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Theorem 3.2.2: Completeness Theorem in R

Let be a Cauchy sequence of real numbers. Then the sequence is bounded.

Let be a sequence of real numbers. The sequence is Cauchy if and only if it converges to some limit a.

Proof:

The proof of the first statement follows closely the proof of the corresponding result for convergent sequences. Can you do it ?

To prove the second, more important statement, we have to prove two parts:

First, assume that the sequence converges to some limit a. Take any > 0. There exists an integer N such that if j > N then | aj - a | < /2. Hence:

| aj - ak | | aj - a | + | a - ak| < 2 / 2 =
if j, k > N. Thus, the sequence is Cauchy.

Second, assume that the sequence is Cauchy (this direction is much harder). Define the set

S = {x R: x < aj for all j except for finitely many}
Since the sequence is bounded (by part one of the theorem), say by a constant M, we know that every term in the sequence is bigger than -M. Therefore -M is contained in S. Also, every term of the sequence is smaller than M, so that S is bounded by M. Hence, S is a non-empty, bounded subset of the real numbers, and by the least upper bound property it has a well-defined, unique least upper bound. Let
a = sup(S)
We will now show that this a is indeed the limit of the sequence. Take any > 0 , and choose an integer N > 0 such that
| aj - ak | < / 2
if j, k > N. In particular, we have:
| aj - aN + 1 | < / 2
if j > N, or equivalently
- / 2 < aj - aN + 1 < / 2
Hence we have:
aj > aN + 1 - / 2
for j > N. Thus, aN + 1 - / 2 is in the set S, and we have that
a aN + 1 - / 2
It also follows that
aj < aN + 1 + / 2
for j > N. Thus, aN + 1 + / 2 is not in the set S, and therefore
a aN + 1 + / 2
But now, combining the last several line, we have that:
|a - aN + 1 | < / 2
and together with the above that results in the following:
| a - aj | < |a - aN + 1 | + | aN + 1 - aj | < 2 / 2 =
for any j > N.

Contributed to this page: Lakshmi Natarajan

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