So every proper subspace of a normed vector space has empty interior. I'm not asking for the proof, my problem is that this seems to me very strange. So if I have a normed vector space, in any proper subspace I can't take any ball inside the subspace?

For example suppose we work on a set with finite measure, $[a,b]$ for example. Let's take $L^{P}$ spaces over $[a,b]$. We know that now $L^{\infty}$ is included in $L^{1}$. So $L^{\infty}$ is a proper subspace of $L^{1}$. Now this means that $L^{\infty}$ is nowhere dense?

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    $\begingroup$ No, it doesn't mean that at all. Saying $L^\infty$ is nowhere dense says that the closure $\overline{L^\infty}$ has empty interior, which is much stronger than saying $L^\infty$ has empty interior. $\endgroup$ – David C. Ullrich Jun 26 at 15:26
  • $\begingroup$ Proper subspaces are like lines and planes through the origin in $\mathbb{R}^3$. They don't contain any open balls of $\mathbb{R}^3$. $\endgroup$ – mechanodroid Jun 26 at 15:30
  • $\begingroup$ @mechanodroid And hence proper subspaces have empty interior. The OP already knows that - what does this have to do with the question of whether $L^\infty$ is nowhere dense? $\endgroup$ – David C. Ullrich Jun 26 at 15:33

What this means is that $L^\infty$, as a subset of $L^1$, has empty interior (since $L^\infty$ is a proper subspace of $L^1$), as you have stated.

But $L^\infty$ is not nowhere dense in $L^1$. In fact, $L^\infty$ is dense in $L^1$. This is because any $L^1$ function can be approximated by simple functions, which are in $L^\infty$. Thus the closure of $L^\infty$ is $L^1$, which certianly has nonempty interior in $L^1$.

  • $\begingroup$ thanks for the answer. So inside $L^{\infty}$ I can't take any ball? $\endgroup$ – banach-alaoglu-zielony Jun 26 at 15:42
  • $\begingroup$ What you can do (and it may be useful to do) is take $B\cap L^\infty$ for any ball $B$ in $L^1$. But this will never be all of $B$. $\endgroup$ – Aweygan Jun 26 at 15:44

In the infinite-dimensional case, subspaces are closed by definition. $L^\infty$ is not closed in $L^1$ (its closure is all of $L^1$), so it is not an example of proper subspace.

$L^\infty$ is dense in $L^1$ (even compactly supported functions are). Of course ir is not nowhere dense, its closure is $L^1$!

And yes, as a subset, $L^\infty$ has empty interior.

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    $\begingroup$ Subspaces are only closed by definition if you define them to be closed, which is certainly not standard. $\endgroup$ – Aweygan Jun 26 at 15:39
  • $\begingroup$ For the theorem stated about subspace not having interior, the term subspace has to to be defined as closed subspace. I think it is pretty common to assume closedness (for ex. In the well known textbook by Fomin and Kolmogorov) $\endgroup$ – GReyes Jun 26 at 15:46
  • $\begingroup$ @GReyes It seems however a lot of times we would want to discuss dense subspaces of say $L^p$, as you said above, but this would be meaningless terminology if we assumed subspaces were closed by default. $\endgroup$ – rubikscube09 Jun 26 at 15:51
  • $\begingroup$ @GReyes: No, it doesn't have to be. I don't even see how it would help. (The converse is true, though: you can certainly show that having nonempty interior implies being closed (in fact, clopen) --- this is true for any subgroup of a topological group. Of course this is vacuous for vector subspaces.) $\endgroup$ – tomasz Jun 26 at 15:53
  • $\begingroup$ @GReyes Any proper subspace (closed or otherwise) has empty interior. Nowhere in the proof is closedness needed. $\endgroup$ – Aweygan Jun 26 at 16:03

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