Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

From Wikipedia

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!

share|cite|improve this question
  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.

share|cite|improve this answer
+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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.