3.2. Cauchy Sequences
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 | < / 2if j, k > N. In particular, we have:
| aj - aN + 1 | < / 2if j > N, or equivalently
- / 2 < aj - aN + 1 < / 2Hence we have:
aj > aN + 1 - / 2for j > N. Thus, aN + 1 - / 2 is in the set S, and we have that
a aN + 1 - / 2It also follows that
aj < aN + 1 + / 2for j > N. Thus, aN + 1 + / 2 is not in the set S, and therefore
a aN + 1 + / 2But now, combining the last several line, we have that:
|a - aN + 1 | < / 2and 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