1. Suppose we have a sequence {xk} with x1 ≤ x2 ≤ x3 ≤ x4 · · · and we also have an upper bound M with xk ≤ M for all xk. Prove that this sequence must have a limit. Hint: you must use the completeness axiom.

2. We say that a number C is a cluster point of a sequence {xk} if any interval (C − , C + ) contains infinitely many sequence elements.

(a). Explain the difference between a limit and a cluster point.

(b). Provide an example of a sequence with more than one cluster point.

3. We will now prove that every Cauchy sequence has a limit on the Real number line. Try this:

(a) Prove that a Cauchy sequence is bounded. The proof is similar to the proof of theorem 1.7, that a convergent sequence is bounded.

(b) Because it is bounded, the Cauchy sequence has a cluster point. Prove that it can’t have more than one. The proof is similar to the proof of theorem 1.6, that a convergent sequence can’t have more than one limit.

Note:

We can prove that a Cauchy sequence has a limit, but this requires us to show that a bounded sequence has a cluster point. This requires us to show that a sequence of shrinking nested closed intervals has a single point as an intersection. That requires us to show that a bounded nondecreasing sequence has a limit, and that requires the completeness axiom. Therefore the completeness of the real number line is a consequence of the completeness axiom—without that axiom, we could prove none of these things.

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