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

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?

share|cite|improve this question
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. – Mark Oct 19 '11 at 20:21
up vote 14 down vote accepted

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.

share|cite|improve this answer
Indeed, any vector space of infinite Hamel dimension less than $2^{\aleph_0}$ cannot be endowed with a complete norm. (This is a stronger statement only if the continuum hypothesis fails.) I found a proof at – Nate Eldredge Oct 20 '11 at 0:27
@Nate: I think this result goes back to Mackey, On infinite-dimensional linear spaces, Trans. Amer. Math. Soc. 57 (1945), 155-207. (see Theorem I-1 on p. 159). Neither Halbeisen-Hungerbühler nor this note they refer to mention this... – t.b. Oct 20 '11 at 3:10
At planetmath and Ask an analyst a one-page proof of the fact about cardinality $\mathfrak c$ was mentioned with reference to H. Elton Lacey, The Hamel Dimension of any Infinite Dimensional Separable Banach Space is c, Amer. Math. Mon. 80 (1973), 298. – Martin Sleziak Oct 20 '11 at 16:35

$\mathbb{R}^\mathbb{N}$ 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.

share|cite|improve this answer
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 $X$). – Nate Eldredge Oct 20 '11 at 0:24
Yes, the question assumes no topolgy. Why was this accepted? – scineram Oct 20 '11 at 5:58
I didn't carefully read this. – Bana Oct 20 '11 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 $\lambda^{\aleph_0}=\lambda$.)

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.