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In a Hilbert space, if we want to show that a sequence $\;(y_n)\;$ hasn't a convergent subsequence , we show that $\;(y_n)\;$ isn't Cauchy sequence. I see this in many proofs but I 'm not sure I totally understand it. I guess it's quite trivial what I am about to ask but why does Cauchy property imply the above?

My approach:

By definition, since $\;(y_n)\;$ isn't Cauchy sequence there are $\;m,n \in \mathbb N \;$ and $\;n_0 \equiv n_0(n,m)\;$ such that $\;\forall m,n \ge n_0\;$ and $\;m \neq n\;$ : $\; \vert \vert y_n - y_m \vert \vert \ge M\;$ for $\;M\;$ positive constant. If I consider $\;n=k_n\;$ and $\;m=k_{n+1}\;$ then it follows $\;(y_n)\;$ cannot have a convergent subsequence.

EDIT: By $\;(y_n)\;$ hasn't a convergent subsequence I mean that any subsequence of $\;y_n\;$ isn't convergent.

Is this right or I missed something? Any help would be valuable.

Thanks in advnace!

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  • $\begingroup$ It's unclear what you're asking. By "hasn't a convergent subsequence" do you mean "there exists a subsequence that does not converge" or "all subsequences do not converge?" $\endgroup$ – user217285 Jun 1 '17 at 21:10
  • $\begingroup$ ??? There are a lot of non-Cauchy sequences that have convergent sub-sequences. Not sure what you're trying to say here.... $\endgroup$ – Paul Jun 1 '17 at 21:10
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    $\begingroup$ Take two elements $x_0 \ne x_1$ and define the sequence as $x_i = x_{\mod(i,2)}$. This is not a Cauchy sequence but there are of course convergent subsequences. $\endgroup$ – Hans Engler Jun 1 '17 at 21:15
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    $\begingroup$ Do you mean if $\| x_n-x_m\| \geq C >0$ then $x_n$ has no convergent subsequence? $\endgroup$ – N. S. Jun 1 '17 at 21:54
  • $\begingroup$ @N.S. Yes! Exactly this... I don't understand it $\endgroup$ – kaithkolesidou Jun 1 '17 at 21:56
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Let $x_n$ be a sequence and $C>0$ be so that for all $m \neq n$ we have $$\| x_n -x_m \| \geq C$$

Then $x_n$ has no converging subsequence.

Proof Assume by contradiction that $x_n$ has a convergent subsequence $x_{k_n}$. Then $x_{k_n}$ is Cauchy.

Pick some $0 < \epsilon <C$. Then, there exists some $N$ so that for all $m,n >N$ we have $$\| x_{k_n} -x_{k_m} \| < \epsilon <C$$

But if $m \neq n$ we also have by assumption $$\| x_{k_n} -x_{k_m} \| \geq C$$ Contradiction.

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  • $\begingroup$ This is exactly what I had in my mind in first place! Thank you very much! So the statement $\;x_n \;$ isn't Cauchy sequence $\; \Rightarrow \;$ $x_n\;$ has no converging subsequence , is true, isn't it? $\; \| x_n -x_m \| \geq C \;$ doesn't imply $\;x_n\;$ isn't Cauchy? I 'm trying to fully understand it, because in the comments section this statement was proven wrong.. $\endgroup$ – kaithkolesidou Jun 1 '17 at 22:06
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    $\begingroup$ @kaithkolesidou No. For example $x_n=(-1)^n$ isn't Cauchy but does have convergent subsequences. The given condition is must stronger than "not Cauchy". $\endgroup$ – N. S. Jun 1 '17 at 22:33
  • $\begingroup$ Oh, I had misunderstood this. Thank you. You've been very helpful!! $\endgroup$ – kaithkolesidou Jun 4 '17 at 7:34
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It is difficult to decipher exactly what you are asking. You should prove that if $(x_n)$ has no convergent subsequences, then $(x_n)$ is not Cauchy. However, in your title and in your attempt, you seem to have these flipped; that is, it seems you are trying to prove that if a sequence isn't Cauchy, then it has no convergent subsequence. The latter is false of course, as can be seen by $x_n = 0$ for $n$ even and $x_n = 1$ for $n$ odd; this sequence is not Cauchy but has convergent subsequences.

The former is true. In a Hilbert space, a sequence is Cauchy iff it is convergent. If a sequence is convergent then every subsequence is also convergent with the same limit. Conversely, if no subsequence converges, then the sequence itself cannot converge and thus cannot be Cauchy.

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