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Every convergent sequence is Cauchy. Every Cauchy sequence in Rk is convergent, but this is not true in general, for example within S= {x:x€R, x>0} the Cauchy sequence (1/n) has no limit in s since 0 is not a member of S.

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Q: Is every cauchy sequence is convergent?
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Any convergent sequence is a Cauchy sequence is converse true?

no converse is not true


Prove that every convergent sequence is a Cauchy sequence?

The limits on an as n goes to infinity is aThen for some epsilon greater than 0, chose N such that for n>Nwe have |an-a| < epsilon.Now if m and n are > N we have |an-am|=|(am -a)-(an -a)|< or= |am -an | which is < or equal to 2 epsilor so the sequence is Cauchy.


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The Cauchy constant, also known as the Cauchy sequence property, tells us that a sequence is convergent if it is a Cauchy sequence. This means that for any arbitrarily small positive number ε, there exists an index after which all elements of the sequence are within ε distance of each other. It is a key property in the study of convergence in mathematics.


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Why set of rational numbers is not complete?

Consider the sequence (a_i) where a_i is pi rounded to the i_th decimal place. This sequence clearly contains only rational numbers since every number in it has a finite decimal expansion. Furthermore this sequence is Cauchy since a_i and a_j can differ at most by 10^(-min(i,j)) or something which can be made arbitrarily small by choosing a lower bound for i and j. Now note that this sequence converges to pi in the reals, so it can not converge in the set of rational numbers. Therefore the rational numbers allow a non-convergent Cauchy sequence and are thus by definition not complete.


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