MHB Is Every Sequence with a Cauchy Subsequence Also Cauchy?

ozkan12
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Let $\left\{{x}_{n}\right\}$ be a sequence...İf $\left\{{x}_{2n}\right\}$ is caucy sequence, can we say that $\left\{{x}_{n}\right\}$ is cauchy sequence ?
 
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ozkan12 said:
Let $\left\{{x}_{n}\right\}$ be a sequence...İf $\left\{{x}_{2n}\right\}$ is caucy sequence, can we say that $\left\{{x}_{n}\right\}$ is cauchy sequence ?

Hi ozkan12,

I don't think so. For example consider the alternating sequence, $\{x_n\}_{n=0}^\infty=\{(-1)^n\}_{n=0}^\infty$. Now $\{x_{2n}\}_{n=0}^\infty$ is a constant sequence which is Cauchy but $x_n$ is not.
 
Hi Sudharaka,

İn some fixed point theorem, to prove that $\left\{{x}_{n}\right\}$ is cauchy sequence, author show that $\left\{{x}_{2n}\right\}$ is cauchy sequence...And in fixed point theorems, we use iteration sequence such that ${x}_{n}=f{x}_{n-1}$...İf we construct $\left\{{x}_{n}\right\}$ in this way, can we say that $\left\{{x}_{n}\right\}$ is cauchy sequence by proving that $\left\{{x}_{2n}\right\}$ is cauchy sequence ?
 
ozkan12 said:
Hi Sudharaka,

İn some fixed point theorem, to prove that $\left\{{x}_{n}\right\}$ is cauchy sequence, author show that $\left\{{x}_{2n}\right\}$ is cauchy sequence...And in fixed point theorems, we use iteration sequence such that ${x}_{n}=f{x}_{n-1}$...İf we construct $\left\{{x}_{n}\right\}$ in this way, can we say that $\left\{{x}_{n}\right\}$ is cauchy sequence by proving that $\left\{{x}_{2n}\right\}$ is cauchy sequence ?

Could you please tell me which book you are referring to so that I can have a look?
 
Sudharaka, I learned this information, Thank you for your attention, best wishes...:)
 
ozkan12 said:
Sudharaka, I learned this information, Thank you for your attention, best wishes...:)

Sure, I have marked the thread as SOLVED. :)
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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