Every real Cauchy sequence is convergent. Proof: Exercise. (2) A Cauchy sequence that has a convergent subsequence is itself convergent. (c)The sequence (a n) converges to A Solution. Note: a sequence s(n) is said to be contractive if there exists a constant k with 0 < k < 1 such that ls(n + 2) - s(n + 1)l <= ls(n+1) - s(n)l for all n E N. So thinking of real numbers in terms of Cauchy sequences really does make sense. A sequence (a n) converges to Aif, for every positive number , there exists an N2N such that whenever n N it follows that ja n Aj< . (ii) Every convergent sequence in Xis Cauchy. Therefore $\left ( \frac{1}{n} \right )$ is a Cauchy sequence. Every convergent sequence (with limit s, say) is a Cauchy sequence, since, given any real number ε > 0, beyond some fixed point, every term of the sequence is within distance ε/2 of s, so any two terms of the sequence are within distance ε of each other. Because the Cauchy sequences are the sequences whose terms grow close together, the fields where all Cauchy sequences converge are the fields that are not ``missing" any numbers. Let the sequence be (a n).By the above, (a n) is bounded.By Bolzano-Weierstrass Proof. Give an example to show that the converse of lemma 2 is false. Theorem 1: Let $(M, d)$ be a metric space. Cauchy sequences are useful because they give rise to the notion of a complete field, which is a field in which every Cauchy sequence converges. Example 5: The closed unit interval [0;1] is a complete metric … A subsequence is a sequence made from selecting certain terms of the sequence to make a new one, like all the odd terms, or all the evens, or every third one, etc. 9.5 Cauchy =⇒ Convergent [R] Theorem. Things that get closer and closer to some flagpole necessarily get closer and closer to each other. (1) Every Cauchy sequence of real or complex numbers is bounded. taking \every Cauchy sequence of real numbers converges" to be the Completeness Axiom, and then proving that R has the LUB Property. (a) Let (an) Be A Cauchy Sequence Such That An # 0 For Every N E N. Is It Always True That (1/an) Is A Cauchy Sequence? (i) Every Cauchy sequence is bounded. (d)The sequence (b n) is a Cauchy sequence Solution. Prove that every contractive sequence is a Cauchy sequence, and hence is convergent.? We will now look at an important result which will say that if $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence then it is bounded. A sequence (xn) in a metric space (X,d) is said to be Cauchy if for every ε>0, there exists Nsuch that d(xn,xm) ≤ ε for all n,m≥ N. Proposition 5.7. Theorem: The normed vector space Rn is a complete metric space. If $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence then $(x_n)_{n=1}^{\infty}$ is also bounded. Let (X,d) be a metric space. Conversely, every real number comes with a Cauchy sequence of rational numbers of which it is the limit (for example, the sequence you get from the decimal expansion of a number, like the one for in the example above, is always a Cauchy sequence). The sequence $((-1)^n)$ provided in Example 2 is bounded and not Cauchy. Assume that (xn) converges to x. Proof of (ii). Definition 5.6. Example 4. (b) Prove That If (an) Is A Cauchy Sequence And There Exists A Real Number M > 0 Such That An > M For Every N E N, Then (1/an) Is A Cauchy Sequence. Give A Proof Or A Counter Example. The converse of lemma 2 says that "if $(a_n)$ is a bounded sequence, then $(a_n)$ is a Cauchy sequence of real numbers."
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