Let $V$ be a real vector space.

Is there always a norm on $V$ such that $V$ is complete with respect to this norm?

If not, is there an easy counterexample?

  • 8
    $\begingroup$ Here's a neat little fact: a Banach space can't have a basis of cardinality $\aleph_0$. Thus the space of polynomials over $\mathbb{R}$, for instance, has no Banach norm. $\endgroup$
    – Mark
    Oct 19, 2011 at 20:21

3 Answers 3


No. Suppose $V$ is a normed space of (Hamel) dimension $\aleph_0$, with basis $\{v_1, v_2, v_3, \ldots \}$ say. Then $V$ is the union of a countable family of finite-dimensional subspaces, namely $\langle v_1 \rangle, \langle v_1, v_2 \rangle, \langle v_1, v_2, v_3 \rangle, \ldots$. Proper closed subspaces of a normed space are nowhere dense, so $V$ is a countable union of nowhere dense sets, and so is incomplete by the Baire category theorem.


$\mathbb{R}^\mathbb{N}$ in the product topology is an example. It's a completely metrisable topological vector space, but there can be no compatible norm because all neighbourhoods of 0 are unbounded.

  • 4
    $\begingroup$ The question didn't ask for a compatible norm. I believe the Hamel dimension of $\mathbb{R}^\mathbb{N}$ is $2^{\aleph_0}$, the same as any separable Banach space $X$ you care to name. Therefore there is a linear isomorphism $T : \mathbb{R}^\mathbb{N} \to X$ (not continuous with respect to the product topology) and so $||x|| := ||Tx||_X$ is a norm on $\mathbb{R}^\mathbb{N}$ that makes it into a Banach space (indeed, one which is isometrically isomorphic to $X$). $\endgroup$ Oct 20, 2011 at 0:24
  • $\begingroup$ Yes, the question assumes no topolgy. Why was this accepted? $\endgroup$
    – scineram
    Oct 20, 2011 at 5:58
  • $\begingroup$ I didn't carefully read this. $\endgroup$
    – Bana
    Oct 20, 2011 at 6:16

Let $\kappa$ be an infinite cardinal.
If $B$ is a Banach space such that the least size of a dense subset of $B$ is $\kappa$, then $B$ is of size $\kappa^{\aleph_0}$. So the only possible sizes of Banach spaces are powers with the exponent $\aleph_0$.

As was pointed out above, every infinite dimensional Banach space has (Hamel) dimension at least $2^{\aleph_0}$. If a Banach space has size $>2^{\aleph_0}$, then its size is actually equal to the dimension. It follows that an infinite cardinal can only be the dimension of a Banach space if it is of the form $\kappa^{\aleph_0}$. But there are many cardinals that are not of that form and for every cardinal $\kappa$ there is a vector space of dimension $\kappa$.

The first infinite cardinal not of the form $\kappa^{\aleph_0}$ is $\aleph_0$, as was pointed out above. The next cardinal is $\aleph_1$, which is of the form $\kappa^{\aleph_0}$ iff the continuum hypothesis holds. The only cardinals for which we can say for sure that they are not of the form $\kappa^{\aleph_0}$ are suprema of increasing chains of cardinals of countable length, like $\aleph_\omega$, the sup of the $\aleph_n$.

On the other hand, any two vector spaces are isomorphic iff they have the same dimension. Also, there are vectorspaces of all dimensions. So, the question whether a vector space has a norm that turns it into a Banach space really only asks which cardinals are dimensions of Banach spaces.

For every cardinal $\kappa$, $\ell^2(\kappa)$ is a Banach space (even Hilbert!) of density $\kappa$ and dimension $\kappa^{\aleph_0}$. It follows that an infinite dimensional vector space is isomorphic to a Banach space if and only if its dimension is of the form $\kappa^{\aleph_0}$ for some $\kappa$. (Actually, $(\kappa^{\aleph_0})^{\aleph_0}=\kappa^{\aleph_0}$, so $\lambda$ is of the form $\kappa^{\aleph_0}$ iff $\lambda^{\aleph_0}=\lambda$.)


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