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Hi I am reading a proof in my functional analysis notes and there is a step I don't really understand;

Since H (infinite dimensional Hilbert space) is separable it contains a countable sense subset $\{g_n\}_{n=1}^{\infty}$ and we have that $span(\{g_n\}_{n=1}^{\infty})=H$

Why does the span equal the whole set? I understand that H is the closure of $\{g_n\}_{n=1}^{\infty}$ so anything not in this set is a limit point of some sequence in the set but since the span is the set of finite linear combinations I don't see how this includes all the limit points. Or is it a typo and should be the closure of the span?

Jonas Meyer
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th0masb
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  • That certainly doesn't have to be the case. Perhaps the notes defined span not only to be the set of linear combinations, but the closure of that set!? – Nathanson Mar 30 '15 at 14:38
  • Ok in that case I think its a typo, should have a bar across to indicate the closure – th0masb Mar 30 '15 at 14:39
  • Damn typos, they are the bane of my life. – th0masb Mar 30 '15 at 14:40
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    Probably. When studying Hilbert spaces, they are going to define Hilbert space basis, as opposed to vector space basis or Hamel basis. For Hilbert space bases the sense in which they generate the space is via $\overline{span()}$. It is possible the note had defined span to take the closure directly. – Nathanson Mar 30 '15 at 14:42

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No, it is not possible that an infinite dimensional Hilbert space is the linear span of a countable subset.

It is obvious that the whole space is the smallest closed subspace containing a dense subset. So as you indicated in a comment, it could be a typo where $\overline{\mathrm{span}}$ was intended instead of $\mathrm{span}$, to indicate that one must take the closure of the linear span. Or the author simply means closed linear span where they write "span." You'll have to consult the notes or the author to know for sure.

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Jonas Meyer
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