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.