There's a popular application of Baire's Category Theorem that shows that $\mathbb{R}[X]$ (the space of all polynomials with real coefficients) is not complete by showing it is a countable union of nowhere dense sets. The sets chosen are all polynomials of degree less than or equal to n.

My question is - why are these sets nowhere dense in any given norm? I can see why this is the case for $l_p$ norms - but why is this generally true for $any$ norm?


  • $\begingroup$ Denseness only depends on the topology, not the norm. If two norms induce the same topology, the same sets are dense in each norm. If two norms do not induce the same topology, all bets are off. $\endgroup$ May 20 '14 at 9:27
  • $\begingroup$ So in this particular statement are we assuming all norms induce the same topology? Is this true in an infinite dimensional normed space (recall not all norms are necessarily equivalent)? $\endgroup$
    – user126743
    May 20 '14 at 9:31
  • $\begingroup$ In infinite dimensional spaces it is not(!) true that all norms are equivalent (and hence induce the same topology). Consider for example the supremum norm and the $L^1$ norm on $C([0,1])$. $\endgroup$
    – PhoemueX
    May 20 '14 at 9:36

This is true, because your subspaces are finite dimensional subspaces. Finite dimensional subspaces are closed with respect to any norm (basically, because all norms on finite dimensional spaces are equivalent, so the finite dimensional subspace will be complete, hence closed).

So the claim left is that the subspace has nonempty interior.

But it is easy to see that any subspace with nonempty interior has to be the whole vector space (take a small $\varepsilon$-Ball around $0$ which is contained in $V$ by assumption and "inflate" that).

EDIT: BTW, this shows that NO vector space with $\dim(V) = \infty$ can be endowed with a norm that makes it complete as long as it admits a countable basis.

  • $\begingroup$ I can understand what you're saying using any reasonable norm, but I'm not sure why this must in fact hold for every possible norm? Can you show that there doesn't exist a norm in which a subspace of finite degree polynomials can contain an open ball? $\endgroup$
    – user126743
    May 20 '14 at 9:38
  • 1
    $\begingroup$ Assume that $\Vert\cdot \Vert$ is a norm on $W$ such that $B_{\varepsilon} (x_0) \subset V$ with $V \lneq W$. This is in your case fulfilled for $V = \Bbb{R}_{\rm{deg} \leq n}$ and $W = \Bbb{R}[X]$, because $V$ is finite dimensional and $W$ is not. Then $x_0 \in V$, so that (by translation) $B_{\varepsilon}(0) \subset V$. By multiplication, we get $B_{n \cdot \varepsilon}(0) \subset V$ for all $n \in \Bbb{N}$ and hence $W \subset V$, a contradiction. $\endgroup$
    – PhoemueX
    May 20 '14 at 9:41
  • $\begingroup$ But since $W$ has infinite diameter, I'm not sure why your statement proves that in fact $W \subset V$. Can you please clearify? $\endgroup$
    – user126743
    May 20 '14 at 9:43
  • $\begingroup$ Take any $w \in W$. Then $\Vert w \Vert < \varepsilon \cdot n$ for some $n \in \mathbb{N}$. Hence $w \in B_{n \cdot \varepsilon}(0) \subset V$. As $w \in W$ was arbitrary, we are done. $\endgroup$
    – PhoemueX
    May 20 '14 at 9:45
  • $\begingroup$ OK thanks a lot! $\endgroup$
    – user126743
    May 20 '14 at 9:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.