Suppose {xn} is a Cauchy sequence and suppose that there exists a real number L

Question

Suppose {xn} is a Cauchy sequence and suppose that there exists a real number L with the property that for each epsilon > 0 the interval (L - epsilon, L + epsilon) contains infinitely many terms of the sequence {xn}. Prove that {xn} rightarrow L. Use only the assumptions of the problem and the definitions of a Cauchy sequence and the limit of a sequence. Do not use any Theorems or the results of any Exercises in the text or elsewhere. Do not use the concept of a subsequence. Do not prove this result using a contradiction argument.

Explanation / Answer

xn is Cauchy.

For each >0 there exist infinitely many terms of the sequence in the interval (L-,L+)

We need to show xn->L

There exists an N1 such that for all n,m > N1, |xn-xm|</2

For this /2 (L-/2,L+/2) contains infinitely many terms of the sequence. There must exist an N such that xN is contained in the set (L-/2,L+/2) and N>N1 as N<N1 are finite number of terms of the sequence.

Therefore it holds in particular that |xn-xN|</2 and xN lies in (L-/2,L+/2) ie |xN-L|</2

For for all n greater than N1 we must have |xn-xN|</2

This implies that |xn-L| <= |xn-xN|+|xN-L|</2+/2 =

Which basically means that xn-> L

What we did was basically show that since the arbitrarily small interval around L (L-e,L+e) contains infinte terms of the sequence xn and the terms of xn themselves get arbitrarily close to each other, they must bunch up at L, ie the sequence converges to L.

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