Consider some countably infinite sequence of elements $f_n$, each belonging to an infinite dimensional Hilbert space $H$, that are all orthogonal to every other member of the sequence. Is this set closed on $H$?

A comment: For a Hilbert space with finite dimension $k$, it appears straightforward to show that an orthogonal sequence of elements of size $k$ is closed in that space. I was wondering if the infinite-dimensional nature of some Hilbert spaces throws a wrench in that.


$f_n$ could very well converge to zero (without $0$ being one of its elements): consider $e_n$ the canonical Hilbert basis of $\ell^2$ and $f_n=2^{-n}e_n$.

Added: With the additional condition that $\inf\limits_{n\in\Bbb N}\lVert f_n\rVert>0$, the support of the sequence is indeed closed: notice that $$\lVert f_n-f_m\rVert=\sqrt{\lVert f_n\rVert^2+\lVert f_m\rVert^2-2\left\langle f_n;f_m\right\rangle}=\sqrt{\lVert f_n\rVert^2+\lVert f_m\rVert^2}\ge \sqrt2 \inf\limits_{n\in\Bbb N}\lVert f_n\rVert$$ Therefore the sequence itself has no Cauchy subsequences, by which it follows that its range must be a discrete closed subset.

  • $\begingroup$ Good counterexample! Are you familiar with any other counterexamples that don’t involve decay to zero of the sequences’s elements? $\endgroup$ – aghostinthefigures Sep 8 '19 at 21:30
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    $\begingroup$ @aghostinthefigures No, because there are not: the subsequent discussion contains basically all the ingredients to prove that either $\overline{\{f_n\,:\,n\in\Bbb N\}}=\{f_n\,:\,n\in\Bbb N\}$ or $\overline{\{f_n\,:\,n\in\Bbb N\}}=\{f_n\,:\,n\in\Bbb N\}\cup\{0\}$. $\endgroup$ – Gae. S. Sep 8 '19 at 21:32

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