Let $X$ be an infinite dimensional vector space over a field $\mathbb{K}$. Suppose that $(X,\|\cdot\|)$ is a complete normed vector space, in the sense that any Cauchy sequence is convergent. Suppose that the closed unit ball of $X$ is compact in the strong topology.

Question 1. Is $X$ necessarily isomorphic to some finite dimensional Banach space ?

Question 2. If the answer for the question 1 is no , can we always find in $\mathcal{L}(X,X)$ an unbounded linear operator ?

Different from this question Is there an infinite-dimensional Banach space with a compact unit ball? I would like to assume the axiom of choice.

Comment. The example I have in mind is $(\mathbb{R}^n,\|\cdot\|_{2})$ as $\mathbb{Q}$ vector space. Clearly $\text{dim}_{\mathbb{Q}}\ \mathbb{R}^n=\infty$ and the unit ball is compact and this space is complete with respect to the standard Euclidean norm $\|\cdot\|_2$, but in this example both questions 1 and 2 are trivial.

  • 2
    $\begingroup$ "would like to assume the" $\endgroup$
    – user5810
    Oct 13, 2011 at 6:49
  • $\begingroup$ @Ricky thanks for point this correction. $\endgroup$
    – Leandro
    Oct 13, 2011 at 7:05
  • $\begingroup$ I think I don't fully understand what a "norm" over a general field $\mathbb K$ is...??? $\endgroup$ Oct 13, 2011 at 10:26
  • $\begingroup$ And, by "strong topology" do you just mean "norm topology"? $\endgroup$ Oct 13, 2011 at 10:42
  • 2
    $\begingroup$ I liked your question, Leandro! I had never thought of $\mathbb{R}$ as a infinite dimensional space with compact unit ball! You opened my mind a bit! :-) +1 $\endgroup$ Oct 13, 2011 at 15:06

1 Answer 1


I think the following meets your setup. Let $\mathbb K = \mathbb Z_2$ with the "absolute value" $|0|=0, |1|=1$ (this is non-Archimedean). See http://en.wikipedia.org/wiki/Absolute_value_%28algebra%29

Set $V = \mathbb Z_2^I$ for some index set $I$, with the trivial norm $\|0\|=0$ and $\|x\|=1$ for all other vectors $x$. This satisfies the usual rules, with $\|kx\| = |k|\|x\|$ for $k\in\mathbb K, x\in V$. Clearly $V$ actually have the discrete metric, and so is complete. Now, the closed unit ball is all of $V$, and not compact if $I$ is infinite.

BUT, I could instead define $\|x\|=2$ for $x\not=0$. Still we have a norm. Now the closed unit ball is $\{0\}$; and so is compact. All non-trivial linear maps have norm $1$.

To me, this seems like a very, very silly example, which perhaps shows that the original question needs tweaking with a bit...

If you start with an Archimedean absolute value, then really you have a subfield of $\mathbb C$ or $\mathbb R$ (which must contain $\mathbb Q$). By continuity, I then think you can turn $V$ into a $\mathbb R$ vector space, with a "norm" in the usual sense. Then if $V$ has a compact unit ball, it must be finite dimensional (over $\mathbb R$). So your example of $\mathbb R^2$ over $\mathbb Q$ is in a sense all that can happen.

  • $\begingroup$ Thanks Matthew for the example. The last paragraph I have to think a bit more. Anyway your example answer the question since I did not made any assumptions on the norm. $\endgroup$
    – Leandro
    Oct 14, 2011 at 2:20

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