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Does anybody has any idea how to proof that a a convergent sequence in Hilbert space is bounded?

Thanks for help, I need this to hopefully get to understand a proof of another theorem.

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Something very similar is true in any metric space. Suppose that $\langle x_n:n\in\Bbb N\rangle$ converges to $x$. Then there is an $m\in\Bbb N$ such that $d(x_n,x)<1$ for all $n\ge m$. Let $u=\max\{d(x_n,x):n<m\}$; then $d(x_n,x)\le u+1$ for all $n\in\Bbb N$. In other words, every term of the sequence lies in the closed ball of radius $u+1$ centred at its limit $x$ and in that sense can be described as bounded.

To finish the argument, just show that in Hilbert space this implies that there is some $M\in\Bbb R$ such that $\|x_n\|\le M$ for all $n\in\Bbb N$; you can describe $M$ in terms of $u$ and $\|x\|$.

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Hint: if $x_n$ is close to $y$, $\|x_n\|$ is close to $\|y\|$. Use the triangle inequality to make this precise.

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