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Let $H$ be a Hilbert space with infinite dimension. Is it possible that there is a finite orthonormal basis $O$ for $H$, i.e. a set whose linear span is dense in $H$? What if $H$ is separable?

This is no homework. Intuitively, I think it is not possible, but could not prove it.

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No. The span of finitely many elements is finite-dimensional and therefore already closed.

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The definition of the "dimension" of Hilbert space is the cardinal number of the maximal orthonormal basis (say basis for simplicity). Every basis has same cardinal number.

If the linear span of some orthonormal subset is $H,$ then this subset is basis.

So your statement would not happen.

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  • $\begingroup$ I meant dimension in the traditional sense. $\endgroup$ – user370967 Mar 28 '18 at 8:22
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As already stated by Andre S., it is not possible that a Hilbert space of infinite dimension contains a finite orthonormal basis. Moreover, an infinite-dimensional Hilbert space is separable if and only if it has a (then necessarily countable) orthonormal basis, see https://en.wikipedia.org/wiki/Hilbert_space#Separable_spaces.

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  • $\begingroup$ This is not true. Also non-separable Hilbert spaces have orthonormal bases (unless you use some non-standard definitions). $\endgroup$ – gerw Mar 27 '18 at 9:33
  • $\begingroup$ Then either we're not talking about the same thing or the quoted Wikipedia site is obviously wrong! ;-) Note that I mentioned that the orthonormal basis is countable. $\endgroup$ – ComplexFlo Mar 27 '18 at 13:32
  • $\begingroup$ The quoted page says "A Hilbert space is separable if and only if it admits a countable orthonormal basis." This is different from your "Moreover, an infinite-dimensional Hilbert space is separable if and only if it has a [...] orthonormal basis [...]" Note that the Wikipedia page also says "every Hilbert space admits an orthonormal basis" (under "Hilbert dimension"). $\endgroup$ – gerw Mar 27 '18 at 15:01

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