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Let $\mathcal{V}$ be a countably infinite dimensional inner product space over $\mathbb{C}$ and Hilbert space $\mathcal{H}$ be the completion of $\mathcal{V}$. Will $\mathcal{H}$ always be a countably infinite dimensional Hilbert space?

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What part of the proof is not clear to you? –  GEdgar Jul 25 '12 at 3:23
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It depends on what you mean. If you mean that the completion will be a separable Hilbert space (which is equivalent to it having a numerable complete orthonormal system, or Hilbert basis), then yes: take any orthonormal basis of $\mathcal{V}$: this will be a complete orthonormal system in $\mathcal{H}$. If you mean: will the underlying complex vector space be countable infinite dimensional, then the answer is no, by Baire property and the fact that finite dimensional subspaces are always closed subspaces (and have empty interior), a Banach space cannot have countable infinite Hamel basis. –  Olivier Bégassat Jul 25 '12 at 3:23
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zji: Will you please clarify whether you mean Hilbert space dimension (cardinality of a maximal orthonormal set) or vector space dimension (cardinality of a maximal linearly independent set)? If you mean the latter, Qiaochu has answered the question, and the following question is closely related: math.stackexchange.com/q/17627. If you mean the former, then Olivier's comment gives a reason it is true. –  Jonas Meyer Jul 25 '12 at 3:53
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Thanks to Olivier for the comment and clarification. I am sorry for the ambiguity in the question. The first interpretation (about separable Hilbert space) is what I meant. Yet, it is still not clear to me why the basis for $\mathcal{V}$ will be a complete basis for $\mathcal{H}$. Any hint? In the standard proof of completion of an inner product space, $\mathcal{H}$ is taken to be some equivalence classes of Cauchy sequences in $\mathcal{V}$, so how is the basis of $\mathcal{H}$ taken to be the basis of $\mathcal{V}$ and why is it complete? –  zji Jul 25 '12 at 4:32
    
@zji: The linear span of an orthonormal basis for $\mathcal V$ is $\mathcal V$, hence dense in $\mathcal H$. In $\mathcal H$, only $0$ is orthogonal to every vector in $\mathcal V$. –  Jonas Meyer Jul 25 '12 at 4:40
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