sequence {𝑥𝑛 } is convergent if and only if it is a Cauchy Sequence

Proof.

Let {𝑥𝑛 } be a real sequence. Then if {𝑥𝑛 } is convergent then by above theorem it is a
Cauchy Sequence.

To prove the converse, let {𝑥𝑛 } be a Cauchy Sequence. Then given 𝛿, there exists a positive
integer 𝑛0 = 𝑛0 (𝛿) such that for all 𝑛 ≥ 𝑛0 and 𝑚 ≥ 𝑛0 , |𝑥𝑛 − 𝑥𝑚 | < 𝛿

In particular for all 𝑛 ≥ 𝑛0 we have |𝑥𝑛 − 𝑥𝑛0 | < 𝛿 or 𝑥𝑛0 − 𝛿 < 𝑥𝑛 < 𝑥𝑛0 + 𝛿 …(1)

Suppose 𝐸 = {𝑥 ∈ ℝ: 𝑇ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝑁 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 ≥ 𝑁, 𝑥 < 𝑥𝑛 }.

Then by (1), we have 𝑥𝑛0 − 𝛿 ∈ 𝐸. Hence 𝐸 is non empty.

We claim that 𝑥𝑛0 + 𝛿 is an upper bound of 𝐸. If not, there exists 𝑥 ∈ 𝐸 such that
𝑥 > 𝑥𝑛0 + 𝛿. Then again by (1), we have 𝑥𝑛 < 𝑥𝑛0 + 𝛿 < 𝑥 for all 𝑛 ≥ 𝑛0 ……….(2)

But as 𝑥 ∈ 𝐸, by definition of 𝐸 there exists some 𝑛1 such that for all 𝑛 ≥ 𝑛1 , 𝑥 < 𝑥𝑛

So that, if 𝑁 = 𝑚𝑎𝑥 {𝑛0 , 𝑛1 } for all 𝑛 ≥ 𝑁

We have, 𝑥𝑛 < 𝑥𝑛0 + 𝛿 by (1) and 𝑥𝑛 < 𝑥𝑛0 + 𝛿 < 𝑥 by (2). This is a contradiction
proving that 𝑥𝑛0 + 𝛿 is an upper bound of 𝐸

Hence the set 𝐸 is non empty and is bounded above. Then 𝐸 has a least upper
bound, say ℓ

We now claim that the sequence {𝑥𝑛 } converges to ℓ. To prove this, as {𝑥𝑛 } is a Cauchy
Sequence, given 𝜖 > 0 there exists a positive integer 𝑛0 = 𝑛0 (𝜖) such that

𝜖 𝜖 𝜖
|𝑥𝑛 − 𝑥𝑛0 | < and hence, 𝑥𝑛0 − 2 < 𝑥𝑛 < 𝑥𝑛0 + 2 for all 𝑛 ≥ 𝑛0
2

𝜖 𝜖
This implies that 𝑥𝑛0 − 2 ∈ 𝐸 so that 𝑥𝑛0 − 2 < ℓ

𝜖
But as 𝑥𝑛0 + 𝛿 is an upper bound of 𝐸, we have ℓ ≤ 𝑥𝑛0 + .
2

𝜖 𝜖 𝜖
So that 𝑥𝑛0 − 2 < ℓ <𝑥𝑛0 + 2. or |𝑥𝑛0 − ℓ| < 2

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