Are there any vector spaces that cannot be given a norm that makes the vector space a complete metric space?

Question

And if so, how can one prove that there is no such norm? I suppose one can use the form of Baire Category Theorem which states that a complete metric space cannot be written as a countable union of nowhere dense subsets to show that the metric space defined by the given vector space and norm is not complete but I have no idea on how to generalise this idea to all possible norms.

Answer 1

If a space admits an enumerable non finite basis $(e_i)_{i\in \bf N}$ it cannot be complete, as it is the enuramble union of finite dimensional vector spaces $V_n= vect (e_1,...e_n)$.

Answer 2

Yes. Take the space $c_{00}$ of all sequences $(a_n)_{n\in\mathbb N}$ of real numbers such that $a_n=0$ if $n\gg1$. For each $n\in\mathbb N$, let $e(n)$ be the sequence such that its $n$^th is $1$, where as all other terms are equal to $0$. Then, for any norm $\lVert\cdot\rVert$ on $c_{00}$, the series$$\sum_{n=0}^\infty\frac{e(n)}{n^2\bigl\lVert e(n)\bigr\rVert}$$is a Cauchy series of elements of $c_{00}$ which doesn't converge in $c_{00}$.

Answer 3

The vector space $\sigma$ of all sequences in $\mathbb{R}^\omega$ with the property that $\{n: x_n \neq 0\}$ is finite (usually taken in the subspace topology induced from the product (the $\sigma$-product this is sometimes called)) is a standard example.

It's a countable union of finite-dimensional subspaces. In any norm topology on $\sigma$ the finite-dimensional subspaces are closed and have empty interior, so in any such norm the space cannot be Baire (and so cannot be complete).