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!
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.