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Let $X$ be a Banach space, $X_n$ be a normed space. Let $(v_n)$ be a Cauchy sequence $X$ which diverges. Prove that every subsequence of $(v_n)$ diverges.

First of all, I have a hard time believing that a Cauchy sequence can be divergent at all. But I have found one post about this (Can we find a divergent Cauchy sequence?) and it is said that if our space is finite dimentional, then we cannot construct any Cauchy sequence that would be divergent. I assume that in a normed space it is possible (right?).

Anyway, how should I prove such preposition?

And for those who downvote this question without explanations: this question is from the book "Lectures in Functional Analaysis" by Roman Vershynin, page 17, Exercise 1.3.7(1).

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Karagum
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1 Answers1

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See Cauchy Sequence that Does Not Converge for some examples.

Anyways, the claim is easy if you just look at the definitions.

Method 1. If $(v_k)$ diverges, by definition, it fails to converge to $x$ for all $x\in X$. Hence, for all $x\in X$, $\exists \epsilon_x>0$ such that $\forall k$, there exists $n_k>k$ such that $||v_{n_k}-x||\geq \epsilon_x$.

The reason the claim holds for subsequences $(v_{m_k})$ is because in the above definition, just replace $k$ with $m_k$. Integers of the form $m_k$ are included in the "$\forall k$" above.

Method 2. Suppose not. That is, there's a subsequence that converges. Then it's an easy exercise (using triangle inequalities) to show that a Cauchy sequence with a convergent subsequence must converge (even in incomplete spaces, where this isn't always the case). Indeed, this very fact is a standard way to show the reals are complete. Typically, one shows that Cauchy sequences are bounded (true in incomplete spaces), then shows bounded sequences have a convergent subsequence (true in the reals, but not necessarily other spaces), then shows that the original sequence converges to the subsequential limit (if it exists, this is true in incomplete spaces).

ProfOak
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