Orthonormal set and Hilbert Spaces I am studying Hilbert spaces for the first time and I would appreciate any help.
Problem statement:

Suppose we are given a finite orthonormal set and we take the span. Is this span (equipped with a usual dot product) a Hilbert space?

If the orthonormal set is instead countable and we take the closure of the span, is this a Hilbert space?
My Question:
If my understanding is correct, I believe the answer is yes to both. My issue is that I have trouble with showing the completeness requirement in both cases.
Request: Can someone demonstrate how to show the completeness? Thank you for the help.
 A: Al finite dimensional nomed linear space are complete and this answers the first question.
For  the second question the answer depends on where you are taking the closure. If you are taking the closure in an incomplete inner product space the answer is negative but if the closure is taken in a Hilbert space then the anser is YES.
Counter-example: Consider the space of sequences $(a_n)$ of real numbers which have at most finitely many non-zero elements. This is an inner product space under the inner product $\langle x, y \rangle =\sum x_ny_n$ where $x=(x_n), y=(y_n)$. Let $e_1,e_2,...$ be the standard orthonormal set: $e_n$ has $1$ in the $n-$th place and $0$ elsewhere. Then the closure of the span of this orthonormal set is not complete.
A: Any finite dimensional inner product space $V$ over $\mathbb{C}$ is isomorphic as an inner product space to $\mathbb{C}^n$. To prove this, send an orthonormal basis of $V$ to an orthonormal basis of $\mathbb{C}^n$. It is an easy consequence of the completeness of $\mathbb{C}^n$ and this result that $V$ is complete, i.e. $V$ is a Hilbert space.
Edit: Kavi Rama Murthy showed in his answer that the closure of the span of a countable orthonormal set in an inner product space $V$ need not be complete. If $V$ is complete, i.e. $V$ is a Hilbert space, then the closure of any subset of $V$ is complete. In fact, if $X$ is a complete metric space and $A \subset X$ is closed, then $A$ is complete. Proof: If $x_n \in A$ and $(x_n)$ is Cauchy, then by completeness of $X$, $x_n \to x$ for some $x \in X$. Since $A$ is closed, $x \in A$.
A: You're correct that the answer is "yes" in both cases.
The easiest way to see this (in my opinion) is that closed subspaces of hilbert spaces are themselves hilbert spaces. This is a fairly easy exercise, and if you've not seen it before, you should try to prove it yourself. For completeness, though (pun intended), I'll include a proof under the fold:

 Every subspace of an inner product space is an inner product space by just restricting the domain of the ambient inner product. This is a purely algebraic fact, with no analytic content. Now, remember that a hilbert space is an inner product space so that the norm induced by the inner product is complete. That is, every cauchy sequence should have a limit. Now it's easy to see that this happens if and only if your subspace is closed topologically. After all, we know (since the ambient space is complete) that a limit exists. So by closedness, that limit is in the subspace too.

Now to answer your precise question, we need to show that

*

*finite dimensional subspaces are closed

*the closures of spans of countable sets are closed

The second assertion is trivial, so let's look at finite dimensional spaces. You can argue this directly, and I'll leave the proof to you. I will leave a hint under the fold, though:

 It'll be useful to fix a finite basis for your subspace, you really don't need much analysis at all.


I hope this helps ^_^
