For subspaces of $\mathbb R^n$ we know that they are compact if and only if they are closed and bounded. Is the same true for all normed finite-dimensional vector spaces perhaps not over $\mathbb R$ or $\mathbb C$? In particular is the unit sphere in them always compact? Is it true even in not-Banach spaces?

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    $\begingroup$ Yes. If a normed space has dimension $n$, then it is isomorphic to Euclidean $n$-space. $\endgroup$ Jan 12, 2015 at 12:54
  • $\begingroup$ See this, e.g., for some insight as to why my claim above holds. $\endgroup$ Jan 12, 2015 at 13:00
  • $\begingroup$ See also en.wikipedia.org/wiki/Riesz%27s_lemma#Converse $\endgroup$
    – Surb
    Jan 12, 2015 at 13:09
  • $\begingroup$ @DavidMitra Could you give me some more information about why this isomorphism exists? $\endgroup$ Jan 12, 2015 at 14:38
  • $\begingroup$ The second answer in the link in my second comment gives a proof. (This is usually the first thing proved in a course on normed spaces. You should be able to find a proof in any introductory text covering normed spaces.) $\endgroup$ Jan 12, 2015 at 14:42

3 Answers 3


If you formulate it that generally, no, this is not true. It holds for finite dimensional vector spaces over $\mathbb{R}$ or $\mathbb{C}$. But in the field itself we already have that closed and bounded implies compact.

So if we work over the field $\mathbb{Q}$, then this is itself a one-dimensional vector space over itself, in the standard norm $|\cdot|$. And there the unit ball $[-1,1] \cap \mathbb{Q}$ is closed and bounded, but not compact: take any sequence of rationals between $[-1,1]$ that converges to an irrational number (e.g. the finite decimal partial expansions); This, or any subsequence, does not converge (as it, and any subsequence, converges in the reals to an irrational, so cannot converge to any other real, in particular no rational). So it is not sequentially compact (which is equivalent to compactness in metric spaces).

So we need the field itself to have the Heine-Borel property that closed and bounded implies compact. This does not hold for the rationals, nor (I think) for $\mathbb{Q}_p$.

  • $\begingroup$ I'm not sure about the Heine-Borel property in vector spaces over $\mathbb Q_p$ but in then or other locally compact vector spaces the unit sphere is compact (as closed unit ball is compact and unit sphere is closed). Perhaps you know whether a finite vector space over a locally compact field is always locally compact? $\endgroup$ Jan 14, 2015 at 20:12
  • $\begingroup$ Sure, it's a finite product of locally compact spaces. $\endgroup$ Jan 14, 2015 at 20:14
  • $\begingroup$ But the topology on the vector space isn't the same as the topology on the product of fields. $\endgroup$ Jan 14, 2015 at 20:17
  • $\begingroup$ I think it is. Do you have an example of this not being the case? Certainly it's true for $n=1$, right? $\endgroup$ Jan 14, 2015 at 20:20
  • $\begingroup$ Perhaps we could prove it for a one good norm and then use that all norms are equivalent. $\endgroup$ Jan 14, 2015 at 20:29

We do have the following result, valid for finite dimensional normed vector spaces over normed and complete fields.

Consider $k$ is a field with a norm $|\cdot | \colon F \to [0, \infty)$ that makes him a complete normed field ( the norm is sub-additive and multiplicative). Moreover, $V$ is a finite dimensional vector space over $k$ that has a norm $||\cdot ||$ with the usual properties of norms. Fact: the norm $||\cdot ||$ is equivalent to the $l^{1}$ norm $$||a_1 e_1 + \ldots + a_d e_d||_1= |a_1| + \ldots + |a_d|$$ (so all the norms on $V$ are equivalent).

This result applies not only to finite dimensional vector spaces over $\mathbb{R}$ or $\mathbb{C}$, but also to finite dimensional vector spaces over $p$-adic fields like $\mathbb{Q}_p$ and finite extensions of them. Moreover, all these fields are locally compact so then $V$ will also be locally compact, as the topology will be the product topology.

For a proof I recommend a resuld in Van der Waerden book on Algebra, chap on normed fields, needs to be slightly adapted.

Note that this is not true for finite dimensional vector spaces say over $\mathbb{Q}$. Here is a family of non-equivalent norms

$$||(a,b)|| = |a \pm b \sqrt{d} |$$


Actually, there is a theorem stating that if $V$ is a finite dimensional vector space over $\mathbf{R}$, then all norms on $V$ are equivalent. This implies that in $V$ the compact sets are exacted the closed and bounded sets, as the unit sphere for instance, yes.


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