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In particular, topological vector spaces are uniform spaces and one can thus talk about completeness, uniform convergence and uniform continuity. (This implies that every Hausdorff topological vector space is completely regular.[2]) The vector space operations of addition and scalar multiplication are actually uniformly continuous. Because of this, every topological vector space can be completed and is thus a dense linear subspace of a complete topological vector space.

  1. I was wondering generally what some conditions are for a uniform space to be completed? I couldn't find relevant information on Wikipedia and likes.
  2. Similar question for a metric space?
  3. Because of what, can every topological vector space be completed?

Thanks and regards!

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2 Answers 2

  1. Every uniform space has a unique (up to uniform isomorphism) completion.

  2. Every metric space can be completed (this is a standard result in analysis).

The book General Topology by Engelking has a whole section on uniform spaces if you are interested in seeing the proofs.

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+1 Thanks! Yes, I am interested in where the completability results for uniform spaces, metric spaces and TVS are in that book and other references? I just searched that book, but might miss things. –  Tim Feb 27 '12 at 15:04
    
Kelley, General Topology also has a chapter on uniform spaces. [edited the answer to correct spelling Engelking] –  GEdgar Feb 27 '12 at 15:22

Let $\hat{X}$ be the completion of the TVS $(X,a,s)$ as a uniform space (as in the mentioned General Topology books). Herein $a$ and $s$ denote the vector space addition and scalar multiplication. As $a$ and $s$ are uniformly continuous maps, they have a unique uniformly continuous extension $\hat{a}$ and $\hat{s}$ from $\hat{X} \times \hat{X}$ resp. $K \times \hat{X}$ to $\hat{X}$ where $K$ denotes the underlying field.

Now you can easily verify that $\hat{a}$ and $\hat{s}$ fulfill the vector space axioms on $\hat{X}$. This is the key argument: The purely non-algebraic uniform completion of a TVS preserves the algebraic vector space structure and turns the uniform completion into a complete TVS.

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