We will now prove that every Cauchy sequence has a limit on the Real number line

Question

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. (c) Prove that if a bounded sequence has only one cluster point C, then it must be a limit. Hint: if you remove all the sequence elements within an interval of C, and you have infinitely many elements left over, isn't that still a bounded infinite sequence?

Explanation / Answer

a)we know that every cauchy sequence is convergent.and all convergent sequences are bounded.hence cauchy sequences is bounded.

b)we know a result that a sequence is convergent iff it is bounded and a unique limit point.all cauchy sequences are convergent.hence all cauchy sequences has a unique limit point.

c)if we remove all sequences elements within an interval of C then we get a bounded infinite sequence by bolzano weistrass theorem every bounde infinite sequence has a unique limit point.therefore the cluster point become limit

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